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(A) The entropy change for a system can be calculated as:$\Delta S = n C_{v} \ln \frac{T_{2}}{T_{1}} + n R \ln \frac{V_{2}}{V_{1}} \dots (I)$For an is

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$10.61$

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(A) The entropy change for a system can be calculated as:$\Delta S = n C_{v} \ln \frac{T_{2}}{T_{1}} + n R \ln \frac{V_{2}}{V_{1}} \dots (I)$For an isochoric process,$V_{2} = V_{1}$,so $\ln \frac{V_{2}}{V_{1}} = \ln(1) = 0$.For a diatomic gas,$C_{v} = \frac{5R}{2}$.Substituting the values into expression $(I)$:$\Delta S = 1 	imes \frac{5 	imes 8.314}{2} 	imes \ln \frac{500}{300}$$\Delta S = 2.5 	imes 8.314 	imes 0.5108$$\Delta S = 10.61\,J/K$

$1$ mole of a diatomic gas is heated through an isochoric process from $300\,K$ to $500\,K$. The change in entropy is: (in $,J/K$)

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