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Q.

One mole of an ideal diatomic gas $(C_{v} = 5 cal)$ was transformed from initial $25 ° C$ 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 : $(R = 2 cal / mol / K)$

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a

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

b

$3 \ln \frac{298}{373} + 2 \ln 10$

c

$7 \ln \frac{373}{298} + 2 \ln \frac{1}{10}$

d

$5 \ln \frac{373}{298} + 2 \ln \frac{1}{10}$

answer is B.

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Detailed Solution

Changes to the entropy formula in general

$Δ S_{s y n} = n C_{v} \ln (\frac{T_{2}}{T_{1}}) + n R \ln (\frac{V_{2}}{V_{1}})$

given value

$\begin{aligned} T_{1} = 25^{\circ} C = 298 k \end{aligned}$

$T_{2} = 100^{\circ} C = 373 K$

$C_{v} = 5 Cal$

$R = 2 Cal / molK$

$\begin{aligned} V_{1} = 1 L \\ V_{2} = 10 L \end{aligned}$

$ΔS = 5 \ln (\frac{373}{298}) + 2 \ln (\frac{10}{1})$

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

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