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PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS SPRING 2006 Class Activity - Class 4 January 27, 2006 Name________SOLUTION__________________ Do problem 4-11 from the textbook. Since many of you do not have the textbook with you, here it is: An ideal diatomic gas, for which cv = 5R/2, occupies a volume of 2 m³ at a pressure of 4 atm, and a temperature of 20ºC. The gas is compressed to a final pressure of 8 atm. Compute the final volume, the final temperature, the work done, the heat released, and the change in internal energy for: (a) A reversible isothermal compression. (b) A reversible adiabatic compression. (a) For an isothermal process, T is constant, so PV is constant. P1V1 = P2V2, so V2 P1 V1 P2 4 atm 2 m3 V2 = 1 m3 8 atm Since the process is isothermal, T does not change. T = 20ºC = 293 K V PV W PdV , and P 1 1 , so V V V2 dV 1 m3 V W P1V1 P1V1 ln 2 (4 atm)(1.01 105 Pa/atm)(2 m3 ) ln 3 V1 V V 2 m 1 V2 2 1 W = – 5.6 ×105 J Q = – 5.6 ×105 J U = 0 Since T is constant, U = 0, so Q = W (b) For an adiabatic process, we need , where Then, cP cV 7 5 2 2 cP , and cP cV R 52 R R 72 R cV 7 1.4 . 5 1/ P P 4 atm 3 P1V1 P2V2 , so V2 1 V1 , and V2 1 V1 2m P 8 atm P2 2 PV T PV (8 atm)(1.22 m3 ) (293 K) PV = nRT, so 1 1 1 , and T2 2 2 T1 P2V2 T2 P1V1 (4 atm)(2 m3 )1 V2 V1 P1V1 1 1 V 1 V V2 1 V2 = 1.22 m3 T2 = 357 K = 84ºC V dV P1V1 . Then, W P1V1 V V V PV 1 1 1 P1V1 P2V2 P1V1 P2V2 1 1 1 1 1 V2 V1 1 V1 1 V2 1 1 W PdV , and PV P1V1 , so P W 1 / 1.4 2 1 [( 4 atm)(2 m3 ) (8 atm)(1.22 m3 )](1.01 105 Pa/atm) 1.4 1 Since the process is adiabatic, Since Q = 0, U = – W = – (– 4.4 ×105 J) W W = – 4.4 ×105 J Q=0 U = 4.4 ×105 J