PHYS1001 PHYSICS 1 REGULAR Module 2 Thermal Physics Chapter 17 First Law of Thermodynamics

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1 PHYS1001 PHYSICS 1 REGULAR Module Thermal Physics Chater 17 First Law of Thermodynamics References: 17.1 to 17.9 Examles: 17.1 to 17.7 Checklist Thermodynamic system collection of objects and fields. If a system is in a state of thermodynamical equilibrium, all objects comrising the system are at the same temerature. The system is surrounded by a boundary through which energy can be exchanged with the surrounding environment by heat and work. Thermodynamics is the study of systems involving energy exchanges via work and heat. Thermodynamics rocesses - changes in,, T by Q added or removed and/or by done on or by system. A fixed quantity of gas (n, N or m tot ) enclosed in a cylinder is the thermodynamic system consider in most of chater. The state of the gas is described by its temerature T (K), ressure (Pa) and volume (m 3 ). Heat: Q > 0 energy added to system Q < 0 energy removed from system Heat refers to the energy transferred from a lace of higher temerature to a lace of lower temerature. ork: > 0 energy removed from system by system doing work on the surroundings eg exanding gas. < 0 energy added to system by work being done on the system by its surroundings eg comressing a gas. ork done by a change in volume equals the area under a - curve = d 1 (17.) a03/1/thermal/th doc :17 PM 1/11/0 1

2 Internal energy U of a system sum of the kinetic energies of all the articles of the system, lus the sum of all the otential energies of interaction between these articles due to the random, chaotic motion of the articles. The value of U is not imortant, what matters is the change in the internal energy in a thermodynamic rocess. U = U U 1 First law of thermodynamics conservation of energy transfer of energy by work and heat between thermodynamics system and surrounding environment. U = Q (17.4) The internal energy of an ideal gas deends only on its temerature, not on its ressure or volume. The internal energy of an isolated system is constant. Internal energy is not a form of energy but a way of describing the fact that the energy in atoms is both stored as otential and kinetic energy. Paths between thermodynamic states Heat Q and work done deend uon the ath taken between two states. The change in internal energy U of a system during any thermodynamic rocess deends only on the initial and final states i.e. indeendent of the ath and does not deend uon the kind of rocess that occurs U is an intrinsic roerty of system. It is meaningful to seak of the internal energy of a system, but not how much heat it contains. Thermodynamics rocesses: in all rocesses energy is conserved First law of thermodynamics. Energy conservation uts limits on what it is ossible, something can t haen where the total energy after the event is greater than before. Energy can be transferred from one location to another, but the total must remains the same. a03/1/thermal/th doc :17 PM 1/11/0

3 Ideal Gas work = d = n R T = N k T k = R / N A m tot = n M T U S m tot N n heat Q Q = n C T C N = n N A U = Q S = dq T C a03/1/thermal/th doc :17 PM 1/11/0 3

4 All rocesses (all aths) reversible rocess U = Q = n C T C = C + R (Eq ) dq S = 1 T (Eq ) Adiabatic (Q = 0) U = - (Eq. 17.8) γ = constant (Eq. 17.4) T γ-1 = constant (Eq. 17.) C 1 = ( 11 ) = ( 11 ) R γ 1 (Eq. 17.6) ratio of secific heats γ = C / C (Eq ) S = 0 Isochoric ( = 0) U = Q = n C T (Eq. 17.9) = 0 Isobaric ( = 0) = (Eq ) Q = n C T Isothermal ( T = 0) U = 0 Q = = n R T ln( / 1 ) reversible rocess S = Q / T Cyclic U = 0 = Q reversible cyclic rocess S = 0 (one cycle) Free exansion U = 0 Q = = 0 T = 0 Not a reversible rocess as a gas suddenly exands into a vacuum. Although the initial and final states are well defined, it is a nonequilibrium rocess and can t be lotted on a - diagram. a03/1/thermal/th doc :17 PM 1/11/0 4

5 Notes ork = F dr = A dx = d( Ax) = r x x ( ) r 1 x 1 x1 d 1 = area under - grah F = A dx A 1, = > 0 > 0 Cyclic: clockwise 1 to > 0 Anticlockwise 1 to < 0 1 < 0 Internal energy For an ideal gas where all the molecules can be treated as oint masses, the internal energy is the sum of all the kinetic energies of all the molecules. Average kinetic energy of a molecule er degree of freedom ½ k T Monatomic gas (N molecules): degrees of freedom, f = 3 average kinetic energy of a molecule ( ) internal energy ( ½ ) 3 ½mv = k T avg U = N mv = NkT = nna kt = nrt = nct avg Molar heat caacity C = (3/) R C = C + R = (5/) R a03/1/thermal/th doc :17 PM 1/11/0 5

6 Diatomic gas (N molecules): degrees of freedom, f = 5 average kinetic energy of a diatomic molecule ( ½ ) 3 5 mv = kt + kt = kt avg internal energy ( ½ ) translation rotation U = N mv = NkT = nna kt = nrt = nct avg f U = nr T = nc T Molar heat caacity C = (5/) R C = C + R = (7/) R Thermodynamic system Ideal Gas work = d = n R T = N k T k = R / N A m tot = n M T U S m tot N n heat Q Q = n C T C N = n N A U = Q S dq = T C Assumtions changes occur slowly system is always in thermal equilibrium,, T same throughout system and do nor change with time. Thermal rocesses All rocesses (all aths) U = Q = n C T C = C + R (17.17) ratio of secific heats γ = C / C (17.18) a03/1/thermal/th doc :17 PM 1/11/0 6

7 Isothermal ( T = 0) U = 0 Boyle s Law ( ) T 1 = T 1 1 = nrt Q = = d = d = nrt 1 1 ln 1 Q = = n R T ln( / 1 ) Isothermal rocess ressure (kpa) Isothermals = constant volume (m 3 ) 100 K 400 K 800 K Isochoric ( = 0) = 0 U = Q = n C T (17.9) Isochoric rocess ressure (kpa) to : Q > 0 T < T T > 0 U > = Isothermals = constant volume (m ) 100 K 400 K 800 K a03/1/thermal/th doc :17 PM 1/11/0 7

8 Isobaric ( = 0) = (17.10) Q = n C T U = nc T Q = nc T = = nrt U = Q = nc T = nc T nr T = nc T C = C + R For a given rise in temerature, heat Q must be larger than in the constant volume case because in the constant ressure case, work is done in rasing the iston during the exansion. Isobaric rocess ressure (kpa) Isothermals = constant volume (m 3 ) 100 K 400 K 800 K Adiabatic (Q = 0) U = - (17.8) γ = C / C (17.8) C = C + R C = (f / ) R C = (f / + 1) R γ = (f + ) / f γ = constant (17.4) T γ-1 = constant (17.) C = R = γ 1 (17.6) 1 ( 1 1 ) ( 1 1 ) An adiabat is always steeer on a - diagram than the near by isotherms since γ > 1. Adiabatic rocesses can occur when the system is well insulated or a very raid a03/1/thermal/th doc :17 PM 1/11/0 8

9 rocess occurs so that there is not enough time for a significant heat to be transferred eg raid exansion of a gas; a series of comressions and exansions as a sound wave roagates through air; atmosheric rocesses changes in atmosheric ressure HIGH ressure cell, falling air is comressed and warmed LO ressure cell, rising air is comressed and cooled condensation and rain. convergence divergence HIGH - more uniform conditions - inhibits cloud formation sunshine divergence convergence LO -less uniform conditions - encourages cloud formation U = nrt nc dt = d = d dt R d T = T 1 T C 1 R R C ln( T / T1) = ln( / 1) = ln( 1/ ) C R C R = C C = 1= γ 1 C C 1 ( γ 1) T 1 ( γ 1) = T = constant T1 11 = T T ( γ 1) T 1 = = T1 1 1 γ 1 γ = = constant 1 00 Adiabatic rocess ressure (k Pa) 180 Isothermals = constant volume (m 3 ) 100 K 400 K 800 K a03/1/thermal/th doc :17 PM 1/11/0 9

10 Problem One mole of oxygen enclosed in a cylinder with a movable iston (assume the gas is ideal) is taken from an initial state A to another state B then to state C and back to state A. Find the values of Q, and U for the aths A to B; B to C; C to A and the comlete cycle A to B to C to A and clearly indicate the sign + or for each rocess. Does this cycle reresent a heat engine? C ressure (kpa) A B 100 K 400 K 800 K volume (m 3 ) Solution Setu n = 1 mol oxygen diatomic f = C = 5/ R C = 7/ R R = 8.31 J.mol -1.K -1 C = 0.8 J.mol -1.K -1 C = 9.1 J.mol -1.K -1 Action A to B is isobaric T A B = T B T A = ( ) K = 300 K A = B = = 40 kpa = Pa A B = ( ) m 3 = 0.06 m 3 A B > 0 since gas exands A B = = ( )(0.06) = J a03/1/thermal/th doc :17 PM 1/11/0 10

11 U A B > 0 since the temerature increases U A B = nc T = (1)(0.8)(300) J = J Q A B > 0 since U A B > 0 and U A B = Q > 0 Q > > 0 Q A B = nc T = (1)(9.1)(300) J = J Check: First law U A B = Q A B - A B = ( ) J = J B to C is isochoric T B C = T C T B = ( ) K = 400 K 40 kpa = Pa C = 80 kpa = B = Pa B C = Pa 4 A B = 0 m 3 B C = 0 since no change in volume B C = 0 U B C > 0 since the temerature increases U B C = nc T = (1)(0.8)(400) J = J Q A B = U A B since = 0 Q A B = J C to A T C A = T A T C = ( ) K = -700 K A = 40 kpa = Pa C = 80 kpa = Pa C A = Pa A = 0.0 m 3 C = 0.08 m 3 C A = 0.06 m 3 C A < 0 since gas is comressed C A = area under curve = area of rectangle + area of triangle C A = - { (0.06)( ) + (½)(0.06)( )} J C A = - J a03/1/thermal/th doc :17 PM 1/11/0 11

12 U C A < 0 since the temerature decreases U C A = nc T = (1)(0.8)(-700) J = J Q C A = U C B + Q C C A A = J = J Comlete cycle U = 0 J A to B B to C C to A cycle (total values) (kj) Q (kj) U (kj) Q (kj) The discreancies between U and Q are due to round off errors. cycle < 0 work is done on the system the system is not a heat engine because a heat engine needs to d o a net amo unt of work on the surroundings each cycle. The net work corresonds to the area unenclosed i.e. the are of the triangle.06)( ) J = cycle = - (1/)(0 J (value agrees with table) a03/1/thermal/th doc :17 PM 1/11/0 1

13 Examle T he ressure decreases as one travels u through the atmoshere and the temerature decreases with height above the earth s surface through the trooshere. hy? Consider a arcel of air of mass m as it r ises through the trooshere. As the air rises, its volume changes because of the ressure changes, while the mass remains constant. The changes in volume occur adiabatically: Q = 0 T γ 1 = constant air, γ = 1.4 (γ - 1) = 0.4 > 0 T e can calculate the decrease in temerature with height using the First Law of Thermodynamics U = Q = - > 0 since the gas exands and does work on the surroundings U < 0 T Assuming the gas is ideal and for an infinitesimal change du = n C dt d = d = n R T d + d = n R dt d = n R dt d du = - d n C d T = - n R dt + d n (C + R) dt = n C dt = d dt/d = / n C heat caa cities : m c dt = n C dt n C P = m c ressu re with height = - ρ g h d /dh = - ρ g dh/d = -1/ (ρ g) density ρ = m / = m / ρ (dt/dh) (dh/d) = (dt/dh)(-1/(ρ g) ) = (m / ρ) / (m c ) dt/dh = - g / c temerature gradient g ~ 10 m.s -1 c (air) ~ 1000 J.kg -1.K -1 dt/dh ~ C.m -1 Dry air adiabatic lase rate ~ -1 C er 100 m or ~ -10 C er km rise in altitude a03/1/thermal/th doc :17 PM 1/11/0 13