One mole of an ideal diatomic gas (C_V = 5 c | vedclass

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(B) The general formula for entropy change is $\Delta S = nC_V \ln \left(\frac{T_2}{T_1}\right) + nR \ln \left(\frac{V_2}{V_1}\right)$.Given:$n = 1 \

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$5 \ln \frac{373}{298} + 2 \ln 10$

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(B) The general formula for entropy change is $\Delta S = nC_V \ln \left(\frac{T_2}{T_1}ight) + nR \ln \left(\frac{V_2}{V_1}ight)$.Given:$n = 1 \ mol$$T_1 = 25 \ ^{\circ}C = 298 \ K$$T_2 = 100 \ ^{\circ}C = 373 \ K$$C_V = 5 \ cal / mol \cdot K$$R = 2 \ cal / mol \cdot K$$V_1 = 1 \ L$$V_2 = 10 \ L$Substituting the values:$\Delta S = 1 	imes 5 \ln \left(\frac{373}{298}ight) + 1 	imes 2 \ln \left(\frac{10}{1}ight)$$\Delta S = 5 \ln \left(\frac{373}{298}ight) + 2 \ln 10$.

One mole of an ideal diatomic gas $(C_V = 5 \ cal)$ was transformed from initial $25 \ ^{\circ}C$ and $1 \ L$ to the state when temperature is $100 \ ^{\circ}C$ and volume $10 \ L$. The entropy change of the process can be expressed as $(R = 2 \ cal / mol \cdot K)$ :-

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