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CAF123

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## Homework Statement

Assume that a gas obeys the VDW Virial expansion Pv = RT + (b-a/RT)P to first order in P and u = 3RT - aP/RT to first order in P, where v and u are molar quantities.

In the following cycle (see attachment), the heat transferred to the gas is transferred by direct thermal contact with a reservoir at T1 (for the const pressure process at P1 and isotherm at T1) and heat removed is transferred via a reservoir in thermal contact at T2 (for the const pressure process P2 and isotherm at T2). Derive an algebraic expression for the following sum around the cycle $$\sum_i \frac{Q_i}{T_{res_i}}$$ and show that the sum is less than zero.

## Homework Equations

First law in terms of change in entropy: du = TdS + Pdv

## The Attempt at a Solution

I need to find 4 expressions for the heat in/out of the working substance. For an isobaric process (P1), ##dQ_{1,in} = \left(\frac{\partial H}{\partial T}\right)_PdT = (4R + \frac{2aP_1}{RT^2})dT,## by differentiating u and v above since H=u+Pv. Integrate this from T2 to T1 gives heat in at constant pressure P1.

For heat in during the isotherm, use the first law so that $$Q_{2,in} = TdS = -\frac{a}{RT_1}(P_2-P_1)-RT_1\ln(P_2/P_1)$$ using ##dS =1/T_1 \left(-\frac{a}{RT_1}(P_2-P_1) - RT_1\ln(P_2/P_1)\right)##that I derived earlier.

For heat out during the constant pressure P2, same as above in the other constant pressure process but reverse the indices and put a negative at the front (since heat in is defined +ve) => $$Q_{3,out} = -4R(T_2-T_1) + \frac{2aP_2}{R}(\frac{1}{T_2}-\frac{1}{T_1})$$

For heat out during the isotherm T2, same as above in the other isotherm process,but reverse indices and put in a negative: ##Q_{4,out} = a/RT_2 (P_1-P_2) + RT_2\ln(P_1/P_2)##

When I sum all these terms and then analyse each term independently, I see that every term is positive and so the sum cannot possibly be less than zero.

Many thanks.