Consider a heat engine that is not reversible. The engine uses $1.000 \mathrm{mol}\( of a diatomic ideal gas. In the first step \)(\mathrm{A})$ there is a constant temperature expansion while in contact with a warm reservoir at \(373 \mathrm{K}\) from \(P_{1}=1.55 \times 10^{5} \mathrm{Pa}\) and $V_{1}=2.00 \times 10^{-2} \mathrm{m}^{3}$ to \(P_{2}=1.24 \times 10^{5} \mathrm{Pa}\) and $V_{2}=2.50 \times 10^{-2} \mathrm{m}^{3} .$ Then (B) a heat reservoir at the cooler temperature of \(273 \mathrm{K}\) is used to cool the gas at constant volume to \(273 \mathrm{K}\) from \(P_{2}\) to $P_{3}=0.91 \times 10^{5} \mathrm{Pa} .$ This is followed by (C) a constant temperature compression while still in contact with the cold reservoir at \(273 \mathrm{K}\) from \(P_{3}, V_{2}\) to \(P_{4}=1.01 \times 10^{5} \mathrm{Pa}, V_{1} .\) The final step (D) is heating the gas at constant volume from \(273 \mathrm{K}\) to \(373 \mathrm{K}\) by being in contact with the warm reservoir again, to return from \(P_{4}, V_{1}\) to \(P_{1}, V_{1} .\) Find the change in entropy of the cold reservoir in step \(\mathrm{B}\). Remember that the gas is always in contact with the cold reservoir. (b) What is the change in entropy of the hot reservoir in step D? (c) Using this information, find the change in entropy of the total system of gas plus reservoirs during the whole cycle.

Step by step solution

01

Calculate the change in entropy of the cold reservoir in step B

The gas is cooled at constant volume in step B from \(P_2\) to \(P_3\) and from \(373K\) to \(273K\). To find the change in entropy of the cold reservoir, we need to find the heat transfer \(Q_{B}\) during the cooling process first. Using the specific heat capacity at constant volume for a diatomic gas, \(C_{v}=\frac{5}{2}R\) and the ideal gas law, the heat transfer can be calculated as: $$Q_{B}=nC_{v}\Delta T = nC_{v}(T_{3}-T_{2})$$ We are given that \(n = 1 \mathrm{mol}\), \(C_{v}=\frac{5}{2}R\), \(T_{2}=373K\), \(T_{3}=273K\), therefore: $$Q_{B} = 1\cdot\frac{5}{2}R(273-373)=-\frac{5}{2}R(100)$$ Now the change in entropy for the cold reservoir during step B can be determined as: $$\Delta S_{cold}=-\frac{Q_B}{T_{cold}}$$ Insert the calculated \(Q_B\) and the temperature of the cold reservoir \(T_{cold}=273K\): $$\Delta S_{cold}=-\frac{-\frac{5}{2}R(100)}{273}=\frac{500R}{546}$$

02

Calculate the change in entropy of the hot reservoir in step D

In step D, the gas is heated at constant volume from \(P_4\) to \(P_1\) and from \(273K\) to \(373K\). To find the change in entropy of the hot reservoir, we need to find the heat transfer \(Q_{D}\) during the heating process first. Using the specific heat capacity at constant volume for a diatomic gas, \(C_{v}=\frac{5}{2}R\), and the ideal gas law, the heat transfer can be calculated as: $$Q_{D}=nC_{v}\Delta T = nC_{v}(T_{1}-T_{4})$$ We are given that \(n = 1\ \mathrm{mol}\), \(C_{v}=\frac{5}{2}R\), \(T_{1}=373K\), \(T_{4}=273K\): $$Q_{D} = 1\cdot\frac{5}{2}R(373-273)=\frac{5}{2}R(100)$$ Now the change in entropy for the hot reservoir during step D can be determined as: $$\Delta S_{hot}=-\frac{Q_D}{T_{hot}}$$ Insert the calculated \(Q_D\) and the temperature of the hot reservoir \(T_{hot}=373K\): $$\Delta S_{hot}=-\frac{\frac{5}{2}R(100)}{373}=-\frac{500R}{746}$$

03

Calculate the total change in entropy for the whole cycle

To find the total change in entropy for the whole cycle (gas plus reservoirs), we need to sum the change in entropy of the cold reservoir from step B and the change in entropy of the hot reservoir from step D: $$\Delta S_{total}=\Delta S_{cold}+\Delta S_{hot}$$ Plugging in our values for \(\Delta S_{cold}\) and \(\Delta S_{hot}\): $$\Delta S_{total}=\frac{500R}{546}-\frac{500R}{746}$$ Now, we can determine a common denominator and simplify the expression: $$\Delta S_{total}=\frac{500R(746-546)}{546\times746}$$ Finally, we can write the total change in entropy: $$\Delta S_{total}=\frac{500R(200)}{546\times746}=\frac{100000R}{407076}$$

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