F = U TS G = H TS. Availability, Irreversibility S K Mondal s Chapter 5. max. actual. univ. Ns =

Transcription

1 Availability, Irreversibility S K Mondal s Chater 5 I = W W 0 max ( S) = Δ univ actual 7. Irreversibility rate = I = rate of energy degradation = rate of energy loss (Wlost) = 0 S gen for all rocesses Sgen = mds 8. W actual dq = du + dw this for closed system act C dq C dw act h + + gz + = h + + gz + this for steady flow dm dm 9. Helmholtz function F = U S 0. Gibb s function G = H S. Entroy generation number (Ns) Ns = s gen mc. o calculate ds P 4

2 Availability, Irreversibility S K Mondal s Chater 5 P () i Use S S = m C ln + C ln P P For closed system ds = du + Pd d P or ds = m C + d d d = mc + mr d d ds = m C + mr For steady flow system or But note that and ds = dh dp d ds = m C dp P = mr P d dp mr ds = m C mr = P P P ds = du + Pd Both valid for closed system only. ds = dh dp 3. In Pie flow Entroy generation rate S gen = Q S sys 0 ( ) mc P = ms ( S) Due to lack of insulation it may be > for hot fluid or < for cold fluid rate of irreversibility ( I ) = 0 0 S gen in Kg/s P, P, 4. Flow with friction Decrease in availability = Δ P mr0 P 5

3 Availability, Irreversibility S K Mondal s Chater 5 5. Second Law efficiency n Min. exergy intake to erform the given tast (X ) min = = / II I carnot η Actual exergy intake to erform the given tast (X) η η X min = W if work is involved 0 = Q if Heat is involved. A common measure on energy use efficiency is the first law efficiency, η. he first law efficiency is defined as the ratio of the outut energy of a device to the inut energy of the device. he first law is concerned only with the quantities of energy, and disregards the forms in which the energy exists. It does not also discriminate between the energies available at different temeratures. It is the second law of thermodynamics which rovides a means of assigning a quality index to energy. he concet of available energy or energy rovides a useful measure of energy quality. With this concet it is ossible to analyze means of minimizing the consumtion of available energy to erform a given rocess, thereby ensuring the most efficient ossible conversion of energy for the required task. he second law efficiency, η, of a rocess is defined as the ratio of the minimum available energy (or energy) which must be consumed to do a task divided by the actual amount of available energy (or energy) consumed in erforming the task. or η η II II = minimum exergy intake to erform the given task actual exergy intake to erform the same task A A = min where A is the availability or energy. A ower lant converts a fraction of available energy A or Wmax to useful work W. For the desired outut of W, Amin = W and A = Wmax, Here, 6

4 S K Mondal s Now Availab bility, Irreversibility W I = W Wand η = max II W W W W max η = = I Q W Q I max = η η II Carnot η I η = II η Carnot 0 Since W = Q (l - ),can also be obtained directly as follows max W ηi η = = II 0 ηcarnot Q If work is involved, Amin = W (desired) and if heat is involved, Am max Q in = 0 Chater 5 he general definition of second law efficiency of a rocess can be obtained in terms of change in availability during the rocess: Produc Second Law Efficiency = η = ction of availability III Law Destruction of availability.. Available Energy Referred to a Cycle Decreasee in Available Energy when Heat is ransferred through a Finite emerature Deference :- Whenever heat is transferred throughh finite temerature deference, there is a decrease in the availability of energy so transferred. Let us consider a reversiblee heat enginee oerating between and 0 (Fig. shown in below) hen and Q = ΔS, Q = 0 ΔS, Δ W = A.E. = ( 0) Δs Let us now assume that heat Q is transferred through a finite temerature difference from the reservoir or source at to the enginee absorbing, heat at, lower turn (Fig.shown in below). 7

5 S K Mondal s Availab bility, Irreversibility Chater 5 he availability of Q as received by the engine at can be found by allowing the engine to oerate reversible in a cycle between and 0, receiving Q and rejecting Q. Now Since Q Q ' = Δs Since and 0 = Δs' 0 ' Q = Δ s = Δs' >, Δ s' > Δs ' Δ s' > Δs W' = Q - Q' W < W, because Q Q' > Q =' Δs' Δs' 0 W = Q Q = Δs Δ s > Q 0 Available energy lost due to irreversiblee heat transfer through finite temerature deference between the source and the working fluid during the heat addition rocess is given by W - W = Q Q = 0(Δs - Δs) or decrease in A.E. = 0(Δs - Δs) he decrease in available energy or exergy is thus the roduct of the lowest feasible temerature of heat rejection and the additional entroy change in the system while receiving heat irreversibly, comared to the case of reversible heat transfer from the same source. he greater is the temerature difference ( - ), the greater is the heat rejection Q and the greater will be the unavailable art of the energy sulied or energy (Fig. above). Energy is said to be degraded each time, it flows through a finite temerature difference. PROBLEMS & SOLUIONS Examle What is the maximum useful work which can be obtained when 00 kj are abstracted from a heat reservoir at 675 K in an environmental at 88 K? What is the loss of useful work if, (a) a temerature dro of 50 C is introducedd between the heat source and the heat engine, on the one hand and the heat engine and the heat sink, on the other. (b) the source temerature dros by 50 C 8

6 S K Mondal s Availab bility, Irreversibility Chater 5 and the sink temerature rises by 50 C during the heat transferr rocess according to the linear law dq d = ± constan nt? Solution: Availability= W o 888 max = Q = 00 = kj Q ΔS = 00 = = 0.6 kj / K 65 ( ) W =ΔS = = 45.9kJ o Loss in availability = =.4 kj. ( ) Average temerature of source 9

7 Availability, Irreversibility S K Mondal s Chater 5 + " a = = 650 K ( ) ( ) 0 Average temerature of sink 0a 33 Q 00 Δ = = kj S" = / a 650 K 0 + " = = W " =ΔS" = 5.83 kj / K Loss in availability = = 5.5 kj. Examle. A lead storage battery used in an automobile is able to deliver 5. MJ of electrical energy. his energy is available for starting the car. Let comressed air be considered for doing an equivalent amount of work in starting the car. he comressed air is to be stored at 7 MPa, 5 C, what is the volume of the tank that would be required to let the comressed air having an availability of 5. MJ? For air v = 0.87, where is in K, in kpa, and v in m 3 /kg. Solution: Availability is given by a = ( u uo) o( s so) + o( v vo) v = = 0.08 m / kg vo = = m / kg a = ln 0.87ln 00( ) = 79 kj / kg 500 For availability of 5. MJ, m = = 8.635kg 79 3 = mv. = = 0.77m Examle 3. Air enters a comressor in steady flow at 40 kpa, 7 C and 70 m/s and leaves it at 350 kpa, 7 C and 0 m/s. he environment is at 00 kpa, 7 C. Calculate er kg of air (a) the actual amount of work required, (b) the minimum work required, and (c) the irreversibility of the rocess. Solution: his is the case of a steady flow system in which the maximum useful work is given by C C Wmax useful = φ φ + C C = ( H os) ( H os) + As air is to be considered as a erfect gas, h = C Hence C C W = C S S + ( ) ( ) max useful o In the above, exression (S - S) is the entroy change. For a erfect gas, when ressure and temerature very, the entroy change is given by the relation K

8 Availability, Irreversibility S K Mondal s Chater 5 P S S = C ln Rln P Hence P C C Wmax useful = C ( ) o C ln Rln + P arious arameters given are: = 90 K; = 400 K; C =.005 kj/kg K P = 40 kpa; P = 350 kpa C = 70m / s; C = 0 m/s; o = 80 K (a) Actual work done in case of a steady flow rocess is given by C 0 Wactual = ( H H) =.005( ) + = 4.kJ 3 0 he negative sign obviously indicates that work is done on the air in comression (b) Minimum work =.005(90-400) ln 0.87ln = kj (c) Irreversibility= = 6.8 kj. Examle 4. By how much is the available energy of 5 kg of air increased by heating reversibly at a constant ressure of.5 atm. from 7 C to 7 C with the lowest available temerature of 0 C? (C for air = 005 J/kg C). Solution: Change in entroy = mc ln 500 = ln =.567 kj / K 300 Availability = Δ S = = 75kJ. 0

9 Availability, Irreversibility S K Mondal s Chater 5 ASKED OBJECIE QUESIONS (GAE, IES, IAS) Previous 0-Years GAE Questions GAE-. A steel billet of 000 kg mass is to be cooled from 50 K to 450 K. he heat released during this rocess is to be used as a source of energy. he ambient temerature is 303 K and secific heat of steel is 0.5 kj/kg K. he available energy of this billet is: [GAE-004] (a) MJ (b) MJ (c) 0.35 MJ (d) 0.0 MJ Availability GAE-. Availability of a system at any given state is: [GAE-000] (a) A roerty of the system (b) he maximum work obtainable as the system goes to dead state (c) he total energy of the system (d) he maximum useful work obtainable as the system goes to dead state GAE-3. A heat reservoir at 900 K is brought into contact with the ambient at 300 K for a short time. During this eriod 9000 kj of heat is lost by the heat reservoir. he total loss in availability due to this rocess is: [GAE-995] (a) 8000 kj (b) 9000 kj (c) 6000 kj (d) None of the above Irreversibility GAE-4. Consider the following two rocesses: [GAE-00] a. A heat source at 00 K loses 500 kj of heat to sink at 800 K b. A heat source at 800 K loses 000 kj of heat to sink at 500 K Which of the following statements is RUE? (a) Process I is more irreversible than Process II (b) Process II is more irreversible than Process I (c) Irreversibility associated in both the rocesses is equal (d) Both the rocesses are reversible Available Energy IES-. IES-. Previous 0-Years IES Questions What will be the loss of available energy associated with the transfer of 000 kj of heat from constant temerature system at 600 K to another at 400 K when the environment temerature is 300 K? [IES-004] (a) 50 kj (b) 50 kj (c) 500 kj (d) 700 kj What is the loss of available energy associated with the transfer of 000 kj of heat from a constant temerature system at 600 K to another at 400 K when the environmental temerature is 300 K? [IES-008]

10 Availability, Irreversibility S K Mondal s Chater 5 (a) 50 kj (b) 50 kj (c) kj (d) 80kJ Available Energy Referred to a Cycle IES-3. Availability IES-4. A heat source H can suly 6000kJ/min. at 300 C and another heat source H can suly kj/min. at 00 C. Which one of the following statements is correct if the surroundings are at 7 C? [IES-006] (a) Both the heat sources have the same efficiency (b) he first heat source has lower efficiency (c) he second heat source has lower efficiency (d) he first heat source roduces higher ower Assertion (A): he change in availability of a system is equal to the change in the Gibbs function of the system at constant temerature and ressure. Reason (R): he Gibbs function is useful when evaluating the availability of systems in which chemical reactions occur. [IES-006] (a) Both A and R are individually true and R is the correct exlanation of A (b) Both A and R are individually true but R is NO the correct exlanation of A (c) A is true but R is false (d) A is false but R is true IES-5. For a steady flow rocess from state to, enthaly changes from h = 400 kj/kg to h = 00 kj/kg and entroy changes from s =. kj/kg-k to s = 0.7 kj/kg-k. Surrounding environmental temerature is 300 K. Neglect changes in kinetic and otential energy. he change in availability of the system is: [IES-003] (a) 40 kj/kg (b) 300 kj/kg (c) 80 kj/kg (d) 90 kj/kg IES-6. Availability function for a closed system is exressed as: [IES-00] (a) φ = u+ ov os (b) φ = du + odv ods (c) φ = du + odv + ods (d) φ = u+ ov+ os IES-7. Consider the following statements: [IES-00]. Availability is the maximum theoretical work obtainable.. Claeyron's equation for dry saturated steam is given by d h s g hf ( g f ) = dq s 3. A gas can have any temerature at a given ressure unlike a vaour which has a fixed temerature at a given ressure. s 4. Joule homson coefficient is exressed as μ = h Of these statements (a), and 3 are correct (b), 3 and 4 are correct (c) and 3 are correct (d), and 4 are correct 3

11 Availability, Irreversibility S K Mondal s Chater 5 IES-8. 0kg of water is heated from 300 K to 350 K in an insulated tank due to churning action by a stirrer. he ambient temerature is 300 K. In this context, match List-I and List-II and select the correct answer using the codes given below the Lists: [IES-000] List-I List-II A. Enthaly change.. kj/kg B. Entroy change/kg. 968 kj C. Availability/kg kj D. Loss of availability J/kg-k Codes: A B C D A B C D (a) 3 4 (b) 4 3 (c) 3 4 (d) 4 3 IES-9. Neglecting changes in kinetic energy and otential energy, for unit mass the availability in a non-flow rocess becomes a = ɸ - ɸo, where ɸ is the availability function of the [IES-998] (a) Oen system (b) Closed system (c) Isolated system (d) Steady flow rocess IES-0. Consider the following statements: [IES-996]. Availability is generally conserved. Availability can either be negative or ositive 3. Availability is the maximum theoretical work obtainable 4. Availability can be destroyed in irreversibility Of these correct statements are: (a) 3 and 4 (b) and (c) and 3 (d) and 4 Irreversibility IES-. he irreversibility is defined as the difference of the maximum useful work and actual work: I = Wmax,useful- Wactual. How can this be alternatively exressed? (a) I = ( Δ S +Δ S ) (b) I = ( ΔS Δ S ) [IES-005] o system surrounding o system surrounding (c) I = ( Δ S + Δ S ) (d) I = ( ΔS Δ S ) o system surrounding o system surrounding IES-. Assertion (A): All constant entroy rocesses are adiabatic, but all adiabatic rocesses are not isentroic. [IES-006] Reason (R): An adiabatic rocess which resists the exchange of energy to the surroundings may have irreversibility due to friction and heat conduction. (a) Both A and R are individually true and R is the correct exlanation of A (b) Both A and R are individually true but R is NO the correct exlanation of A (c) A is true but R is false (d) A is false but R is true IES-3. Which of the following statement is incorrect? [IES-99] (a) he greater the ressure difference in throttling the lesser the irreversibility (b) he rimary source of internal irreversibility in ower is fluid friction in rotary machines. (c) he greater the irreversibility, the greater the increase in adiabatic rocess 4

12 Availability, Irreversibility S K Mondal s Chater 5 (d) he entroy of the universe is continually on the increase. Previous 0-Years IAS Questions Available Energy IAS-. IAS-. What will be the loss of available energy associated with the transfer of 000 kj of heat from constant temerature system at 600 K to another at 400 K when the environment temerature is 300 K? [IAS-995] (a) 50 kj (b) 50 kj (c) 500 kj (d) 700 kj An inventor claims that heat engine has the following secifications: [IAS-00] Power develoed = 50 kw; Fuel burned er hour = 3 kg Heating value of fuel =75,000 kj er kg; emerature limits = 67 C and 7 C Cost of fuel = Rs. 30/kg; alue of ower = Rs. 5/kWh (a) Possible (b) Not ossible (c) Economical (d) Uneconomical IAS-3. For a reversible ower cycle, the oerating temerature limits are 800 K and 300 K. It takes 400 kj of heat. he unavailable work will be: [IAS-997] (a) 50 kj (b) 50 kj (c) 0 kj (d) 00 kj Quality of Energy IAS-4. Increase in entroy of a system reresents [IAS-994] (a) Increase in availability of energy (b) Increase in temerature (c) Decrease in ressure (d) Degradation of energy Availability IAS-5. If u,, v, s, hand refer to internal energy, temerature, volume, entroy, enthaly and ressure resectively; and subscrit 0 refers to environmental conditions, availability function for a closed system is given by: [IAS-003] (a) u + Po v o s (b) u Po v+ o s (c) h + Po v os (d) h Po v + o s IAS-6. Match List-I with List-II and select the correct answer using the codes given below the lists: List-I List-II A. Irreversibility. Mechanical equivalent B. Joule homson exeriment. hermodynamic temerature scale C. Joule's exeriment 3. hrottling rocess D. Reversible engines 4. Loss of availability Codes: A B C D A B C D (a) 3 4 (b) 4 3 (c) 4 3 (d) 4 3 5

13 Availability, Irreversibility S K Mondal s Chater 5 Irreversibility IAS-7. he loss due to irreversibility in the exansion valve of a refrigeration cycle shown in the given figure is reresented by the area under the line. (a) GB (b) AG (c) AH (d) BH [IAS-999] IAS-8. Assertion (A): When a gas is forced steadily through an insulated ie containing a orous lug, the enthaly of gas is the same on both sides of the lug. [IAS-997] Reason (R): he gas undergoes an isentroic exansion through the orous lug. (a) Both A and R are individually true and R is the correct exlanation of A (b) Both A and R are individually true but R is NO the correct exlanation of A (c) A is true but R is false (d) A is false but R is true Second Law efficiency IAS-9. Assertion (A): he first-law efficiency evaluates the energy quantity utilization, whereas the second-law efficiency evaluates the energy quality utilization. Reason (R): he second-law efficiency for a rocess is defined as the ratio of change of available energy of the source to the change of available energy of the system. [IAS-998] (a) Both A and R are individually true and R is the correct exlanation of A (b) Both A and R are individually true but R is NO the correct exlanation of A (c) A is true but R is false (d) A is false but R is true 6

14 Availability, Irreversibility S K Mondal s Chater 5 Answers with Exlanation (Objective) Previous 0-Years GAE Answers o A.E = mc d = mc oln GAE-. Ans. (a) ( ) 50 = ( ) 303ln = 490MJ 450 GAE-. Ans. (d) Maximum useful work, i.e. total work minus dv work. Not maximum work. GAE-3. Ans. (d) he availability of a thermal reservoir is equivalent to the work outut of a Carnot heat engine oerating between the reservoir and the environment. Here as there is no change in the temeratures of source (reservoir) or the sink (atmoshere), the initial and final availabilities are same. Hence, there is no loss in availability. GAE-4. Ans. (b) Previous 0-Years IES Answers IES-. Ans. (b) Loss of available energy = o ( Δ S) = 300 kj 50kJ univ. = IES-. Ans. (b) Loss of Availability = 0 ΔS 0 = 300K Q Q Δ S = Loss of Availability = 300 = 50 kj ource IES-3. Ans. (c) η = η> η surroundings IES-4. Ans. (b) IES-5. Ans. (c) U.E. = o (s s ) = 300 (. 0.7) = 0 kj/kg Change in availability = (h h ) (U.E.) = (400 00) 0 = 80 kj/kg IES-6. Ans. (a) IES-7. Ans. (c) he availability of a given system is defined as the maximum useful work that can be obtained in a rocess in which the system comes to equilibrium with the surroundings or attains a dead state. Clearly, the availability of a system deends on the condition of the system as well as those of the surroundings. 7

15 Availability, Irreversibility S K Mondal s Chater 5 IES-8. Ans. (c) IES-9. Ans. (a) IES-0. Ans. (a) Availability is the maximum theoretical work obtainable and it can be destroyed in irreversibility. IES-. Ans. (a) I= o ( Δ S) = universe o Δ Ssystem +ΔS surrounding IES-. Ans. (d) A is false, For a rocess due to irreversibility entroy will increase and actual rocess may be ' but due to heat loss to the surroundings, may ' coincide with but the rocess not adiabatic. So, all isentroic rocess is not adiabatic. IES-3. Ans. (a) Previous 0-Years IAS Answers IAS-. Ans. (b) Loss of available energy = o ( Δ S) = 300 kj 50kJ univ. = IAS-. Ans. (b) Maximum ossible efficiency (ηmax) = 300 = = Maximum ossible Power outut with this machine (Wmax) = Q ηmax = kw KW So above demand is imossible. IAS-3. Ans. (b) Available art of the heat (WE) = Q 300 = = 50 kj Unavailable work (Wu) = = 50 kj 50 = ( ) 303ln = 490MJ 450 s IAS-4. Ans. (d) IAS-5. Ans. (a) IAS-6. Ans. (d) 8

16 Availability, Irreversibility S K Mondal s Chater 5 IAS-7. Ans. (d) Entroy will increase in the rocess AH is BH. herefore Irreversibility (I) = o Δ S i.e. area under the line BH. IAS-8. Ans. (c) Exansion through the orous lug is adiabatic as no heat added or rejected to the system. It is not reversible, due to large irreversibility entroy increases so it is not an isentroic rocess. mimimum energy intake to erform the given task IAS-9. Ans. (c) η II = actual energyintake to erform the same task 9

17 hermodynamic Relations S K Mondal s Chater 6 6. hermodynamic Relations heory at a Glance (For GAE, IES & PSUs) Adiabatic index (γ) =+ N where N is degrees of freedom of molecules N=3 for monatomic gas N=5 for diatomic gas N=6 for try atomic gas. Some Mathematical heorem heorem. If a relation exists among the variables x, y and z, then z may be exressed as a function of x and y, or z z dz = dx + dy x y y x then dz = M dx + N dy. where z, M and N are functions of x and y. Differentiating M artially with resect to y, and N with resect to x. M z = y x. y x N z = x y y. x M N = y x x y his is the condition of exact (or erfect) differential. heorem. If a quantity f is a function of x, y and z, and a relation exists among x, y and z, then f is a function of any two of x, y and z. Similarly any one of x, y and z may be regarded to be a function of f and any one of x, y and z. hus, if x = x (f, y) x x dx = df + dy f y y f Similarly, if y = y (f, z) y y dy = df + dz f z z f Substituting the exression of dy in the receding equation 30

18 hermodynamic Relations S K Mondal s Chater 6 x x y y dx = df + df + dz f y f z y f z f x x y x y = + df + dz f y f y z y f z f f Again x x dx = df + dz f z z f x x y = z f y z f f x y z = y z f x f f heorem 3. Among the variables x, y, and z any one variable may be considered as a function of the other two. hus x = x(y, z) x x dx = dy + dz y z y Similarly, z z z dz = dx + dy x y y x x x z z dx = dy + dx + dy y z y x y y z x x x z x z = + dy + dx y z z y y x z y x y x x z = + dy + dx y z y y z x x z x + = 0 y y z z x y x z y = y x yz z x Among the thermodynamic variables, and. he following relation holds good = v Maxwell s Equations:- 3

19 hermodynamic Relations S K Mondal s Chater 6 A ure substance existing in a single hase has only two indeendent variables. Of the eight quantities,,, S, U, H, F (Helmholtz function), and G (Gibbs function) any one may be exressed as a function of any two others. For a ure substance undergoing an infinitesimal reversible rocess (a) du = ds - d (b) dh = du + d + dp = ds + d (c) df = du - ds - Sd = - d - Sd (d) dg = dh - ds - Sd = d - Sd Since U, H, F and G are thermodynamic roerties and exact differentials of the tye dz = M dx + N dy, then M N = y x x y Alying this to the four equations = s S v P = s S S = S = P hese four equations are known as Maxwell s equations. (i) Derive: d ds = C + d [IAS ] v v Let entroy S be imagined as a function of and. 3

20 hermodynamic Relations S K Mondal s Chater 6 ( ) hen S = S, S S or ds = d d + multilying both side by S S ds = d + d S Since = C, heat caacity at constant volume S and = by Maxwell's equation ds C d d = + dividing both side by d ds C d roved = + (ii) Derive: ds = C d d Let entroy S be imagined as a function of and. ( ) hen S = S, S S or ds = d + d multilying both side by S S ds = d + d S Since = C, heat caacity at constant ressure and S = = by Maxwell's equation ds C d d roved. [IES-998] (iii) Derive: 33

21 hermodynamic Relations S K Mondal s Chater 6 β k kc d ds = C d + d = C d β d = + d β C β We know that volume exansivity (β) = [IES-00] and isothermal comressibility (k) = From first ds equation ds = C d + d β = = k As = = or β = k β ds = C d + d k roved From second ds relation ds C d d as = β= = β ds = C d βd roved Let S is a function of, S = S(, ) S S ds = d d + 34

22 hermodynamic Relations S K Mondal s Chater 6 Multily both side by S S ds = d d + or S S ds = d d + or S S ds = d d + S S C = and C = ds = C d C d + From first β k = or k = β k ds = C d + C β d β = = β C kd C = + β β ds d roved. (iv) Prove that C C = v We know that [IAS-998] 35

23 hermodynamic Relations S K Mondal s Chater 6 ds = C d d C d d = + or ( C C v) d = d d + d d = + or d i C C C C sin ce is a function of, =, ( ) or d = d d ii + comaring () i & ( ii) we get = and = C C C C both these give C C = () ( ) Here = or = C C = roved....equation(a) his is a very imortant equation in thermodynamics. It indicates the following imortant facts. (a) Since is always ositive, and for any substance is negative. (C Cv) is always ositive. herefore, C is always greater than Cv. (b) As 0 K, C C or at absolute zero, v C = Cv. (c) When = 0 (e.g for water at 4ºC, when density is maximum. Or secific volume minimum). C = Cv. (d) For an ideal gas, = mr 36

24 hermodynamic Relations S K Mondal s Chater 6 mr = = P mr and = C C = mr or c c = R v v Equation (A) may also be exressed in terms of volume exansively (β) defined as β = and isothermal comressibility (k) defined as k = C Cv = β C Cv = k Ratio of heat caacities At constant S, the two ds equations become Cd s = ds Cd v s = ds v C S = γ C v = = v s since γ >, > s herefore, the sloe of an isentroic is greater than that of an isotherm on -v diagram (figure below). For reversible and adiabatic comression, the work done is s s s W = h h = vd = Area - s

25 S K Mondal s herm modynamic Relations Chater 6 (Fig. Comression Work in Different Reversible Process) For reversible and isothermal comression, the work done would be W = h = Area W < W For olytroic comression with < n < γ. the work done will be between these two values. So isothermal comression minimum work. he adiabatic comressibility (ks) is defined as or ks = s C = = γ C v s k γ = ks Energy Equation h = s vd For a system undergoing an infinitesimal reversible rocess between two equilibrium states, the change of internal energy is du = ds - d Substituting the first ds equation du = Cd v + d d = Cd v + d if U = (, ) U U du = d + d U = his is known as energy equation. wo alication of the equation are given below- 38

26 hermodynamic Relations S K Mondal s Chater 6 (a) For an ideal gas, nr = nr = = U. 0 = = U does not change when changes at = C. U U = U U 0 = = U since 0, 0 = U does not change either when changes at = C. So the internal energy of an ideal gas is a function of temerature only. Another imortant oint to note is that for an ideal gas = nr and = 0 v herefore du = Cv d holds good for an ideal gas in any rocess (even when the volume changes). But for any other substance du = Cv d is true only when the volume is constant and d = 0 Similarly dh = ds + d and ds C d d = dh = Cd + d H = As shown for internal energy, it can be similarly roved from Eq. shown in above that the enthaly of an ideal gas is not a function of either volume or ressure. H H ie. = 0 and = 0 but a function of temerature alone. Since for an ideal gas, = nr and 0 = the relation dh = C d is true for any rocess (even when the ressure changes.) 39

27 hermodynamic Relations S K Mondal s Chater 6 However, for any other substance the relation dh = C d holds good only when the ressure remains constant or d = 0. (b) hermal radiation in equilibrium with the enclosing walls rocesses an energy that deends only on the volume and temerature. he energy density (u), defined as the ratio of energy to volume, is a function of temerature only, or U u = = f( ) only. he electromagnetic theory of radiation states that radiation is equivalent to a hoton gas and it exerts a ressure, and that the ressure exerted by the black body radiation in an enclosure is given by u = 3 Black body radiation is thus secified by the ressure, volume and temerature of the radiation. since. u U = u and = 3 U du uand = = 3 d By substituting in the energy Eq. du u u = 3 d 3 du d = 4 u or ln u = ln 4 + lnb or u = b 4 where b is a constant. his is known as the Stefan - Boltzmann Law. 4 Since U = u = b U 3 Cv 4b = = du 4 3 and b = = 3d 3 From the first ds equation ds = Cvd + d v = 4 b d + b. d 3 For a reversible isothermal change of volume, the heat to be sulied reversibly to kee temerature constant. 4 4 Q = b Δ 3 For a reversible adiabatic change of volume 40

28 hermodynamic Relations S K Mondal s Chater b d = 4b d 3 d d or = 3 3 or = const If the temerature is one-half the original temerature. he volume of black body radiation is to be increased adiabatically eight times its original volume so that the radiation remains in equilibrium with matter at that temerature. (v) Prove that β U = and C C = + k Hence show that β C C = [ IES- 003] v k Here β= k = β = = k we know that = or β = k = roved. From ds relations 4

29 hermodynamic Relations S K Mondal s Chater 6 ds = C d dp = C d + d ( C C v) d = dp d + v or d = dp + d i C C C C (, ) () Since is a function of = (, ) d = d d ii + ( ) ( ) Comairing i & ii we get = and = C C C C = C C ( ) as du = dq d du = ds d U S or = U S or + = From Maxwell' s hird relations S = U C C = = + (vi) Prove that Joule homson co-efficient μ = = C h [IES-00] he numerical value of the sloe of an isenthalic on a diagram at any oint is called the Joule Kelvin coefficient. 4

30 hermodynamic Relations S K Mondal s Chater 6 μ = h Here dh = ds + d as ds C d d = dh = Cd d d Cd d + = if H = cos t. dh = 0 so C ( d) ( d) 0 h h = = = C h C = C (vii) Derive clausius claeyron equation d hfg d hfg and d d = = v v R ( g f) [IES-000] S Maxwells equation = When saturated liquid convert to saturated vaour at constant temerature. During the evaoration, the r. & is indeendent of volume. d sg sf d = sat vg vf hfg sg sf = sfg = d hfg or d = v v ( ) sat g f It is useful to estimate roerties like h from other measurable roerties. At a change of hage we may find h i.e. latent heat. fg 43

31 hermodynamic Relations S K Mondal s Chater 6 At very low ressure vg vf g as vf very small R vg = R or vg = d hfg hfg hfg = = = d v R g R or d hfg d = R h fg or ln = R Knowing vaour ressure at temerature, we may find out at temerature. Joule-Kelvin Effect or Joule-homson coefficient he value of the secific heat c can be determined from v data and the Joule homson coefficient. he Joule homson coefficient μ is defined as J μ J = h Like other artial differential coefficients introduced in this section, the Joule homson coefficient is defined in terms of thermodynamic roerties only and thus is itself a roerty. he units of μ J are those of temerature divided by ressure. A relationshi between the secific heat c and the Joule homson coefficient μ can be established J to write h h h = he first factor in this exression is the Joule homson coefficient and the third is c. hus 44

32 S K Mondal s herm modynamic Relations c = μ J h ( ) / Chater 6 Wi ith ( h ) / ( h) = / / he artial derivative ( h / he following exression results: ) this can be written as ), called the constant-temerature coefficient, can be eliminated. v c = μ v J allows the value of c at a state to be determinedd using v data and the value of the Joule homson coefficient at that state. Let us consider next how the Joule homson coefficient can be found exerimentally. he numerical value of the sloe of an isenthalic on a - diagram at any oint is called the Joulezero is the Kelvin coefficient and is denoted by μ. hus the locus of alll oints at which μ is J J inversion curve. he region inside the inversion curve where μ is ositive is called the cooling J region and the region outside where is negative is called the heating region. So, μ J c h = μ J μ = J h Gibbs Phase Rule Gibbs Phase Rule determines what is exected to define the state of a system F = C P + 45

33 hermodynamic Relations S K Mondal s Chater 6 F = Number of degrees of freedom (i.e.., no. of roerties required) C = Number of comonents P = Number of hases e.g., Nitrogen gas C = ; P =. herefore, F = o determine the state of the nitrogen gas in a cylinder two roerties are adequate. A closed vessel containing water and steam in equilibrium: P =, C = herefore, F =. If any one roerty is secified it is sufficient. A vessel containing water, ice and steam in equilibrium P = 3, C = therefore F = 0. he trile oint is uniquely defined. 46

34 hermodynamic Relations S K Mondal s Chater 6 ASKED OBJECIE QUESIONS (GAE, IES, IAS) Previous 0-Years GAE Questions Maxwell's Equations GAE-. Which of the following relationshis is valid only for reversible rocesses undergone by a closed system of simle comressible substance (neglect changes in kinetic and otential energy? [GAE-007] (a) δq = du + δw (b) ds = du + d (c) ds = du + δw (d) δq = du + d GAE-. Considering the relationshi ds = du + d between the entroy (S), internal energy (U), ressure (), temerature () and volume (), which of the following statements is correct? [GAE-003] (a) It is alicable only for a reversible rocess (b) For an irreversible rocess, ds > du + d (c) It is valid only for an ideal gas (d) It is equivalent to law, for a reversible rocess Difference in Heat Caacities and Ratio of Heat Caacities GAE-3. he secific heats of an ideal gas deend on its [GAE-996] (a) emerature (b) Pressure (c) olume (d) Molecular weight and structure GAE-4. he secific heats of an ideal gas deends on its [GAE-996] (a) emerature (b) Pressure (c) olume (d) Molecular weight and structure GAE-5. For an ideal gas the exression s s v is always equal to: c ( a) zero ( b) ( c) R ( d) R c v [GAE-997] GAE-6. A kw, 40 litre water heater is switched on for 0 minutes. he heat caacity C for water is 4. kj/kg K. Assuming all the electrical energy has gone into heating the water, increase of the water temerature in degree centigrade is: [GAE-003] (a).7 (b) 4.0 (c) 4.3 (d) 5.5 Joule-Kelvin Effect or Joule-homson coefficient GAE-7. Which combination of the following statements is correct? [GAE-007] 47

35 hermodynamic Relations S K Mondal s Chater 6 P: A gas cools uon exansion only when its Joule-homson coefficient is ositive in the temerature range of exansion. Q: For a system undergoing a rocess, its entroy remains constant only when the rocess is reversible. R: he work done by a closed system in an adiabatic rocess is a oint function. S: A liquid exands uon freezing when the slo of its fusion curve on Pressure emerature diagram is negative. (a) R and S (b) P and Q (c) Q, R and S (d) P, Q and R GAE-8. A ositive value to Joule-homson coefficient of a fluid means [GAE-00] (a) emerature dros during throttling (b) emerature remains constant during throttling (c) emerature rises during throttling (d) None of these GAE-9. A gas having a negative Joule-homson coefficient (µ < 0), when throttled, will: [GAE-00] (a) Become cooler (b) Become warmer (c) Remain at the same temerature (d) Either be cooler or warmer deending on the tye of gas GAE-0. Match 4 correct airs between List-I and List-II for the questions For a erfect gas: List-I List-II (a) Isobaric thermal exansion coefficient. 0 (b) Isothermal comressibility. (c) Isentroic comressibility 3. /v (d) Joule homson coefficient 4. / 5. / 6. /γ [GAE-994] Previous 0-Years IES Questions Some Mathematical heorems IES-. Given: [IES-993] = ressure, = emerature, v = secific volume Which one of the following can be considered as roerty of a system? d. dv d v. d ( a) dv ( b) vd ( c) + ( d) v Maxwell's Equations IES-. Which thermodynamic roerty is evaluated with the hel of Maxwell equations from the data of other measurable roerties of a system? [IES 007] (a) Enthaly (b) Entroy (c) Latent heat (d) Secific heat 48

36 hermodynamic Relations S K Mondal s Chater 6 IES-3. Consider the following statements ertaining to the Claeyron equation:. It is useful to estimate roerties like enthaly from other measurable roerties. [IES-006]. At a change of hase, it can be used to find the latent heat at a given ressure. s 3. It is derived from the relationshi = v Which of the statements given above are correct? (a), and 3 (b) Only and (c) Only and 3 (d) Only and 3 IES-3a Claeyron s equation is used for finding out the [IES-0] (a) Dryness fraction of steam only (b) Entroy of suerheater vaour only (c) Secific volume at any temerature and ressure (d) otal heat of suerheated steam only ds Equations IES-4. ds equation can be exressed as: [IES-00] β dv dv (a) ds = Cvd + (b) ds = Cvd + k k k β (c) ds = Cvd + dv (d) ds = Cvd + d β k IES-5. Which one of the following statements alicable to a erfect gas will also be true for an irreversible rocess? (Symbols have the usual meanings). [IES-996] (a) dq = du + d (b) dq = ds (c) ds = du + d (d) None of the above IES-6. Consider the following thermodynamic relations: [IES-000]. ds = du + dv. ds = du dv 3. ds = dh+ vd 4. ds = dh vd Which of these thermodynamic relations are correct? (a) and 3 (b) and 4 (c) and 3 (d) and 4 Difference in Heat Caacities and Ratio of Heat Caacities IES-7. Match List-l (erms) with List-II (Relations) and select the correct answer using the codes given below the Lists: [IES-003] List-I (erms) List-II (Relations) A. Secific heat at constant volume, Cv v. B. Isothermal comressibility k. v v v 49

37 hermodynamic Relations S K Mondal s Chater 6 C. olume exansivity β s 3. v D. Difference between secific heats at constant v ressure and at constant C Cv 4. v Codes: A B C D A B C D (a) 3 4 (b) 4 3 (c) 3 4 (d) 4 3 IES-8. IES-9. IES-0. Assertion (A): Secific heat at constant ressure for an ideal gas is always greater than the secific heat at constant volume. [IES-00] Reason (R): Heat added at constant volume is not utilized for doing any external work. (a) Both A and R are individually true and R is the correct exlanation of A (b) Both A and R are individually true but R is NO the correct exlanation of A (c) A is true but R is false (d) A is false but R is true An insulated box containing 0.5 kg of a gas having Cv = 0.98 kj/kgk falls from a balloon 4 km above the earth s surface. What will be the temerature rise of the gas when the box hits the ground? [IES-004] (a) 0 K (b) 0 K (c) 40 K (d) 60 K As comared to air standard cycle, in actual working, the effect of variations in secific heats is to: [IES-994] (a) Increase maximum ressure and maximum temerature (b) Reduce maximum ressure and maximum temerature (c) Increase maximum ressure and decrease maximum temerature (d) Decrease maximum ressure and increase maximum temerature IES-. he number of degrees of freedom for a diatomic molecule [IES-99] (a) (b) 3 (c) 4 (d) 5 C IES-. he ratio for a gas with n degrees of freedom is equal to: [IES-99] C v (a) n + I (b) n I (c) n (d) + n IES-3. Molal secific heats of an ideal gas deend on [IES-00] (a) Its ressure (b) Its temerature (c) Both its ressure and temerature (d) he number of atoms in a molecule IES-4. Assertion (A): Ratio of secific heats C Cv decreases with increase in temerature. [IES-996] Reason (R): With increase in temerature, C decreases at a higher rate than Cv. (a) Both A and R are individually true and R is the correct exlanation of A (b) Both A and R are individually true but R is NO the correct exlanation of A 50

38 hermodynamic Relations S K Mondal s Chater 6 (c) A is true but R is false (d) A is false but R is true IES-5. It can be shown that for a simle comressible substance, the relationshi P C Cv = exists. [IES-998] v Where C and C v are secific heats at constant ressure and constant volume resectively. is the temerature is volume and P is ressure. Which one of the following statements is NO true? (a) C is always greater than C v. (b) he right side of the equation reduces to R for ideal gas. P (c) Since can be either ositive or negative, and v P must have a sign that is oosite to that of v (d) Is very nearly equal to for liquid water. must be ositive, Joule-Kelvin Effect or Joule-homson coefficient IES-6. Joule-homson coefficient is defined as: [IES-995] h h (a) (b) (c) (d) h h IES-7. he throttling of certain gasses may be used for getting the refrigeration effect. What is the value of Joule homson coefficient (µ) for such a throttling rocess? [IES-007] (a) µ = 0 (b) µ = (c) µ < (d) µ > IES-8. Which one of the following is correct? [IES 007] When a real gas undergoes Joule-homson exansion, the temerature (a) May remain constant (b) Always increases (c) May increase or decrease (d) Always decreases IES-9. IES-0. Assertion (A): hrottling rocess for real gases at initial temerature higher than maximum inversion temerature is accomanied by decrease in temerature of the gas. [IES-003] Reason (R): Joule-Kelvin coefficient μj is given ( / ) and should have a h ositive value for decrease in temerature during throttling rocess. (a) Both A and R are individually true and R is the correct exlanation of A (b) Both A and R are individually true but R is NO the correct exlanation of A (c) A is true but R is false (d) A is false but R is true Match List-I (Name of entity) with List-II (Definition) and select the correct answer using the codes given below the lists: [IES-00] List-I (Name of entity) List-II (Definition) 5

39 hermodynamic Relations S K Mondal s Chater 6 A. Comressibility factor. B. Joule homson coefficient. v v h C. Constant ressure secific heat 3. h D. Isothermal comressibility 4. v R Codes: A B C D A B C D (a) 4 3 (b) 4 3 (c) 3 4 (d) 4 3 IES-. Joule homson coefficient is the ratio of [IES-999] (a) Pressure change to temerature change occurring when a gas undergoes the rocess of adiabatic throttling (b) emerature change to ressure change occurring when a gas undergoes the rocess of adiabatic throttling (c) emerature change to ressure change occurring when a gas undergoes the rocess of adiabatic comression (d) Pressure change to temerature change occurring when a gas undergoes the rocess of adiabatic comression IES-. he Joule homson coefficient is the [IES-996] (a) h of ressure-temerature curve of real gases (b) v of temerature-entroy curve of real gases h (c) of enthaly-entroy curve of real gases s (d) of ressure-volume curve of real gases IES-3. Match the following: [IES-99] List-I List-II A. Work. Point function B. Heat. ds C. Internal energy 3. Code: A B C D A B C D 5 u D. Joule homson Coefficient 4. dv h

40 hermodynamic Relations S K Mondal s Chater 6 (a) 4 3 (b) 4 3 (c) 4 3 (d) 4 3 Clausius-Claeyron Equation IES-4. Consider the following statements in resect of the Clausius Claeyron equation: [IES-007]. It oints to one ossible way of measuring thermodynamic temerature.. It ermits latent heat of vaorization to be estimated from measurements of secific volumes of saturated liquid, saturated vaour and the saturation temeratures at two nearby ressures. 3. It does not aly to changes from solid to the liquid hase and from solid to the aour hase. Which of the statements given above are correct? (a), and 3 (b) and only (c) and 3 only (d) and 3 only IES-5. he equation relating the following measurable roerties: [IES-005] (i) he sloe of saturation ressure temerature line (ii) he latent heat, and (iii) he change in volume during hase transformation; is known as: (a) Maxwell relation (b) Joules equation (c) Claeyron equation (d) None of the above IES-6. IES-7. IES-8. he variation of saturation ressure with saturation temerature for a liquid is 0. bar/k at 400 K. he secific volume of saturated liquid and dry saturated vaour at 400 K are 0.5 and 0.00 m 3 /kg What will be the value of latent heat of vaorization using Clausius Claeyron equation? [IES-004] (a) 6000 kj/kg (b) 600 kj/kg (c) 000 kj/kg (d) 60 kj/kg If h,, and v refer to enthaly, ressure, temerature and secific volume resectively and subscrits g and f refer to saturation conditions of vaour and liquid resectively then Clausius-Claeyron equation alied to change of hase from liquid to vaour states is: [IES-996, 006] d ( hg hf ) d ( hg hf) (a) = (b) = dt ( v v ) dt ( v v ) (c) d dt g ( hg hf ) f = (d) d ( vg vf ) = dt ( h h ) Which one of the following functions reresents the Claeyron equation ertaining to the change of hase of a ure substance? [IES-00] (a) f (,, hfg) (b) f (,, hfg, vfg) (c) f (,, hfg, sfg) (d) f (,, hfg, sfg, vfg) g g f f IES-9. he Claeyron equation with usual notations is given by: [IES-000] d hfg dp hfg d hfg dp hfg ( a) = ( b) = ( c) = ( d) = dp v d v dp v d v sat fg sat fg sat fg sat fg IES-30. Clausius-Claeyron equation gives the 'sloe' of a curve in [IES-999] (a) v diagram (b) h diagram (c) diagram (d) S diagram 53

41 hermodynamic Relations S K Mondal s Chater 6 IES-3. he thermodynamic arameters are: [IES-997] I. emerature II. Secific olume III. Pressure I. Enthaly. Entroy he Claeyron Equation of state rovides relationshi between: (a) I and II (b) II, III and (c) III, I and (d) I, II, III and I Gibbs Phase Rule IES-3. Number of comonents (C), hase (P) and degrees of freedom (F) are related by Gibbs-hase rule as: [IES-00] (a) C P F = (b) F C P = (c) C + F P = (d) P + F C = IES-33. As er Gibb's hase rule, if number of comonents is equal to then the number of hases will be: [IES-00] (a) (b) 3 (c) 4 (d) 5 IES-34. Gibb's hase rule is given by: [IES-999] (F = number of degrees of freedom; C = number of comonents; P = number of hases) (a) F = C + P (b) F = C + P (c) F = C P (d) F = C P + IES-35. Gibb's free energy 'c' is defined as: [IES-999] (a) G = H S (b) G = U S (c) G = U + (d) G = H + S IES-36. Which one of the following relationshis defines the Helmholtz function F? [IES-007] (a) F = H + S (b) F = H S (c) F = U S (d) F = U +S IES-37. Assertion (A): For a mixture of solid, liquid and vaour hases of a ure substance in equilibrium, the number of indeendent intrinsic roerties needed is equal to one. [IES-005] Reason(R): he hree hases can coexist only at one articular ressure. (a) Both A and R are individually true and R is the correct exlanation of A (b) Both A and R are individually true but R is NO the correct exlanation of A (c) A is true but R is false (d) A is false but R is true IES-38. Consider the following statements: [IES-000]. Azeotroes are the mixtures of refrigerants and behave like ure substances.. Isomers refrigerants are comounds with the same chemical formula but have different molecular structures. 3. he formula n + + q = m is used for unsaturated chlorofluorocarbon comounds (m, n, and q are the numbers atoms of carbon, hydrogen, fluorine and chlorine resectively). Which of these statements are correct? (a) and 3 (b) and 3 (c) and (d), and 3 54

42 hermodynamic Relations S K Mondal s Chater 6 Previous 0-Years IAS Questions Maxwell's Equations IAS-. According to the Maxwell relation, which of the following is/are correct? v s s P (a) = (b) = [IAS-007] P v (c) v P s = v ds Equations IAS-. (d) All of the above Which one of the following exressions for ds is true for a simle comressible substance? (Notations have the usual meaning) [IAS-998] (a) dh vd (b) dh + vd (c) dh dv (d) dh + dv Difference in Heat Caacities and Ratio of Heat Caacities IAS-3. he secific heat C is given by: [IAS-000] (a) v (b) s (c) v s (d) v IAS-4. IAS-5. IAS-6. s s For an ideal gas the exression is always equal to: v c ( a) zero ( b) ( c) R ( d) R [IAS-003] c v Assertion (A): Secific heat at constant ressure for an ideal gas is always greater than the secific heat at constant volume. [IAS-000] Reason (R): Heat added at constant volume is not utilized for doing any external work. (a) Both A and R are individually true and R is the correct exlanation of A (b) Both A and R are individually true but R is NO the correct exlanation of A (c) A is true but R is false (d) A is false but R is true Match List-I with List-II and select the correct answers using the codes given below the lists. [IAS-00] List-I List-II A. Joule homson co-efficient. 5 R B. C for monatomic gas. Cv 55

43 hermodynamic Relations S K Mondal s Chater 6 C. C Cv for diatomic gas 3. R D. U v 4. P Codes: A B C D A B C D (a) 3 4 (b) 4 3 (c) 3 4 (d) 4 3 h IAS-7. Ratio of secific heats for an ideal gas is given by (symbols have the usual meanings) [IAS-999] (a) R C (b) C R (c) + C R (d) R + C Joule-Kelvin Effect or Joule-homson coefficient IAS-8. Which one of the following roerties remains unchanged for a real gas during Joule-homson rocess? [IAS-000] (a) emerature (b) Enthaly (c) Entroy (d) Pressure Clausius-Claeyron Equation IAS-9. If h,, and v refer to enthaly, ressure, temerature and secific volume resectively and subscrits g and f refer to saturation conditions of vaour and liquid resectively then Clausius-Claeyron equation alied to change of hase from liquid to vaour states is: [IAS-003] d ( hg hf ) d ( hg hf) (a) = (b) = dt ( v v ) dt ( v v ) (c) d dt g ( hg hf ) f = (d) d ( vg vf) = dt ( h h ) IAS-0. Which one of the following is the correct statement? [IAS-007] Claeyron equation is used for: (a) Finding secific volume of vaour (b) Finding secific volume of liquid (c) Finding latent heat of vaorization (c) Finding sensible heat g g f f IAS-. Assertion (A): Water will freeze at a higher temerature if the ressure is increased. [IAS-003] Reason (R): Water exands on freezing which by Claeyron's equation gives negative sloe for the melting curve. (a) Both A and R are individually true and R is the correct exlanation of A (b) Both A and R are individually true but R is NO the correct exlanation of A (c) A is true but R is false (d) A is false but R is true 56

44 hermodynamic Relations S K Mondal s Chater 6 IAS-. Match List-I with List-II and select the correct answer using the codes given below the lists [IAS-994] List-I List-II A. Mechanical work. Clausius-Claeyron equation dq B. 0. Gibb's equation C. Zeroth Law 3. High grade energy D. H S 4. Concet of temerature Codes: A B C D A B C D (a) 3 4 (b) 3 4 (c) 3 (d) 3 4 Gibbs Phase Rule IAS-3. Which one of the following relationshis defines Gibb's free energy G? [IAS-007] (a) G = H + S (b) G = H S (c) G = U + S (d) G = U S IAS-4. he Gibbs free-energy function is a roerty comrising [IAS-998] (a) Pressure, volume and temerature (b) Ethaly, temerature and entroy (c) emerature, ressure and enthaly (d) olume, ethaly and entroy 57

45 hermodynamic Relations S K Mondal s Chater 6 Answers with Exlanation (Objective) Previous 0-Years GAE Answers GAE-. Ans. (d) GAE-. Ans. (d) GAE-3. Ans. (a) GAE-4. Ans. (d) S S dq GAE-5. Ans. (c) = = = CP P P P S S dq = = = C S S = CP C = R P GAE-6. Ans. (c) Heat absorbed by water = Heat sulied by heater. o mwcw ( Δ ) = P t or ( Δ ) = 0 60 or ( Δ ) = 4.3 C w w w GAE-7. Ans. (b) GAE-8. Ans. (a) μ = i,e. μ > o, P is ( ive) so must be ive. P h GAE-9. Ans. (b) Joule-homson co-efficient. Here,, ive and P, -ive so must be h P h +ive so gas will be warmer. GAE-0. Ans. (a) 4, (b) 5, (c) 6, (d) Previous 0-Years IES Answers IES-. Ans. (d) P is a function of v and both are connected by a line ath on and v coordinates. hus dv and vd are not exact differentials and thus not roerties. If X and Y are two roerties of a system, then dx and dy are exact differentials. If the M N differential is of the form Mdx + Ndy, then the test for exactness is y = x x y Now alying above test for d. dv + v, ( / ) ( / v) ( R / v ) R = = or 0 = v v v v his differential is not exact and hence is not a oint function and hence d +. dv v is not a oint function and hence not a roerty. 58

46 hermodynamic Relations S K Mondal s Chater 6 And for hus d v. d ( / ) ( v/ ) ( R/ P) or 0 0 = = = P P d v. d is exact and may be written as ds, where s is a oint function and hence a roerty IES-. Ans. (a) From Maxwell relation Claeyron equation comes. IES-3. Ans. (b) 3 is false. It is derived from the Maxwell s 3 rd relationshi s = v IES-3a Ans. (c) IES-4. Ans. (a) IES-5. Ans. (c) he relations in (a) and (b) are alicable for a reversible rocesses and (c) ds = du + d is a relation among roerties which are indeendent of the ath. IES-6. Ans. (b) IES-7. Ans. (c) IES-8. Ans. (a) IES-9. Ans. (c) Potential energy will converted to heat energy. gh mgh = mcvδ or Δ = = = 40K c 980 v IES-0. Ans. (b) IES-. Ans. (d) A diatomic gas (such as that of oxygen) has six degrees of freedom in all-three corresonding to translator motion, two corresonding to rotatory motion and one corresonding to vibratory motion. Exeriments have shown that at ordinary temeratures, the vibratory motion does not occur. Hence, at 7 C, an oxygen molecule has just five degrees of freedom. IES-. Ans. (d) v IES-3. Ans. (b) For ideal gas CP and C are constant but mole is deends on the number of atoms in a molecule. IES-4. Ans. (c). A is correct but R is false. We know that C = a+k+k +K 3 C v = b+ K+K +K 3 See C and C v both increase with temerature and by same amount. As C > C v then C ercentage increase of C is less than C v. So decreases with temerature. C IES-5. Ans. (c) Sign of must be ositive P v v is always negative. IES-6. Ans. (a) IES-7. Ans. (d) Actually Joule homson coefficient will be ositive. IES-8. Ans. (c) For ideal gas µ = 0 and for real gas µ may be ositive (N, O, CO etc.) or negative (H). 59

47 S K Mondal s IES-9. Ans. (d) When a real gas which is initially at a temerature lower than the maximum inversion temerature is throttled, its temerature decreases. herm modynamic Relations Chater 6 IES-0. Ans. (b) IES-. Ans. (b) Joule homson coefficient is the ratio of temerature change to ressure change when a gas undergoes adiabatic throttling. IES-. Ans. (a) he sloe of the isenthalic curve at any oint is known as Joule-homson coefficient and is exressed as, μ = h IES-3. Ans. (a) IES-4. Ans. (b) IES-5. Ans. (c) dp hfg IES-6. Ans. (c) = d sat ( g f) dp 5 or hfg = ( g f ) = 400 ( ) 0. 0 J / kg = 000kJ / kg d sat IES-7. Ans. (b) IES-8. Ans. (b) IES-9. Ans. (b) IES-30. Ans. (c) IES-3. Ans. (d) Claeyron equation state rovides relationshi between temerature, secific volume, ressure and enthaly. IES-3. Ans. (d) IES-33. Ans. (c) IES-34. Ans. (d) F = C P + IES-35. Ans. (a) Gibb's free energy 'G' is defined as G = H S. IES-36. Ans. (c) IES-37. Ans. (d) F = C P + C =, P = 3 or F = 3 + = 0 IES-38. Ans. (d) Isomers: : Comounds with the same chemical formula but different molecular structure. 60

48 hermodynamic Relations S K Mondal s Chater 6 Previous 0-Years IAS Answers IAS-. Ans. (c) P S = P see highlights. IAS-. Ans. (a) dq = dh d or ds = dh d dq s C = = dq = ds IAS-3. Ans. (c) [ ] to memorize Maxwell s relation remember P S, ive and S S IAS-4. Ans. (c) S S dq = = = CP P P P S S dq C = = = S S CP C R = = P IAS-5. Ans. (a) Both A and R correct and R is the correct exlanation of A IAS-6. Ans. (b) C Cv for all ideal gas is R, So C-3, (a) & (c) out. A automatically match 4, and C γ = R γ for monatomic gas γ = 5. So, γ = 5. 3 R C C IAS-7. Ans. (a) C Cv = R and γ = = = C R v C R C IAS-8. Ans. (b) IAS-9. Ans. (b) IAS-0. Ans. (c) IAS-. Ans. (a) IAS-. Ans. (d) IAS-3. Ans. (b) IAS-4. Ans. (b) 6

49 S K Mondal s Pure Substances Chater 7 7. Pure Substances heory at a Glance (For GAE, IES & PSUs) A ure substance is defined as one that is homogeneous and invariable in chemical comosition throughout its mass. he relative roortions of the chemical elements constituting the substance are also constant. Atmosheric air, steam water mixture and combustion roducts of a fuel are regarded as ure substances. But the mixture of air and liquid air is not a ure substance, since the relative roortions of oxygen and nitrogen differ in the gas and liquid hases in equilibrium. Mixtures are not ure substances. (e.g., Humid air) Excetion!! Air is treated as a ure substance though it is a mixture of gases. In a majority of cases a minimum of two roerties are required to define the state of a system. he best choice is an extensive roerty and an intensivee roerty. he state of a ure substance of given mass can be fixed by secifying two indeendent intensive roerties, rovided the system is in equilibrium. his is known as the 'two-roerty rule'. he state can thus be reresented as a oint on thermodynamic roerty diagrams. Once any two roerties of a ure substance are known, other roerties can be determined from the available thermodynamic relations. -v Diagram for a Pure Substance 63

50 S K Mondal s Pure Substances Chater 7 rile oint rile oint is a oint where all the three hases exist in equilibrium. u = 0 for water at trile oint. (You can assign any number you like instead of 0). Since, and v for water at trile oint are known you can calculate h for water at trile oint (it will not be zero). he entroy of saturated water is also chosen to be zero at the trile oint. Gibbs Phase Rule at rile Point If C =, φ = 3, then f = 0. he state is thus unique for a substance; and refers to the trile oint where all the three hases exist in equilibrium. ln = A + B/ valid only near the trile oint.(called Antoine s equation) 64

51 S K Mondal s Pure Substances Chater 7 - Diagram for a Pure Substance 65

52 S K Mondal s -v- Surface Pure Substances Chater 7 -s Diagram for a Pure Substance Critical Point A state at which a hase change begins or ends is called a saturation state. he dome shaed region comosed of the two-hase liquid vaor states is called the vaor dome. he lines bordering the vaor dome are called saturated liquid and saturated vaor lines. At the to of the dome, where the saturated liquid and saturated vaor lines meet, is the critical oint. he critical temerature c of a ure substance is the maximum temeraturee at which liquid and vaor hases can coexist in equilibrium. he ressure at the critical oint is called the critical ressure, c. he secific volume at this state is the critical secific volume. 666

53 S K Mondal s he isotherm assing through the critical oint is called the critical isotherm, and the corresonding temerature is knownn as the critical temerature (tc). he ressure and volume at the critical oint are known as the critical ressure (Pc) and the critical volume (vc) resectively. For water Pc =. bar tc = C vc = m 3 /kg Pure Substances Chater 7 Characteristics of the critical oint: Fig. Critical Point on P v, S, h s diagram It is the highest temerature at which the liquid and vaour hases can coexist. At the critical oint hfg, h ufg and vfg are zero. Liquid vaour meniscus will disaear. Secific heat at constant ressure is infinite. A majority of engineering alications (e.g., steam based ower generation; Refrigeration, gas liquefaction) involve thermodynamic rocesses close to saturation. For saturated hase often it enthaly is an imortant roerty. Enthaly-ressure charts are used for refrigeration cycle analysis. Enthaly-entroyy charts for water are used for steam cycle analysis. Note: Unlike ressure, volume and temerature whichh have secified numbers associated with it, in the case of internal energy, enthaly (and entroy) only changes are required. Consequently, a base (or datum) is definedd - as you have seen in the case of water. For examle for NIS steam tables u = 0 for water at trile oint. (You can assign any number you like instead of 0). [Don t be surrised if two different sets of steam tables give different values for internal energy and enthaly]. 67

54 S K Mondal s Since, and v for water at trile oint are known you can calculate h for water at trile oint (it will not be zero). From the first and second laws of thermodynamics, the following roerty relation was obtained. or ds = dh - vd Pure Substances h-s Diagram or Mollier Diagram for a Pure Substance h s = Chater 7 his equation forms the basis of the h-s diagram of a ure substance, also called the Mollier diagram. he sloe of an isobar on the h-s coordinates is equal to the absolute saturation temerature (tsat + 73) at that ressure. If the temerature remains constant the sloe will remain constant. If the temerature increases, the sloe of the isobar will increase. Quality or Dryness Fraction he zone between the saturated liquid and the saturated vaour region is called the two hase region - where the liquid and vaour can co-exist in equilibrium. Dryness fraction: It is the mass fraction of vaour in the mixture. Normally designated by x. On the saturated liquid line x = 0 On the saturated vaour line x = x can have a value only between 0 and Data tables will list roerties at the two ends of saturation. o calculate roerties in the two-hase region:, will be the same as for saturated liquid or saturated vaour. 68

55 Pure Substances S K Mondal s Chater 7 AND g g g ( ) v = x v + x v ( ) h = x h + x h ( ) u = x u + x u v = v f + xvfg u = u f + xufg h = h f + xhfg s = s f + xsfg f f f Saturation States When a liquid and its vaour are in equilibrium at a certain ressure and temerature, only the ressure or the temerature is sufficient to identify the saturation state. If the ressure is given, the temerature of the mixture gets fixed, which is known as the saturation temerature, or if the temerature is given, the saturation ressure gets fixed. Saturated liquid or the saturated vaour has only one indeendent variable, i.e, only one roerty is required to be known to fix u the state. a Steam able(saturated state) give the roerties of sutured liquid and saturated vaour. In steam table the indeendent variable is temerature. At a articular temerature, the values of saturation ressure, and states,,h and S g g g refer to the saturated vaour state; and v,h, and s fg fg fg refer to the changes in the roerty values during evaoration (or condensation) at the temerature. where v fg = v g v f and s fg = s g s f. If steam able the indeendent variable is ressure. At a articular ressure, the values of saturation temerature t, and v f, v g, h f, h fg, s f, and s g are given. whether the ressure or the temerature is given, either steam able can be conveniently used for comuting the roerties of saturation states. 69

56 S K Mondal s Pure Substances Chater 7 If dataa are required for intermediate temerature or ressures, linear interolation is normally accurate. he reason for the two tables is to reduce the amount of interolation required. Liquid vaour Mixtures: Let us consider a mixture of saturated liquid water and water vaour in equilibrium at ressure and temerature t. he comosition of the mixture by mass will be within the vaour dome figure. he roerties of the mixture are as given in Article. i.e Fig. where v, v, h, s and s fg are the saturation roerties at the given ressure and temerature. f fg fg f If or t and the quality of the mixture are given, the roerties of the mixture (v, u, h and s) can be evaluated from the above equations. Sometimes instead of quality, one of the above equations say secific volume v and ressure or temerature are given. In that case, the quality of the mixture x has to be calculated from the given v and or t and then x being known, other roerties are evaluated. If or t and the quality of the mixture are given, the roerties of the mixture (v, u, h and s) can be evaluated from the above equations. Sometimes, instead of quality, one of the above roerties, say, secific volume v and ressure or temerature are given in that case, the quality of the mixture x has to be calculated from the given v and or t and then x being known, other roerties are evaluated. Suerheated aour: When the temerature of the vaour is greater than the saturation temerature corresonding to the given ressure, the vaour is said to be suerheated (state in figure). he difference between the temerature of the suerheated vaour and the saturation temerature at that ressure is called the suerheat or the degree of suerheat. As shown in figure below, the differencee (t - tsat) is the suerheat. (Fig. - Suerheat and Sub-cooling.) 70

57 Pure Substances S K Mondal s Chater 7 In a suerheated vaour at a given ressure, the temerature may have different values greater than the saturation temerature. Steam able gives the values of the roerties (volume, enthaly, and entroy) of suerheated vaour for each tabulated air of values of ressure and temerature, both of which are now indeendent. Interolation or extraolation is to be used for airs of values of ressure and temerature not given. Comressed Liquid: When the temerature of a liquid is less than the saturation temerature at the given ressure, the liquid is called comressed liquid (state in figure above). he ressure and temerature of comressed liquid may vary indeendently, and a table of roerties like the suerheated vaour table could be arranged, to give the roerties at any and t. However, the roerties of liquids vary little with ressure. Hence the roerties are taken from the saturation tables at the temerature of the comressed liquid. When a liquid is cooled below its saturation temerature at a certain ressure it is said to be sub-cool. he difference in saturation temerature and the actual liquid temerature is known as the degree of Sub-cooling, or simly, Sub-cooling. Charts of hermodynamic Proerties One of the imortant roerties is the change in enthaly of hase transition hfg also called the latent heat of vaorisation or latent heat of boiling. It is equal to hg - hf. Similarly u - internal energy change due to evaoration and v - volume change due to fg fg evaoration can be defined (but used seldom). he saturation hase deicts some very interesting roerties: he following saturation roerties deict a maximum:. ρ. ( ρ ρg) 3. h 4. f f fg c c v v 7. ( ρ ρ ρ ) 8. h ( ) ( ) c g f c f g g ( ) he equation relating the ressure and temerature along the saturation is called the vaour ressure curve. Saturated liquid hase can exist only between the trile oint and the critical oint. Measurement of Steam Quality hrottling calorimeter: hrottling calorimeter is a device for determining the quality of a twohase liquid vaor mixture. 7

58 S K Mondal s Pure Substances Chater 7 Intermediate states are non-equilibrium states not describable by thermodynamic coordinates. he initial state (wet) is given by and x, and the final state by and t (suerheated). Now since h = h H + x h = h f fg h h f or x = h fg With and t being known, h can be found out from the suerheated steam table. he values of hf and hfg are taken from the saturated steam table corresonding to ressure. herefore, the quality of the wet steam x can be calculated. he hrottling Calorimeter Examle A suly line carries a two-hase liquid vaor mixture of steam at 0 bars. A small fraction of the flow in the line is diverted throughh a throttling calorimeter and exhausted to the atmoshere at bar. he temerature of the exhaust steam is measured as 0ºC. Determine the quality of the steam in the suly line. Solution Known: Steam is diverted from a suly line through a throttling calorimeter and exhausted to the atmoshere. Find: Determine the quality of the steam in the suly line. Schematicc and Given Data: 7

59 S K Mondal s Pure Substances Chater 7 Assumtions:.. he control volume shown on the accomanying figure is at steady state. he diverted steam undergoes a throttling rocess. Analysis: For a throttling rocess, the energy and mass balances reduce to give h = h,, hus, with state fixed, the secific enthaly in the suly line is known, and state is fixed by the known values of and h. As shown on the accomanying v diagram, state is in the two-hase liquid vaor region and state is in the suerheated vaor region. hus Solving for x h = h = h + x h _ h f g f x h = h ( h f h g f From steam table at 0 bars, hf = kj/kg and hg = kj/kg. At bar and 0ºC, h = kj/kg from steam table into the above exression, the quality in the line is x = (95.6%). ) For throttling calorimeters exhausting to the atmoshere, the quality in the line must be greater than about 94% to ensure that the steam leaving the calorimeterr is suerheated. hrottling When a fluid flows through a constricted assage, like a artially oened valve, an orifice, or a orous lug, there is an areciable dro in ressure, and the flow is said to be throttled. Figure shows the rocess of throttling by a artially oened valve on a fluid flowing in an insulated ie. In the steady-flow energy equation. 73

60 S K Mondal s dq dm = Pure Substances 0, dw x dm = and the changes in P.E. are very small and ignored. hus, the S. F.E.E. reduces to C C h + = h + 0 Chater 7 Often the ie velocities in throttling are so low thatt the K.E. terms are also negligible. So h = h or the enthaly of the fluid before throttling is equal to the enthaly of the fluid after throttling. Consider a throttling rocess (also referred to as wire drawing rocess) here is no work done (rising a weight) If there is no heat transfer Conservation of mass requires that Since and are at the same level W = 0 Q = 0 C = C Z = Z C 74

61 Pure Substances S K Mondal s Chater 7 From SFEE it follows that h = h Conclusion: hrottling is a constant enthaly rocess (isenthalic rocess) 75

62 Pure Substances S K Mondal s Chater 7 Proerties of Pure Substances Highlights. rile Point On P diagram it is a Point. On P diagram it is a Line. Also in S diagram it is a Line. On U diagram it is a riangle.. rile oint of water = 73.6 K P = bar Entroy (s) = 0 Internal energy (u) =0 = 0.0ºC = mm of Hg Enthaly = u + P = slightly ositive. 3. rile oint of CO P 5 atm and = 6.55K = C that so why sublimation occurred. 4. Critical oint For water c =. bar 5.5 kgf/cm tc = 374.5ºC K vc = m 3 /kg At critical oint h = 0 ; v = 0 ; s = 0 fg fg fg 5. Mollier Diagram Basis of the h s diagram is ds = dh vd h h s = = P s P he sloe of an isobar on the h s co-ordinates is equal to the absolute saturation temerature at that ressure. And for that isobars on Mollier diagram diverges from one another. 6. Dryness fraction 76

63 Pure Substances S K Mondal s Chater 7 x = mv m + m v L 7. v = ( x) vf + xvg v = vf + xvfg u = ( x) uf + xug u = uf + xufg h = ( x) hf + xhg and h = hf + xhfg s = ( x) sf + xsg s = sf + xsfg 8. Suer heated vaour: When the temerature of the vaour is greater than the saturation temerature corresonding to the given ressure. 9. Comressed liquid: When the temerature of the liquid is less than the Saturation temerature at the given ressure, the liquid is called comressed liquid. 0. In combined calorimeter x = x x x = from throttle calorimeter. x = from searation calorimeter. 77

64 Pure Substances S K Mondal s Chater 7 ASKED OBJECIE QUESIONS (GAE, IES, IAS) Previous 0-Years GAE Questions Common data for Question Q Q3 In the figure shown, the system is a ure substance ket in a iston-cylinder arrangement. he system is initially a two-hase mixture containing kg of liquid and 0.03 kg of vaour at a ressure of 00 kpa. Initially, the iston rests on a set of stos, as shown in the figure. A ressure of 00 kpa is required to exactly balance the weight of the iston and the outside atmosheric ressure. Heat transfer takes lace into the system until its volume increases by 50%. Heat transfer to the system occurs in such a manner that the iston, when allowed to move, does so in a very slow (quasi-static I quasi-equilibrium) rocess. he thermal reservoir from which heat is transferred to the system has a temerature of 400 C. Average temerature of the system boundary can be taken as 7 C. he heat transfer to the system is I kj, during which its entroy increases by 0 J/K. Atmosheric ressure. Secific volumes of liquid (vf) and vaour (vg) hases, as well as values of saturation temeratures, are given in the table below. Pressure (kpa) Saturation temerature, sat ( C) vf(m 3 /kg) vg(m 3 /kg) GAE-. At the end of the rocess, which one of the following situations will be true? (a) Suerheated vaour will be left in the system [GAE-008] (b) No vaour will be left in the system (c) A liquid + vaour mixture will be left in the system (d) he mixture will exist at a dry saturated vaour state GAE-. he work done by the system during the rocess is: [GAE-008] (a) 0. kj (b) 0. kj (c) 0.3 kj (d) 0.4kJ GAE-3. he net entroy generation (considering the system and the thermal reservoir together) during the rocess is closest to: [GAE-008] (a) 7.5 J/K (b) 7.7 J/K (c) 8.5 J/K (d) 0 J/K 78

65 Pure Substances S K Mondal s -s Diagram for a Pure Substance Common Data for Questions GAE-4- GAE-5 A thermodynamic cycle with an ideal gas as working fluid is shown below. Chater 7 GAE-4. he above cycle is reresented on -S lane by [GAE-007] GAE-5. If the secific heats of the working fluid are constant and the value of secific heat ratio γ is.4, the thermal efficiency (%) of the cycle is: [GAE-007] (a) (b) 40.9 (c) 4.6 (d) 59.7 GAE-6. he sloes of constant volume and constant ressure lines in the -s diagram are.. and.. resectively. [GAE-994] 79

66 Pure Substances S K Mondal s Chater 7 h-s Diagram or Mollier Diagram for a Pure Substance GAE-7. Constant ressure lines in the suerheated region of the Mollier diagram have what tye of sloe? [GAE-995] (a) A ositive sloe (b) A negative sloe (c) Zero sloe (d) May have either ositive or negative sloes Quality or Dryness Fraction GAE-8. Consider a Rankine cycle with suerheat. If the maximum ressure in tile cycle is increased without changing the maximum temerature and the minimum ressure, the dryness fraction of steam after the isentroic exansion will increase. [GAE-995] hrottling Statement for Linked Answer Questions Q9 & Q0: he following table of roerties was rinted out for saturated liquid and saturated vaour of ammonia. he titles for only the first two columns are available. All that we know is that the other columns (columns 3 to 8) contain data on secific roerties, namely, internal energy (kj/kg), enthaly (kj/kg) and entroy (kj/kgk) [GAE-005] GAE-9. he secific enthaly data are in columns [GAE-005] (a) 3 and 7 (b) 3 and 8 (c) 5 and 7 (d) 5 and 8 GAE-0. When saturated liquid at 40 C is throttled to -0 C, the quality at exit will be [GAE-005] (a) 0.89 (b) 0. (c) 0.3 (d) GAE-. When wet steam flows through a throttle valve and remains wet at exit (a) its temerature and quality increases [GAE-996] (b) its temerature decreases but quality increases (c) its temerature increases but quality decreases (d) its temerature and quality decreases GAE-. When an ideal gas with constant secific heats is throttled adiabatically, with negligible changes in kinetic and otential energies [GAE-000] ( a) Δ h = 0, Δ = 0 ( b) Δ h > 0, Δ = 0 ( c) Δ h > 0, Δ S > 0 ( d) Δ h = 0, Δ S > 0 Where h, and S reresent resectively, enthaly, temerature and entroy, temerature and entroy GAE-3. One kilomole of an ideal gas is throttled from an initial ressure of 0.5 MPa to 0. MPa. he initial temerature is 300 K. he entroy change of the universe is: [GAE-995] 80

67 S K Mondal s (a) 3.38 kj/k Pure Substances Chater 7 ( b)40.3 kj/k (c) kj/k (d) kj/k Previous 0-Years IES Questions IES-. IES-. Assertion (A): Water is not a ure substance. [IES-999] Reason (R): he term ure substance designates a substance which is homogeneous and has the same chemical comosition in all hases. (a) Both A and R are individually true and R is the correct exlanation of A (b) Both A and R are individually true but R is NO the correct exlanation of A (c) A is true but R is false (d) A is false but R is true he given diagram showss an isometric cooling rocess of a ure substance. he ordinate and abscissa are resectively (a) Pressure and volume (b) Enthaly and entroy (c) emerature and entroy (d) Pressure and enthaly [IES-998] IES-3. he ordinate and abscissa in the given figure showing the saturated liquid and vaour regions of a ure substance reresent: (a) emerature and ressure (b) Enthaly and entroy (c) Pressure and volume (d) Pressure and enthaly IES-4. he given diagram shows the throttling rocess of a ure substance. he ordinate and abscissa are resectively (a) Pressure and volume (b) Enthaly and entroy (c) emerature and entroy (d) Pressure and enthaly [IES-997] [IES-995] IES-5. Which one of the following systems can be considered to be containing a ure substance? [IES-993] 8

68 S K Mondal s Pure Substances Chater 7 IES-6. IES-7. IES-8. IES-9. Consider the following: [IES-009]. Air. Gaseous combustion roducts 3. Steam Which of thesee are ure substances, assuming there is no hase change? (a) and only (b) and 3 only (c) and 3 only (d), and 3 Assertion (A): At a given temerature, the enthaly of suer-heated steam is the same as that of saturated steam. [IES-998] Reason (R): he enthaly of vaour at lower ressures is deendent on temerature alone. (a) Both A and R are individually true and R is the correct exlanation of A (b) Both A and R are individually true but R is NO the correct exlanation of A (c) A is true but R is false (d) A is false but R is true Considers the following roerties of vaour: [IES-009]. Pressure 3. Dryness fraction. emerature 4. Secific volume Which of these two roerties alone are not sufficient to secify the condition of a vaour? (a) and ( b) and 3 (c) and 3 (d) 3 and 4 Which one of the following is correct? [IES-008] he secific volume of water when heated from 0 C (a) First increases and then decreases (b) First decreases and then increases (c) Increases steadily (d) Decreases steadily v Diagram for a Pure Substance IES-0. Which v diagram for steam illustrates correctly the isothermal rocess undergone by wet steam till it becomes suerheated? [IES 995, 007] 8

69 S K Mondal s Pure Substances Chater 7 Diagram for a Pure Substance IES-. Consider the hase diagram of a certain substance as shown in the given figure. Match List-I (Process) with List-II ( Curves/lines) and select the correct answer using the codes given below the lists: List-I (Process) A. aorization B. Fusion C. Sublimation Codes: A (a) (c) 3 List-II (Curves/lines). EF. EG 3. ED B C 3 (b) (d) A B 3 C 3 [IES-00] 83

70 Pure Substances S K Mondal s Chater 7 -v- Surface IES-. he -v- surface of a ure substance is shown in the given figure. he two-hase regions are labelled as: (a) R, and X (b) S, U and W (c) S, W and (d) R, and -s Diagram for a Pure Substance IES-3. [IES-999] he conversion of water from 40 C to steam at 00 C ressure of bar is best reresented as [IES-994] 84

71 S K Mondal s IES-4. Critical Point IES-5. Pure Substances he following figure shows the -s diagram for steam. With resect to this figure, match List I with List II and select the correct answer using the codes given below the Lists: List-I A. Curve I B. Curve II C. Curve III D. Curve I List-II. Saturated liquid line. Saturated vaour line 3. Constant ressure line 4. Constant volume line Codes: A B C D A (a) 4 3 (b) (c) 3 4 (d) Which one of the following is correct? At critical oint the enthaly of vaorization is (a) deendent on temerature only (b) maximum (c) minimum (d) zero B C D Chater 7 [IES-994] [IES-008] IES-6. Consider the following statements about critical oint of water: [IES-993]. he latent heat is zero.. he liquid is denser than its vaour. 3. Steam generators can oerate above this oint. Of these statements (a), and 3 are correct (b) and are correct (c) and 3 are correct (d) and 3 are correct IES-7. IES-8. Which one of the following statements is correct when saturation ressure of a vaour increases? [IES 007] (a) Saturation temerature decreases (b) Enthaly of evaoration decreases (c) Enthaly of evaoration increases (d) Secific volume change of hase increases Match List I with List III and select the correct answer using the code given below the Lists: [IES-005] List-I List-III A. Critical oint. All the three hases - solid, liquid and vaour co-exists in equilibrium B. Sublimation. Phase change form solid to liquid C. rile oint 3. Proerties of saturated liquid and saturated vaour are identical D. Melting 4. Heating rocess where solid gets directly transformed to gaseous hase Codes: A B C D A B C D (a) 4 3 (b) 3 4 (c) 4 3 (d)

72 Pure Substances S K Mondal s Chater 7 IES-9. With increase of ressure, the latent heat of steam [IES-00] (a) Remains same (b) Increases (c) Decreases (d) Behaves unredictably IES-0. List-I gives some rocesses of steam whereas List-II gives the effects due to the rocesses. Match List I with List II, and select the correct answer using the codes given below the lists: [IES-995] List-I List-II A. As saturation ressure increases. Entroy increases. B. As saturation temerature increases. Secific volume increases. C. As saturation ressure decreases 3. Enthaly of evaoration decreases. D. As dryness fraction increases 4. Saturation temerature increases. Code: A B C D A B C D (a) 3 4 (b) 4 3 (c) 4 3 (d) 4 3 h-s Diagram or Mollier Diagram for a Pure Substance IES-. Which one of the following reresents the condensation of a mixture of saturated liquid and saturated vaour on the enthaly-entroy diagram? [IES-004] (a) A horizontal line (b) An inclined line of constant sloe (c) A vertical line (d) A curved line Measurement of Steam Quality IES-. Saturated liquid at a high ressure P having enthaly of saturated liquid 000 kj/kg is throttled to a lower ressure P. At ressure enthaly of saturated liquid and that of the saturated vaour are 800 and 800 kj/kg resectively. he dryness fraction of vaour after throttling rocess is: [IES-003] (a) 0. (b) 0.5 (c) 8/8 (d) 0.8 IES-3. IES-4. Consider the following statements regarding the throttling rocess of wet steam: [IES-00]. he steam ressure and temerature decrease but enthaly remains constant.. he steam ressure decreases, the temerature increases but enthaly remains constant. 3. he entroy, secific volume, and dryness fraction increase. 4. he entroy increases but the volume and dryness fraction decrease. Which of the above statements are correct? (a) and 4 (b) and 3 (c) and 3 (d) and 4 Match List-I (Aaratus) with List-II (hermodynamic rocess) and select the correct answer using the code given below the Lists: [IES-006] List-I List-II A. Searating calorimeter. Adiabatic rocess B. hrottling calorimeter. Isobaric rocess C. Sling sychrometer 3. Isochoric rocess 86

73 S K Mondal s D. Gas thermometer Codes: A B C D (a) 3 4 (c) 4 3 Pure Substances 4. Isenthalic rocess A B C D (b) 4 3 (d) 3 4 Chater 7 IES-5. Select the correct answer using the codes given below the Lists: List-I List-III A. Bomb calorimeter B. Exhaust gas calorimeterr C. Junker gas calorimeter D. hrottling calorimeter. Pressure. Enthaly 3. olume 4. Secific heats Code: A B C D A B C D (a) 3 4 (b) 4 3 (c) 3 4 (d) 4 3 [IES-998] hrottling IES-6. In a throttling rocess, which one of the following constant? (a) emeraturee ( b) Pressure (c) Enthaly arameters remains [IES-009] (d) Entroy IES-7. IES-8. Consider the following statements: [IES-000] When dry saturated steam is throttled from a higher ressure to a lower ressure, the Pressure decreases and the volume increases emerature decreases and the steam becomes suerheated emerature and the dryness fraction increase Entroy increases without any change in enthaly Which of thesee statements are correct? (a) and 4 (b), and 4 (c) and 3 (d) and 4 he rocess - for steam shown in the given figure is (a) Isobaric (b) Isentroic (c) Isenthalic (d) Isothermal IES-9. [IES-000] A fluid flowing along a ie line undergoes a throttling rocess from 0 bar to Bar in assing through a artially oen valve. Before throttling, the secific volume of the fluid is 0.5 m 3 /kg and after throttling is.0 m 3 / kg. What is the Change in secific internal energy during the throttling rocess? [IES 007] (a) Zero (b) 000 kj/kg (c) 00 kj/ /kg (d) 300 kj/kg 87

74 Pure Substances S K Mondal s Chater 7 IES-30. he throttling rocess undergone by a gas across an orifice is shown by its states in the following figure: IES-3. In the figure shown, throttling rocess is reresented by (a) a e (b) a d (c) a c (d) a b [IES-996] [IES-99] IES-3. Match List-l with List-Il and select the correct answer using the code given below the lists: [IES-009] List-l List-lI A. Isolated system. Energy is always constant B. Nozzle. Increase in velocity at the exense of its C. hrottling device ressure dro D. Centrifugal comressor 3. Areciable dro in ressure without any change in energy 4. Enthaly of the fluid increases by the amount of work inut Codes: A B C D A B C D (a) 4 3 (b) 3 4 (c) 4 3 (d)

75 S K Mondal s Pure Substances Previous 0-Years IAS Questions Chater 7 IAS-. Which one of the following systems can be considered to be containing a ure substance? [IAS 998] IAS-. IAS-3. IAS-4. IAS-5. Assertion (A): On the enthaly-entroy diagram of a ure substance the constant dryness fraction lines startt from the critical oint. [IAS-00] Reason (R): All the three hases co-exist at the critical oint. (a) Both A and R are individually true and R is the correct exlanation of A (b) Both A and R are individually true but R is NO the correct exlanation of A (c) A is true but R is false (d) A is false but R is true Assertion (A): Air, a mixture of O and N, is a ure substance. [IAS-000] Reason(R): Air is homogeneous in comosition and uniform in chemical aggregation. (a) Both A and R are individually true and R is the correct exlanation of A (b) Both A and R are individually true but R is NO the correct exlanation of A (c) A is true but R is false (d) A is false but R is true If a ure substance contained in a rigid vessel asses through the critical state on heating, its initial state should be: [IAS-998] (a) Subcooled water ( b) Saturated water (c) Wet steam (d) Saturated steam Assertion (A): Air is a ure substance but a mixture of air and liquid air in a cylinder is not a ure substance. [IAS-996] Reason (R): Air is homogeneous in comosition but a mixture of air and liquid air is heterogeneous. (a) Both A and R are individually true and R is the correct exlanation of A (b) Both A and R are individually true but R is NO the correct exlanation of A (c) A is true but R is false (d) A is false but R is true 89

76 S K Mondal s IAS-6. v Diagram for a Pure Substance IAS-7. Pure Substances Assertion (A): emerature and ressure are sufficient to fix the state of a two hase system. [IAS-995] Reason(R): wo indeendent and intensive roerties are required to be known to define the state of a ure substance. (a) Both A and R are individually true and R is the correct exlanation of A (b) Both A and R are individually true but R is NO the correct exlanation of A (c) A is true but R is false (d) A is false but R is true wo-hase regions in the given ressure-volume diagram of a ure substance are reresented by (a) A, E and F (b) B, C and D (c) B, D and F (d) A, C and E Chater 7 [IAS-999] IAS-8. A cyclic rocess ABC is shown on a diagram in figure. he same rocess on a P diagram will be reresent as: [IAS-996] 90

77 Pure Substances S K Mondal s Chater 7 IAS-9. he network done for the closed shown in the given ressure-volume diagram, is (a) 600kN-m (b) 700kN-m (c) 900kN-m (d) 000kN-m rile oint [IAS-995] IAS-0. rile oint temerature of water is: [IAS-000] (a) 73 K (b) 73.4 K (c) 73.5K (d) 73.6 K Diagram for a Pure Substance IAS-. In the following P- diagram of water showing hase equilibrium lines, the sublimation line is: (a) (b) q (c) r (d) S -s Diagram for a Pure Substance IAS-. [IAS-998] Entroy of a saturated liquid at 7 C is.6 kj/kgk. Its latent heat of vaorization is 800 kj/kg; then the entroy of saturated vaour at 7 C would be: [IAS-00] (a).88 kj/kg K (b) 6. kj/kg K (c) 7.93 kj/kg K (d) 0.53 kj/kg K 9

78 Pure Substances S K Mondal s Chater 7 IAS-3. wo heat engine cycles (l and l' - ' - 3' - l ) are shown on -s co-ordinates in [IAS-999] IAS-4. he mean effective ressure of the thermodynamic cycle shown in the given ressurevolume diagram is: (a) 3.0 bar (b) 3.5 bar (c) 4.0 bar (d) 4.5 bar [IAS-999] 9

79 Pure Substances S K Mondal s Chater 7 IAS-5. he given figure shows a thermodynamic cycle on -s diagram. All the rocesses are straight times. he efficiency of the cycle is given by (a) (0.5 h e)/ h (b) 0.5 (h e)/ h (c) (h e)/ 0.5 h (d) (h 0.5 e)/ h [IAS-996] h-s Diagram or Mollier Diagram for a Pure Substance IAS-6. Constant ressure lines in the suerheated region of the Mollier diagram have what tye of sloe? [IAS-007] (a) A ositive sloe (b) A negative sloe (c) Zero sloe (d) May have either ositive or negative sloes IAS-7. Assertion (A): In Mollier chart for steam, the constant ressure lines are straight lines in wet region. Reason (R): he sloe of constant ressure lines in wet region is equal to. (a) Both A and R are individually true and R is the correct exlanation of A (b) Both A and R are individually true but R is not the correct exlanation of A (c) A is true but R is false [IAS-995] (d) A is false but R is true Quality or Dryness Fraction IAS-8. Dryness fraction of steam means the mass ratio of [IAS-00] (a) Wet steam, to dry steam (b) Dry steam to water articles in steam (c) Water articles to total steam (d) Dry steam to total steam hrottling IAS-9. Assertion (A): hrottle governing is thermodynamically more efficient than nozzle control governing for steam turbines. [IAS-000] Reason (R): hrottling rocess conserves the total enthaly. (a) Both A and R are individually true and R is the correct exlanation of A (b) Both A and R are individually true but R is NO the correct exlanation of A (c) A is true but R is false (d) A is false but R is true 93

80 S K Mondal s Answers Pure Substances Chater 7 with Exlanation (Objective) Previous 0-Years GAE Answers GAE-. Ans. (a) Initial olume () = m 3 = m 3 Let dryness fraction = x herefore = ( x) x hat gives an absurd value of x = 8.65 (It must be less than equal to unity). So vaour is suerheated. GAE-. Ans. (d) Work done = first constant volume heating + = 0 + P (-) = 00 ( ) = 0. 4 kj GAE-3. Ans. (c) ( ΔS = (ΔΔ S ) + ( Δ = 0 syatem GAE-4. Ans. (c) GAE-5. Ans. (b) GAE-6. Ans. Higher, Lower GAE-7. Ans. (a) Mollier diagram is a h-s lot. ds= dh -υ d or S) surroundings h = = sloe s is always + ive so sloe always +ive. Not only this if then sloe GAE-8. Ans. False P dv 000 ( ) = 8.5 J/K GAE-9. Ans. (d) GAE-0. Ans. (b) h = h = x h 40 0 ( ( ) ) f 0+ xhg or = x x 48.0 or x = 0. 94

81 Pure Substances S K Mondal s Chater 7 GAE-. Ans. (b) GAE-. Ans. (d) Δ h = o Δ s > 0 Δ < 0 P GAE-3. Ans. (a) S S = Cav ln Ru ln P P Change in entroy of the universe = Ru ln P 0. kj = 8.34 ln = K For an ideal gas change in enthaly is a function of temerature alone and change in enthaly of a throttling rocess is zero. Previous 0-Years IES Answers IES-. Ans. (d) Water for all ractical urose can be considered as ure substance because it is homogeneous and has same chemical comosition under all hases. IES-. Ans. (d) IES-3. Ans. (d) he ordinate and abscissa in given figure are ressure and enthaly. Such diagram is common in vaour comression refrigeration systems. IES-4. Ans. (d) he throttling rocess given in figure is on ressure-enthaly diagram. IES-5. Ans. (d) IES-6. Ans. (b) A ure substance is one whose chemical comosition does not change during thermodynamic rocesses. Pure Substance is one with uniform and invariant chemical comosition. Eg: Elements and chemical comounds are ure substances. (water, stainless steel) Mixtures are not ure substances. (eg: Humid air) Excetion!! Air is treated as a ure substance though it is a mixture of gases. Gaseous combustion roducts are a mixture of gases and not a ure substance. IES-7. Ans. (d) IES-8. Ans. (a) IES-9. Ans. (b) he largest density of water near atmosheric ressure is at 4 c. IES-0. Ans. (c) U to saturation oint ressure must be constant. After saturation its sloe will be ive, as v = R or v = const. or vd + dv = 0 or d = dv v 95

82 Pure Substances S K Mondal s Chater 7 IES-. Ans. (c) IES-. Ans. (c) IES-3. Ans. (a) IES-4. Ans. (c) IES-5.Ans.(d) Characteristics of the critical oint. It is the highest temerature at which the liquid and vaour hases can coexist.. At the critical oint hfg, ufg and vfg are zero. 3. Liquid vaour meniscus will disaear. 4. Secific heat at constant ressure is infinite. IES-6. Ans. (d) At critical oint, the latent heat in zero and steam generators can oerate above this oint as in the case of once through boilers. he density of liquid and its vaour is however same and thus statement is wrong. IES-7. Ans. (b) IES-8. Ans. (b) IES-9. Ans. (c) IES-0. Ans. (c) IES-. Ans. (b) h ds = dh d or = s P he sloe of the isobar on the h s diagram is equal to the absolute tem, for condensation is cost so sloe is const, but not zero so it is inclined line. IES-. Ans. (a) For throttling rocess ( ), h = h h = hf = 000 kj/kg at ressure P h = hf + x ( hg hf) at ressure P 000 = x( ) or x = 0. IES-3. Ans. (c) IES-4. Ans. (c) IES-5. Ans. (a) IES-6. Ans. (c) Consider a throttling rocess (also referred to as wire drawing rocess) here is no work done (rising a weight) W = 0 If there is no heat transfer Q = 0 Conservation of mass requires that C = C Since and are at the same level Z = Z 96

83 Pure Substances S K Mondal s Chater 7 From SFEE it follows that h = h Conclusion: hrottling is a constant enthaly rocess (isenthalic rocess) IES-7. Ans. (b) emerature decreases and the steam becomes suerheated. IES-8. Ans. (d) IES-9. Ans. (d) hrottling is a isenthalic rocess h = h or u + v = u + v or u u = v v = = 300 kj/kg IES-30. Ans. (d) he throttling rocess takes laces with enthaly remaining constant. his rocess on S diagram is reresented by a line starting diagonally from to to bottom. IES-3. Ans. (b) IES-3. Ans. (d) Previous 0-Years IAS Answers IAS-. Ans. (d) IAS-. Ans. (c) Only two hase liquid-vaour is co-exists at the critical oint, but at trile oint-all three hase are co-exists. IAS-3. Ans. (a) A ure substance is a substance of constant chemical comosition throughout its mass. IAS-4. Ans. (c) IAS-5. Ans. (a) IAS-6. Ans. (d) A is false but R is true. IAS-7. Ans. (c) IAS-8. Ans. (d) IAS-9. Ans. (d) Network done is area of closed loo ABCD = Area of traezium AB3 + Area BC63 Area CD56 Area AD = ( 3 ) ( 6 3) ( 6 5) ( 5 ) = = 0 bar m 5 N 3 6 = 0 0 m = 0 Nm = 000 knm m IAS-0. Ans. (d) Remember: rile oint temerature of water = 73.6 K = 0.0 C 97

84 Pure Substances S K Mondal s Chater 7 IAS-. Ans. (a) IAS-. Ans. hfg 800 Sg = S f + =.6 + = 6. kj / kgk 500 sat IAS-3. Ans. (d) IAS-4. Ans. (a) Work (W) = ( ) ( ) + ( ) ( ) = 6kJ W 6 W = Δ m or = m kpa 3bar Δ = = ( ) Heat added = Area under = h( S S) ( h c)( S S ) η = = 0.5( h c) / h S S IAS-5. Ans. (b) Work outut = Area 3 = ( ) ( S S ) ( ) h IAS-6. Ans. (a) Mollier diagram is a h-s lot. h c h ds= dh υ d or = = sloe s P is always + ive so sloe always +ive. Not only this if then sloe IAS-7. Ans. (a) Both A and R are true and R is the correct exlanation of A IAS-8. Ans. (d) IAS-9. Ans. (d) If throttle governing is done at low loads, the turbine efficiency is considerably reduced. he nozzle control may then be a better method of governing. 98

85 S K Mondal s Proerties of Gassess & Gas Mix. Chater 8 8. Proerties of Gasses and Gas Mixture heory at a Glance (For GAE, IES & PSUs) he functional relationshi among the indeendent roerties, ressure P, molar or secific volume v, and temerature, is known as Equation of state i.e. P = R for gases. A hyothetical gas which obeys the law P = R at all temeratures and ressures is called an ideal gas An ideal gas has no forces of intermolecular attraction. he secific heat caacities are constant. Real gas does not conform to equation of state with comlete accuracy. As P 0 or, the real gas aroaches the ideal gas behaviour. Joule s law states that the secific internal energy of a gas deends only on the temerature of the gas and is indeendent of both ressure and volume. PROCESS Constant olume Process: he rocess is reresented on a v diagram as shown in Figure. Here or herefore or work transfer v dv = = 0 v dv = 0 W = 0 Fig. 000

86 S K Mondal s or Q- = Cv ( ) (i.e.,) during constant volume rocess, the heat transfer is the change in internal energy. From gas equation v v = We get, = In herefore Constant Pressure Process: his rocess is reresented on a v diagram as shown in Figure. this rocess. = From gas equation we get v v = Proerties of Gassess & Gas Mix. d Q = de = du = Cv ( ) = Chater 8 Fig.- (or) v = v d W = dv = (v v) d Q = de + d W = du + d W = d(u + v) = d (h) = CP ( ) (i.e.,) heat transfer is the change in enthaly in the case of a constant ressure rocess. Constant temerature (or) Isothermal rocess:- he rocess is reresented in figure(below). Here = d W = dv dv W = dv = C ; since v = C 0

87 S K Mondal s Proerties of Gassess & Gas Mix. Chater 8 v W = v ln = vl n v v where r = v Since du = ( = and d = 0. u ) = ( u C = 0. v ( r) ) Fig. herefore, by the First Law of hermodynamics Q = W = v ln (r) Adiabatic rocess: his is also called isentroic rocess as entroy in this rocess will remain constant and heat transfer in this rocess is also zero. he rocess is reresented in v diagram shown in Figure below. Fig.- he equation for this rocess becomes, γ γ v = v where γ = ratio of secific heats CP (i.e.,) = Cv For this rocess Q = 0 Hence 0 = d E + d W = du + d W or d W = du = mcv d d W = dv or dv = du = mcv d or dv + mcvd = 0 But v = mr differentiating; dv + vd = mr d From the above equation, dv + vd d = mr 0 (equation...)

88 Proerties of Gasses & Gas Mix. S K Mondal s Chater 8 In equation-, value of d is substituted. dv + vd dv + mc v = 0 mr R.dv + Cv (dv + vd) = 0 But R = (CP Cv) herefore (CP Cv) dv + Cv (dv + vd) = 0 or Cv vd + CP dv = 0 Dividing by (Cv v) we get d CP dv. 0 Cv But CP = γ Cv d dv herefore + γ = 0 Integration and rearranging we get, ln + γ ln v = C Or v γ = C (i.e.) γ γ v = v From the gas equation v = mr, we can obtain the relationshis between ressure and temerature and volume and temerature as given below. γ γ γ v = = v Work transfer d W = dv W = dv γ v = C C = γ v herefore dv W = C γ γ + γ + v v = C - γ + γ γ But C = v = v v v W = γ + v v W = γ Also first law becomes W = ΔE = ΔU 03

89 Proerties of Gasses & Gas Mix. S K Mondal s Chater 8 General rocess or olytroic rocess: In the adiabatic rocess the index for is γ. In a most general case this γ can be relaced by n. he values of n for different rocesses are indicated below : if n = 0 the rocess is constant ressure if n = the rocess is isothermal if n = the rocess is constant volume if n = γ the rocess is adiabatic. and for any rocess other than the above, n becomes a general value n. All the equations of adiabatic rocess can be assumed for this rocess with γ relaced by n. n n n v = = v v v W = n But W = is not true as Q is not zero. By the first law of hermodynamics, Q = dv + ( u u ) v v mc ( ) = + v n But by gas equation v = mr and v = mr mr( ) ( ) so, Q = mcv + n R = m ( ) Cv + n CP Cv But, Q = m( ) Cv + n CP Cv = mcv + ( ) n CP ncv Q = ( ) n Substituting C P= γ Cv γ n Q = Cv ( ) n Hence the secific heat for a olytroic rocess is γ n = Cn n ( ) C n = C v 4. For minimum work in multistage comression P = PP3 a. Equal ressure ratio i.e. P P3 = P P 04

90 S K Mondal s Proerties of Gassess & Gas Mix. b. Equal discharge temerature i.e. =3 c. Equal work required for both the stages. Chater 8 5. Equation of statess for real gas a a. an der waals equation ( + )(v-b) ) = R v he coefficient a is introduced to account for the existence of mutual attraction between the molecules. he term a/v is called the force of cohesion. he coefficient b is introduced to account for the volumes of the molecules, and is known as co-volume. b. Beattie Bridgeman equation R ( e)( v + B) A a = - Where A = A0 (- v v ) v b B = B0 (- ) v c e = 3 v his equation does not give satisfactory results in the critical oint region. 6. P he ratio R is called the comressibility factor. 7. At (i) (ii) (iii) (iv) 8. alue of comressibility factor (Z) at critical oint is for an der waals gas. For ideal gas z = Critical Proerties: a =3Pcc c 8 P, b=, and R= cc 3 3 c Where Pc, c and c are critical oint ressure, volume and temerature resectively. Critical Point hree real roots of ander Waal equation coincide. = 0 i.e. Sloe of -v diagram is zero. v c v c 3 3 v = 0 i.e. Chan c < 0 nge of sloe also zero. i.e. negative, and equal to -9c a Boyle s emerature (B) = br Fig. Critical diagram roerties on v 05