When one mole of an ideal gas is compressed to half its initial volume and simultaneously heated to twice its initial temperature, the change in entropy $(△ S)$ is :

C_{V} l n 2

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C_{P} l n 2

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R l n 2

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(C_{V} - R) l n 2

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Solution

The correct option is C $(C_{V} - R) l n 2$
$△ S = n C_{v} l n (\frac{T_{2}}{T_{1}}) + n R l n (\frac{V_{2}}{V_{1}})$
$= C_{v} l n (2) + R l n (1 / 2)$
$= (C_{v} - R) l n (2)$

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When one mole of an ideal gas is compressed to half its initial volume and simultaneously heated to twice its initial temperature, the change in entropy (△S) is :

The correct option is C (CV−R)ln2△S=nCvln(T2T1)+nRln(V2V1)=Cvln(2)+Rln(1/2)=(Cv−R)ln(2)

When one mole of an ideal gas is compressed to half its initial volume and simultaneously heated to twice its initial temperature, the change in entropy $(△ S)$ is :

The correct option is C $(C_{V} - R) l n 2$
$△ S = n C_{v} l n (\frac{T_{2}}{T_{1}}) + n R l n (\frac{V_{2}}{V_{1}})$
$= C_{v} l n (2) + R l n (1 / 2)$
$= (C_{v} - R) l n (2)$