HEAT, WORK, AND THE FIRST LAW OF THERMODYNAMICS

Transcription

1 HET, ORK, ND THE FIRST L OF THERMODYNMIS 8 EXERISES Section 8. The First Law of Thermodynamics 5. INTERPRET e identify the system as the water in the insulated container. The roblem involves calculating the work done to raise the temerature of a system, so the first law of thermodynamics comes into lay. DEELOP ecause the container is a erfect thermal insulator, no heat enters or leaves the water inside of it. Thus, Q 0 and the first law of thermodynamics in Equation 8. gives U Q+, where is the work done by shaking the container. The change in the internal energy of the water is determined from its temerature rise, U mc T (see comments in Section 6. on internal energy). ombining these two exressions for the internal energy change allows us to find the work done. ELUTE The work done on the water is U mc T.0 kg 4.84 kj/ kg K 7.0 K 9.3 kj SSESS ccording to the convention adoted for the first law of thermodynamics, ositive work signifies that work is done on the water. 6. INTERPRET This roblem involves finding the mechanical work that is required to raise the given mass of water by 3.0, which we can calculate by using the first law of thermodynamics. DEELOP The change in the internal energy of the water is U mc T, where T K and the work done on it is 9.0 kj. Inserting this into the first law of thermodynamics (Equation 8.) gives Q U mc T. For art (b), note that if the water had been in erfect thermal isolation, no heat would have been transferred, so Q 0 and we can solve for the work as for art (a). ELUTE (a) Inserting the given quantities into the exression for the heat transferred gives (b) If Q 0, then the work needed is Q mc T 0.50 kg 4.84 kj/ kg K 3.0 K 9.0 kj.7 kj mc T 0.50 kg 4.84 kj/ kg K 3.0 K 6.3 kj SSESS The fact that Q < 0 for art (a) indicates that heat is transferred from the water to the environment, as exected. 7. INTERPRET The system of interest is the gas that undergoes exansion. The roblem involves calculating the change in internal energy of a system, so the first law of thermodynamics comes into lay. DEELOP The heat added to the gas is Q Pt (40 )(5 s) 000 J. In addition, the system does 750 J of work on its surroundings, so the work done by the surroundings on the system is 750 J. The change in internal energy can be found by using the first law of thermodynamics given in Equation 8.. ELUTE Using Equation 8. we find U Q+ 000 J 750 J 50 J SSESS Since U > 0, we conclude that the internal energy has increased. 8- oyright 06 Pearson Education, Inc. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may

2 8- hater 8 8. INTERPRET This roblem involves calculating the rate of heat flow into the system given the rate at which the internal energy is increasing and the rate at which the system is doing work on its surroundings. DEELOP The dynamic form of the first law of thermodynamics (Equation 8.) is dq du d dt dt dt The roblem states that du/dt 45 and d/dt 65 (where the negative sign is aroriate because the system is doing work). ELUTE Inserting the given quantities gives dq dt SSESS The rate of heat flow into the system is ositive because the system is doing work on its surroundings and increasing its internal energy. 9. INTERPRET This roblem is about heat and mechanical energy, which are related by the first law of thermodynamics. The system of interest is the automobile engine. DEELOP Since we are dealing with rates, we make use of the dynamic form of the first law of thermodynamics, Equation 8.: du dq d + dt dt dt If we assume that the engine system oerates in a cycle, then du/ dt 0. The engine s mechanical ower outut d/dt can then be calculated once the heat outut is known. ELUTE The above conditions yield ( dq/ dt ) out 68 k and ( d dt ) ( dq dt ) in gives or d dq dq dq dq d dt dt dt dt dt 0.7 dt d dt ( dq dt) out in out / out 68 k 4 k (0.7) 0.7 / 0.7 /. Equation 8. then SSESS e find the mechanical ower outut d/ dt to be roortional to the heat outut, ( dq/ dt ) out. In addition, d/ dt also increases with the ercentage of the total energy released in burning gasoline that ends u as mechanical work. The mechanical outut is negative because the system is doing work on its surroundings. Section 8. Thermodynamic Processes 0. INTERPRET This roblem involves an ideal gas that changes from an initial ressure-volume state to a final ressure-volume state by traversing the given ressure-volume curve. From this, we are to find the work done by the gas during this rocess. DEELOP The work done by the gas is given by the negative of Equation 8.3, which is by gas d This is the area under the line segment in Figure 8.9. e can find this area from simle geometrical arguments, without having to invoke calculus. ELUTE The work done by the gas is the area of the traezoid under line, which is bygas ( P + P )( ) ( P + P )( ) 3 P oyright 06 Pearson Education, Inc. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may

3 Heat, ork, and The First Law of Thermodynamics 8-3 SSESS so The work can also be obtained from Equation 8.3. On the ath, P+ ( )( P P) ( ) P P 3 bygas d P ( ) + ( ) P.. INTERPRET The exansion of the ideal gas involves two stages: an isochoric (constant-volume) rocess and an isobaric (constant-ressure) rocess. e are asked to find the total work done by the gas. DEELOP For an isochoric rocess, 0 so 0 (see Equation 8.3 and Table 8.). On the other hand, for an isobaric rocess, the work done is which is the negative of Equation 8.7 because we are interested in the work done by the gas (Equation 8.7 gives the work done on the gas). ELUTE Summing the two contributions to find the total work gives + tot SSESS In the diagram of Fig. 8.9, the area under is zero, and that under, a rectangle, is. The work done by the gas is the area under the curve.. INTERPRET This roblem involves the isothermal (i.e., the temerature is constant) exansion of a balloon as it rises. e are asked to find the work done by the helium as the balloon rises. DEELOP During an isothermal exansion, the work done by a given amount of ideal gas is nrt ln which is the negative of Equation 8.3 because that form exresses the work done on the gas. ELUTE SSESS Inserting the given quantities yields 0.30 mol 8.34 J/ mol K 300 K ln(5.0). kj The result is ositive because ositive work is done by the gas to exand its volume. 3. INTERPRET The constant temerature of 300 K indicates that the rocess is isothermal. e are to find the increase in volume of the balloon and the work done by the gas, given the ressure difference in exeriences. DEELOP e assume the gas to be ideal and aly the ideal-gas law given in Equation 7.: nrt. For an isothermal rocess, T constant, so we obtain, from which we can find the fractional volume increase. The total work done by the gas can be calculated using the negative of Equation 8.4: nrtln because we are interested in the work done by the gas, not on the gas. ELUTE (a) For the isothermal exansion rocess, the volume increases by a factor of 00 kpa 4 75 kpa 3 (b) Using Equation 8.4, the work done by the gas is 4 nrt ( ) 3 ln 0.30 mol 8.34 J/ mol K 300 K ln 0 J to two significant figures. SSESS ecause >, we find the work to be ositive, > 0. This makes sense because the gas inside the balloon must do ositive work to exand outward. 4. INTERPRET This roblem involves an ideal gas that we comress in an isothermal rocess to half its volume. e are to find the work needed to do this. oyright 06 Pearson Education, Inc. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may

4 8-4 hater 8 DEELOP In an isothermal comression of a fixed quantity of ideal gas, work is done on the gas so we use Equation 8.4, nrtln( / ), where and T 300 K for this roblem. ELUTE Inserting the given quantities into Equation 8.4 gives nrtln /.5 mol 8.34 J/ mol K 300 K ln / kj. SSESS The work is ositive because ositive work must be exended by an external agent other than the gas to comress the gas. 5. INTERPRET The thermodynamic rocess here is adiabatic, so no heat flows between the system (the gas) and its environment. e are to find the volume change needed to double the temerature. DEELOP In an adiabatic rocess, Q 0, so the first law of thermodynamics (Equation 8.) becomes U. The temerature and volume are related by Equation 8.b: T constant T The temerature doubles, so T T and.4, so we can solve for the fractional change in volume. ELUTE From the equation above, we have T T Thus, for the temerature to double, the volume change must be T T /( ) /(.4 ) /( ) T 0.77 T SSESS e see that increasing the temerature along the adiabat is accomanied by a volume decrease. In addition, since constant, the final ressure is also increased: INTERPRET The thermodynamic rocess here is adiabatic, so no heat flows between the system (the gas) and its environment. e are to find the temerature increase as the gas volume is reduced. DEELOP In an adiabatic rocess, Q 0, so the first law of thermodynamics (Equation 8.) becomes U. The temerature and volume are related by Equation 8.b: T T The final volume is /4 of the initial, so /4 and.4, so we can solve for the change in temerature. ELUTE From the equation above, we have Thus, the temerature change is.4 T T T T (9K)(4) T T T 507 K 9K 6 K K SSESS e see that decreasing the volume along the adiabat is accomanied by a temerature increase. In addition, since constant, the final ressure is also increased: ( 4) INTERPRET This roblem involves an ideal gas that we comress in an isothermal rocess to half its volume. e are to find the work needed to do this. DEELOP In an isothermal comression of a fixed quantity of ideal gas, work is done on the gas so we use Equation 8.4, nrtln( / ), where / 0.05 for this roblem. oyright 06 Pearson Education, Inc. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may

5 Heat, ork, and The First Law of Thermodynamics 8-5 ELUTE Inserting the given quantities into Equation 8.4 gives nrt ln / ln / (.03 0 Pa) 6.8 m ln J. SSESS The work is ositive because ositive work must be exended by an external agent other than the gas to comress the gas. Section 8.3 Secific Heats of an Ideal Gas 8. INTERPRET This roblem involves finding the molar secific heat at constant volume and constant ressure for the given gas mixture, which contains the given quantities of diatomic O and monatomic r. DEELOP This roblem in similar to Examle 8.5, so we will use the same aroach as outlined there. ELUTE ecause O has 5 degrees of freedom, so its average energy er molecule is 5kT/. For r, the average energy er molecule is 3kT/ because it is monatomic. The total internal energy is therefore U no RT n r RT +.5 mol 3.0 mol RT 0.75 mol RT + From Equation 8.6, the molar secific heat at constant volume is ( 0.75 mol) ( + ) du R.0R ndt mol to two significant figures. From Equation 8.9, the molar secific heat at constant ressure is P + R 3.0R SSESS e find that the molar secific heat at constant volume is between that of a diatomic molecule (.5 R) and a monatomic molecule (.5 R), as exected since we have a mixture of these two tyes of molecules. 9. INTERPRET The roblem is about the secific heat of a mixture of gases. e want to know what fraction of the molecules is monatomic, given the its secific-heat ratio. DEELOP The internal energy of a mixture of two ideal gases is U fne+ fne where f i is the fraction of the total number of molecules, N, of tye i, and E i is the average energy of a molecule of tye i. lassically, E g( kt ), where g is the number of degrees of freedom (the equiartition theorem). From Equation 8.6, the molar secific heat at constant volume is du d nr f g T f g T R f g f g ndt dt + + Suose that the temerature range is such that g 3 for the monatomic gas, and g 5 for the diatomic gas, as discussed in Section 8.3. Then R f + f R f ( 3 5 ) (.5 ) where f f since the sum of the fractions of the mixture is unity. Now, can also be secified by the ratio P + R R + R Equating the two exressions allows us to solve for f. ELUTE Solving, we find SSESS.5 f f 57.7% 0.5 From the equation above, we see that the secific-heat ratio can be written as +.5 f oyright 06 Pearson Education, Inc. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may

6 8-6 hater 8 In the limit where all the gas molecules are monatomic, f, and.67. On the other hand, if all the molecules are diatomic, then f 0 and the secific-heat ratio is.4. The equation yields the exected results in both limits. 30. INTERPRET For this roblem, we are to find the secific-heat ratio of a gas consisting of the given ratio of different gas molecules. DEELOP is y generalizing the result of the revious roblem, the molar secific heat of a mixture of three gases f f f For each gas and the mixture, use R ( ) to obtain ( ) f ( ) + f ( ) + f ( ) which we can evaluate to find. ELUTE 3 3 hen the data secified in the roblem is inserted, one gets SSESS ecause we have some triatomic molecules, the secific-heat ratio is less than that for a ure diatomic gas (see revious roblem). 3. INTERPRET The thermodynamic rocess is adiabatic, and we want to know the temerature change when work is done on a monatomic gas and a diatomic gas. DEELOP In an adiabatic rocess, Q 0, so from the first law of thermodynamics (Equation 8.) U, where is the work done on the gas. From Equation 8.6, U n T, the change in temerature is PROLEMS U T n n If the work done er mole on the gas is n.5 kj/mol, then T (.5 kj/mol) 3 3 ELUTE (a) For an ideal monatomic gas, R 8.34 J/ ( mol K ), so T 00 K. 5 (b) For an ideal diatomic gas (with five degrees of freedom), R so T 0 K. SSESS Since the diatomic gas has a greater secific heat, its temerature change is less than that of the monatomic gas. 3. INTERPRET The constant temerature of 440 K indicates that the rocess is isothermal. e are given the amount of work done by the gas, and are asked to find the heat it absorbs and how many moles of gas there are. DEELOP ly the ideal-gas law given in Equation 7.: nrt. For an isothermal rocess, T constant, so we obtain. Since U 0 for an isothermal rocess, the heat absorbed is the negative of the total work done on the gas (Equation 8.4). Q nrtln For this roblem, the gas does 3.3 kj of work on its surroundings, so the surroundings do 3.3 kj of work on the gas, so 3.3 kj. ELUTE (a) Using the equation above, the heat absorbed is Q 3.3 kj. (b) Solving the exression above for the number n of moles gives 3.3 kj n 0.39 mol RT ln ( / ) 8.34 J/ ( mol K) ( 440 K) ln( 0) SSESS The heat absorbed by the gas is equal to the work done by the gas on its surrounding as it exands, and there is no change in temerature. oyright 06 Pearson Education, Inc. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may

7 Heat, ork, and The First Law of Thermodynamics INTERPRET e re asked to find the rate that heat is roduced in the body when cycling. This involves the first law of thermodynamics. DEELOP e re dealing with the rates of work and heat roduction, so we ll use Equation 8.: du dq + d. dt dt dt If the body is releasing stored food energy, that corresonds to a decrease in the body s internal energy: du / dt 500. Likewise, the mechanical ower quoted is for work done by the body, so d / dt 0. e are looking for the rate at which the body roduces heat, which is technically heat that it is losing ( dq). ELUTE The rate of heat roduction is the negative heat absorbed, so dq du d + ( 500 ) + ( 0 ) 380 dt dt dt SSESS ll the signs can be confusing, but essentially the body burns stored energy and some of it is used to do work and the rest is released as heat. 34. INTERPRET This roblem deals with isothermal comression, so the temerature is constant in this rocess. e are to find the work done to comress the given quantity of ideal gas from 3.5 L to 3.0 L. DEELOP ly Equation 8.4, Q nrtln where is the work done by the gas, 3.5 L, and 3.0 L. Since we do 6 J of work on the gas to comress it, 6 J, so we can solve this exression for the temerature T. ELUTE SSESS The temerature of the ideal gas is 6 J T 90 K nrln( ) ( 0.5 mol) 8.34 J/ ( mol K) ln( 3.0/3.5) This gas is about 00 K below room temerature. 35. INTERPRET ssume the air inside the sherical bubble behaves like an ideal gas at constant temerature, so the rocess is isothermal. e need to find the diameter of the bubble at maximum ressure and the work done on the gas in comressing it. DEELOP ly the ideal-gas law given in Equation 7.: nrt. For an isothermal rocess, T constant, which leads to. Since the volume of a sherical bubble of diameter d is the relationshi between the diameter and the ressure is 3 3 4π d πd πd πd 6 6 d d ELUTE (a) Using the equation above, we find the diameter at the maximum ressure to be /3 ( + ) ( + ) mm of Hg d d (.5 mm ).49 mm mm of Hg (b) The work done on the air is given by Equation 8.4, or SSESS on air nrtln ln ln 0.3 kpa π mm of Hg ( 840 mm of Hg) (.5 mm) ln 760 mm of Hg mm of Hg 0.7 μj Positive work is done by the blood in comressing the air bubble. /3 /3 oyright 06 Pearson Education, Inc. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may

8 8-8 hater INTERPRET This roblem involves comressing a gas in an isothermal rocess, so the temerature is constant. e are to find how much work it takes to reduce the volume by a factor of given the work it takes to decrease the volume by a factor of. DEELOP ly Equation 8.4 to both situations. To comress a gas by a factor of two, we have and to comress it by a factor of gives nrtln nrtln Given that T is constant, we can take the ratio of these exressions to find the work needed to comress the gas by a factor of. ELUTE SSESS The work needed to comress the gas by a factor of is ln ln (.5 kj) 6.7 kj ln ln Note that the work required is logarithmic in volume. 37. INTERPRET The thermodynamic rocess here is adiabatic, with no heat flowing between the system (the gas) and its environment. e are to find the secific-heat ratio given the fraction increase in ressure of the gas. DEELOP In an adiabatic rocess, Q 0 and the first law of thermodynamics becomes U. The ressure and volume are related by Equation 8.a: P constant. This imlies so we can solve for. ELUTE Taking the natural logarithm on both sides of the above to solve for, we obtain ( ) ln ln.55 ln ln ln ln.35 SSESS The value of indicates that gas consists of olyatomic molecules (see Problems 9 and 30). 38. INTERPRET The roblem involves comressing a gas adiabatically. DEELOP For an adiabatic rocess, the initial ressure and volume are related to the final ressure and volume through Equation 8.a:. e can use this to rewrite Equation 8. for the work done on the gas in going from the initial to the final volume: ELUTE (a) The final ressure in this case is L ( 98.5 kpa) 73 kpa 4.8 L (b) The work done on the gas in comressing it from 6.5 L to 4.8 L is kpa 6.5 L 6.5 L 69 J L SSESS The volume decreases by a factor of.5, but the ressure increases by a factor of.76. The ressure increases due to both the increase in the density (N/) and an increase in the internal energy of the gas ( U n T). 39. INTERPRET This roblem involves a cyclic rocess. The three rocesses that make u the cycle are: isothermal (), isochoric (), and isobaric (). e are given the ressure at oint and are to find the ressure at oint and the net work done on the gas. oyright 06 Pearson Education, Inc. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may

9 Heat, ork, and The First Law of Thermodynamics 8-9 DEELOP long the isotherm T constant, so the ideal-gas law (Equation 7., nrt) gives For an isothermal rocess, the work done on the gas is (Equation 8.4):. nrtln ELUTE (a) If is an isotherm, with 5 L and L, then the ideal-gas law gives 5 P P ( 60 kpa) 300 kpa (b) The work done on the gas in the isothermal rocess is.0 L nrt ln ln ( 300 J) ln 483 J 5.0 L The rocess is isochoric (constant volume) so 0. The rocess is isobaric (constant ressure) so the work done by the gas is (see Table 8.) 60 kpa 5.0 L.0 L 40 J or the work done on the gas is 40 J. The total work done by the gas is the sum of these three contributions, or J J 43 J 40 J to two significant figures. SSESS Since the rocess is cyclic, the system returns to its original state, there s no net change in internal energy, so U 0. This imlies that Q 40 J. That is, 40 J of heat must come out of the system. 40. INTERPRET This roblem is similar to the revious one, excet that the isotherm curve is now to be considered as an adiabat. e are given the secific-heat ratio and are to find the ressure change in going from oint to oint, and the net work done by the cycle. DEELOP For an adiabat, Equation 8.a is valid. lying this to oints and, and taking the ratio, gives which allows us to find the ressure at oint. The calculation for art (b) then roceeds as for Problem 8.39, with the hel of Equation 8., which gives the work done on the gas in an adiabatic rocess. ELUTE (a) The ressure at oint is to two significant figures. (b) The work done on the gas between oints and is 5.0 L ( 60 kpa) 570 kpa.0 L ( 57 kpa)(.0 L) ( 60 kpa)( 5.0 L) P P 678 J 0.4 The rocess is isochoric (constant volume), so 0. The rocess is the same as for Problem 8.37, so 40 J. Summing all three contributions gives a net work done on the gas of to two significant figures. SSESS J J 440 J. ecause the net work done on the gas is negative, the gas does work on its environment. 4. INTERPRET e identify the thermodynamic rocess here as adiabatic comression. DEELOP In an adiabatic rocess, Q 0, and the first law of thermodynamics becomes U. The temerature and volume are related by Equation 8.b: T constant.4 oyright 06 Pearson Education, Inc. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may

10 8-0 hater 8 From the equation above, we obtain T T T T where / is the comression ratio (for T and at maximum comression). ELUTE Inserting the values given gives T T K K 440 SSESS Note that the temerature T aearing in the gas laws is the absolute temerature. The higher the comression ratio /, the greater the temerature at the maximum comression, and hence a higher thermal efficiency. 4. INTERPRET This roblem involves an adiabatic rocess that results in a change of ressure and volume. e are to find the fractional change in volume given that the ressure increases by a factor of. DEELOP ly Equation 8.a, constant. Using subscrits and to indicate before and after the rocess, resectively, this gives constant which we can solve for the fractional change in volume, using.4 and. ELUTE Inserting the given quantities into the exression above gives.4 ( 0.5) 0.6 SSESS Thus, the new volume is almost 50% of the original volume, which seems reasonable given that the ressure has doubled. 43. INTERPRET e identify the thermodynamic rocess here as adiabatic comression. e are to find the temerature and ressure in the cylinder when it is at maximum comression. DEELOP In an adiabatic rocess, Q 0, and the first law of thermodynamics becomes U. ecause we are dealing with an adiabatic rocess, the temerature and volume are related by Equation 8.b: T constant lying this exression before (subscrit ) and after (subscrit ) comression gives T T T T where 0. is the comression ratio (for T and at maximum comression) and.4. In addition, since constant (Equation 8.a), the final ressure is ( ). ELUTE (a) Substituting the values given in the roblem statement, we find the air temerature at the maximum comression to be (b) The corresonding ressure is T T (30 K)(0.) 80 K 0.3 kpa 0..6 MPa 5.8 atm SSESS The higher the comression ratio /, the greater the temerature and ressure at the maximum comression, and hence, a higher thermal (fuel) efficiency. oyright 06 Pearson Education, Inc. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may

11 Heat, ork, and The First Law of Thermodynamics INTERPRET The thermodynamic rocess here is adiabatic, so no heat flows between the system (the gas) and its environment. e are to find the change in temerature and volume as the ressure is reduced. DEELOP In an adiabatic rocess, Q 0, so the first law of thermodynamics (Equation 8.) becomes U. The temerature, ressure and volume are related by Equation 8.: T T, The initial ressure is.0 atm, and the final ressure is atm. ith 5/3, we can solve for the new volume and temerature. ELUTE (a) From the equation above, we have 3/5 / atm m m (b) The final temerature is 0.34 atm (5/3) m m T T (85 K) 85 K 88 SSESS e see that decreasing the ressure along the adiabat is accomanied by a temerature decrease. In addition, since constant, the final volume increases. 45. INTERPRET This roblem involves an adiabatic comression of a gas, for which we know the initial and final temeratures. e are to find the ratio of the final ressure to the initial ressure of the gas. DEELOP For an adiabatic rocess, the temerature and ressure are related by Equation 8.a: constant Rearranging this equation and using the ideal-gas law (Equation 7., nrt) gives ELUTE constant /( ) /( ) nrt T T Inserting the given quantities into the exression above for temerature gives 5/3 (5/3) T 93 K 354 T 8 K SSESS The ressure increase is enormous in order to accomany the increase in temerature along the adiabat. The corresonding change in volume change is 3/ The final volume is only about 3% of the original. 46. INTERPRET This roblem exlores how different ways of comressing volume (isothermal, adiabatic, or 5 isobaric) affects the internal energy of the system. The internal energy of n moles of diatomic gas is U nrt, by equiartition theorem. DEELOP In an isothermal rocess, the temerature T is ket constant, so U 0. In an isobaric rocess, constant, and T / T /. Finally, in an adiabatic rocess, Q 0 and T T ELUTE (a) For isothermal comression, T 0, so U 0. There is no change in internal energy. (b) For isobaric comression, U 5 / nrt T. U 5 nrt / T (c) For adiabatic comression, 7/5 for diatomic gas, U 5 nrt / T.4.3 U 5 nrt/ T oyright 06 Pearson Education, Inc. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may

12 8- hater 8 SSESS s the gas volume is halved, the internal energy of the gas decreases if comression is isobaric, but increases if the comression is adiabatic. There is no change if the comression is isothermal. 47. INTERPRET In the roblem we comare the work done in adiabatic comression to the work done in isothermal comression. DEELOP For an adiabatic rocess, the initial ressure and volume are related to the final ressure and volume through Equation 8.a:. e can use this to rewrite Equation 8. for the work done on the gas in going from the initial to the final volume: adiabatic On the other hand, for an isothermal rocess, the work done on the gas is (Equation 8.4): isothermal nrtln ln ln ELUTE (a) For / /, and 5/3, from the above exressions, we find the ratio of the work done for the two rocesses to be adiabatic /3.7 isothermal ( )ln( / ) (/3)ln() (b) The extra work goes into internal energy, raising the temerature of the gas. SSESS For diatomic gas where 5/3, the ratio would be adiabatic isothermal ( )ln( / ) (0.4)ln() 48. INTERPRET This roblem involves first an isothermal exansion, then an isobaric comression, then finally an isochoric rocess to return to the original state. e are to draw a diagram of this rocess, then calculate the work done over the entire cycle, then finally, find the heat transfer between the given stages. DEELOP The isothermal exansion is along the curve in the figure below, which is given by P constant. ecause the gas exands, it starts at the lower volume at oint, and exands to occuy the greater volume at oint. For the isobaric comression, the gas follows the horizontal line, with the ressure remaining constant at P P. Finally, for the isochoric rocess, the gas follows the vertical line, with the volume remaining constant at. To find the total work done on or by the gas during this cycle, sum the work done over each interval, tot + + where we consider work done on the system to be ositive. For the isotherm (rocess ), U 0 and the heat transferred to the system is Q 35 J. From the first law of thermodynamics (Equation 8.), this gives Q 35 J. For rocess, the work done on the gas is J, and 0 because the volume of the gas does not change (see Equation 8.7). To find the heat that enters or leaves the system during the stages, recall that the change in internal energy is zero for the whole cycle, so and that the total heat is comosed of two arts: so we can solve for Q. Utot Qtot + tot 0 Qtot Q + Q oyright 06 Pearson Education, Inc. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may

13 Heat, ork, and The First Law of Thermodynamics 8-3 ELUTE (a) See figure above. (b) Summing the various contributions to the work gives so the gas does 3 J of work on its environment. (c) The heat transferred to the system is tot J+ J+ 0 J 3 J Q Q + Q tot tot Q tot Q 3 J 35 J J so J of heat is transferred out of the gas during the rocess. SSESS The total heat transferred to the system over the cycle is Q tot 35 J J 3 J, which rovides the energy for the 3 J of work done by the system over this same cycle. 49. INTERPRET This roblem exlores how different ways of adding heat (isothermal, isochoric, or isobaric) affects the final temerature of the system. Starting with the given quantity of gas at the given temerature, we are to find the work done by the gas uon adding heat to the gas via these different rocesses. DEELOP In an isothermal rocess, the temerature T is ket constant, so U 0. The first law of thermodynamics (Equation 8.) therefore gives Q. In an isochoric rocess, 0 and 0 (see Equation 8.7), so the first law of thermodynamics gives Q U n T. Finally, in an isobaric rocess, 0 and Q n T n( + R) T Use these exressions to solve for T and for each case. P ELUTE (a) ith U 0, Q.75 kj. For an isothermal rocess, the temerature is constant so T 55 K. (b) Solving the exression above for T, we find for an isochoric rocess Q.75 kj T 4 K n ( 3.5 mol)( 5R ) Therefore, T 55 K + T 79 K. ecause the volume does not change, 0. (c) In an isobaric rocess, Q Q.75 kj T 7 K n n( + R) 3.5 mol 5R + R and T 55 K + 7 K 7 K. The work done by the gas is SSESS ( R ) R R.75kJ nr T Q Q 500 J 7 7 omaring all three cases, we find The results agree with that illustrated in Table 8.. T : isothermal < isobaric < isochoric : isochoric < isobaric < isothermal 50. INTERPRET This roblem is an exercise to show the functional relationshi in terms of ressure and volume between an adiabatic rocess and an isothermal rocess. DEELOP The equations of an adiabat and an isotherm, assing through the oint ( 0, 0) in the diagram are and resectively. Differentiate these exressions and evaluate the results at ( 0, 0 ) to find the desired relationshi. oyright 06 Pearson Education, Inc. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may

14 8-4 hater 8 ELUTE Differentiating the above exression for an adiabat and evaluating the result at 0 and 0 gives d d adiabat ( ) Differentiating the exression above for an isotherm and evaluating the result at 0 gives omaring the two results gives d d isotherm 0 0 d d d d adiabat SSESS e have found the exression given in the roblem statement. ecause > 0, we see that the sloe of the curve for an adiabat is greater than that for an isotherm, so the ressure of an adiabatic rocess changes more raidly with volume comared with an isothermal rocess, as shown in Fig INTERPRET The roblem involves a cyclic rocess with three searate stages: adiabatic, isochoric, and isothermal. e are given the initial volume and ressure of the gas, and its adiabatic exonent, and are to find the ressure at oints and and the net work done on the gas. DEELOP For the adiabatic rocess, Q 0 and the first law of thermodynamics becomes U. The ressure and volume are related by Equation 8.a: constant, which gives isotherm Point lies on an isotherm (constant temerature) with, so the ideal-gas law (Equation 7.) yields To find the net work done on the gas, sum the contributions from each stage of the cycle. The contribution may be found from Equation 8. for an adiabatic rocess: For an isochoric (constant volume) rocess 0 (see Equation 8.7), and for the isothermal rocess ln nrt ELUTE (a) From the equation above, the ressure at oint is (b) The ressure at oint is (c) The net work done by the gas is The work for the adiabatic segment is kpa 40 kpa 50 kpa kpa. ( 39.9 kpa)( 3.0 m 3 ) ( 50 kpa)(.00 m 3 ) 0.67 is for an isochoric rocess and equals zero. Finally, nrt is for an isothermal rocess and is ln 50 kj ln 75 kj 3 94 kj oyright 06 Pearson Education, Inc. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may

15 Summing these contributions gives Heat, ork, and The First Law of Thermodynamics kj kj 80 kj so the work done on the gas 80 kj. SSESS Since the rocess is cyclic, the system returns to its original state, there s no net change in internal energy, so U 0. This imlies that Q 80 kj. That is, 80 kj of heat must come out of the system. 5. INTERPRET This roblem involves a cyclic rocess with an adiabatic comression, an isobaric comression, and an isothermal exansion. e are to find the net work done on the gas and its minimum volume. DEELOP The diagram for this roblem is different than that of Figure 8.4; only oint is the same (see figure below). Note that 400 J and. To find the net work done on the gas, sum u the contributions from each stage. For the adiabatic stage, the work done on the gas is given by Equation 8.: For the isobaric rocess, the work done on the gas is given by Equation 8.7: ia the adiabatic rocess, we can find the roduct (using Equation 8.), which gives ia the isothermal rocess, we know that is at the same temerature as. e also know that because is isobaric. Thus, from the ideal-gas law (Equation 7.), we have Therefore, the work is The work done on the gas for the isothermal rocess is given by Equation 8.4, which gives (using the idealgas law nrt) ln The minimum volume of the gas occurs at oint, and is given by oyright 06 Pearson Education, Inc. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may