hw15_sol_s11

# We emphasize that in order to obtain work and energy

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where we have used the ideal gas law in the last step. We emphasize that, in order to obtain work and energy in Joules, pressure should be in Pascals (N / m 2 ) and volume should be in cubic meters. The degrees of freedom for a diatomic gas is f = 5. (a) The internal energy change is

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  3 3 3 3 int int 3 5 5 2.0 10 Pa 4.0m 5.0 10 Pa 2.0m 2 2 5.0 10 J. c a c c a a E E p V p V   (b) The work done during the process represented by the diagonal path is  3 3 diag = 3.5 10 Pa 2.0m 2 a c c a p p W V V which yields W diag = 7.0×10 3 J. Consequently, the first law of thermodynamics gives 3 3 3 diag int diag ( 5.0 10 7.0 10 ) J 2.0 10 J. Q E W     (c) The fact that E int only depends on the initial and final states, and not on the details of the ―path‖ between them, means we can write 3 int int int 5.0 10 J c a E E E   for the indirect path, too. In this case, the work done consists of that done during the constant pressure part (the horizontal line in the graph) plus that done during the constant volume part (the vertical line):  3 3 4 indirect 5.0 10 Pa 2.0m 0 1.0 10 J. W Now, the first law of thermodynamics leads to 3 4 3 indirect int indirect ( 5.0 10 1.0 10 ) J 5.0 10 J. Q E W     11. Suppose the gas expands from volume V i to volume V f during the isothermal portion of the process. The work it does is i f V V V V V V nRT V dV nRT pdV W f i f i ln , where the ideal gas law pV = nRT was used to replace p with nRT / V . Now V i = nRT / p i and V f = nRT / p f , so V f / V i = p i / p f . Also replace nRT with p i V i to obtain ln . i i i f p W pV p Since the initial gauge pressure is 1.03 10 5 Pa, p i = 1.03 10 5 Pa + 1.013 10 5 Pa = 2.04 10 5 Pa. The final pressure is atmospheric pressure: p f = 1.013 10 5 Pa. Thus  5 5 3 4 5 2.04 10 Pa 2.04 10 Pa 0.14m ln 2.00 10 J. 1.013 10 Pa W During the constant pressure portion of the process the work done by the gas is W = p f ( V i V f ). The gas starts in a state with pressure p f , so this is the pressure throughout this portion of the process. We also note that the volume decreases from V f to V i . Now V f = p i V i / p f , so . 10 44 . 1 ) 14 . 0 )( 10 04 . 2 10 013 . 1 ( ) ( 4 3 5 5 J m Pa Pa V p p p V p V p W i i f f i i
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