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Unformatted text preview: Quiz 2 for Physics 176: Answers Professor Greenside Problems That Require Writing 1. (8 points) A liter of an ideal gas is cooled at constant pressure until its volume is halved. The air is then allowed to expand isothermally back to its original volume. Derive expressions for the total work W done on the gas and the total heat Q that is transferred to the gas as a result of this process. Make sure to indicate whether W and Q have positive, negative, or zero values. Answer: This process involves an isobaric compression followed by an isothermal expansion as shown in this figure: 0.0 0.2 0.4 0.6 0.8 1.0 1.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2 Let the initial volume be V (1 liter = 10 3 m 3 ) and the initial pressure be P . (Note that we can’t deduce a numerical value for P from the available information, we would have to know N and T .) The lower curve is a piece of the isothermal hyperbola P = P ( V ) = ( NkT ) /V = c/V where c is some positive constant. We can deduce the value of c by requiring that the hyperbola continue from the isobaric curve at V = V / 2, i.e., the point ( V / 2 , P ) must lie on the hyperbola which gives us c = P V / 2. Then the pressure at the end of the isothermal expansion must be c/V = ( P V / 2) / V = P / 2, i.e., the final pressure is half the initial pressure. You could also reach this conclusion by realizing that, for, an ideal gas satisfying P V = NkT one has T = ( P/ ( Nk )) V . So an isobaric compression from V to V / 2 with P and N constant must correspond to decreasing the initial temperature T to T / 2. If the temperature is now kept constant with value T / 2 by an isothermal expansion back to the original volume, the pressure must decrease to P / 2. Since this twostep process is not cyclic (the end point ( V , P / 2) is not the same as the starting point ( V , P )), the total change in energy Δ U is not zero. A straightforward way to get the total work done on the gas and total heat added to the gas is then to compute Δ U , the work done on the gas W = R P ( V ) dV , and the heat Q = Δ U W at each step and then combine the values of Δ U , W , and Q at each step to get the final values. Before doing any calculation, let’s figure out qualitatively the signs of Δ U total , W total , and Q total . We know from the equipartition theorem for an ideal gas that U = Nf ( kT/ 2) = ( f/ 2) P V so Δ U = ( f/ 2)Δ( P V ). We easily see from the graph that the final pressure is less than the initial pressure while the final volume is the same as the initial volume. So Δ U = ( f/ 2)Δ( P V ) = ( f/ 2) V ( P final P initial ) < 0, the internal energy has decreased by the end. Another way to see this is that the energy must decrease during the isobaric compression since Δ( P V ) < 0, and then the energy stays constant during any 1 isothermal process (Δ U = ( Nfk/ 2)Δ T ). So the total decrease in energy just corresponds to the decrease in energy associated with the isobaric compression....
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 Spring '08
 Behringer
 Physics, Thermodynamics, Entropy, Mean free path

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