QuestionJune 19, 2025

2 mol of an ideal diatomic gas are cooled at constant volume until the pressure is reduced to 1/3 the initial pressure . What is the entropy change of the gas? A. -63.9J/K B +63.9J/K C. -9.13J/K D. 0 E. -45.6J/k

2 mol of an ideal diatomic gas are cooled at constant volume until the pressure is reduced to 1/3 the initial pressure . What is the entropy change of the gas? A. -63.9J/K B +63.9J/K C. -9.13J/K D. 0 E. -45.6J/k
2 mol of an ideal diatomic gas are cooled at constant volume until the pressure is reduced to 1/3 the initial pressure . What is the entropy change of the gas? A. -63.9J/K B +63.9J/K C. -9.13J/K D. 0 E. -45.6J/k

Solution
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Answer

E. -45.6 \, \text{J/K} Explanation 1. Identify the formula for entropy change For an ideal gas, the change in entropy at constant volume is given by \Delta S = nC_v \ln\left(\frac{T_2}{T_1}\right), where C_v is the molar heat capacity at constant volume. 2. Relate pressure and temperature At constant volume, P \propto T. Therefore, \frac{T_2}{T_1} = \frac{P_2}{P_1} = \frac{1}{3}. 3. Calculate C_v for diatomic gas For a diatomic gas, C_v = \frac{5}{2}R, where R = 8.314 \, \text{J/mol K}. 4. Substitute values into the entropy change formula \Delta S = 2 \times \frac{5}{2} \times 8.314 \ln\left(\frac{1}{3}\right) 5. Compute the result \Delta S = 5 \times 8.314 \times \ln\left(\frac{1}{3}\right) = 41.57 \times (-1.0986) = -45.6 \, \text{J/K}

Explanation

1. Identify the formula for entropy change<br /> For an ideal gas, the change in entropy at constant volume is given by $\Delta S = nC_v \ln\left(\frac{T_2}{T_1}\right)$, where $C_v$ is the molar heat capacity at constant volume.<br /><br />2. Relate pressure and temperature<br /> At constant volume, $P \propto T$. Therefore, $\frac{T_2}{T_1} = \frac{P_2}{P_1} = \frac{1}{3}$.<br /><br />3. Calculate $C_v$ for diatomic gas<br /> For a diatomic gas, $C_v = \frac{5}{2}R$, where $R = 8.314 \, \text{J/mol K}$.<br /><br />4. Substitute values into the entropy change formula<br /> $\Delta S = 2 \times \frac{5}{2} \times 8.314 \ln\left(\frac{1}{3}\right)$<br /><br />5. Compute the result<br /> $\Delta S = 5 \times 8.314 \times \ln\left(\frac{1}{3}\right) = 41.57 \times (-1.0986) = -45.6 \, \text{J/K}$
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