When 1 mol of an ideal gas is compressed to half of its volume,its temperature becomes double. Then the change in entropy $(△ S)$ would be:

C_{v} l n 4

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C_{p} l n 2

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C_{V} R l n 4

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(C_{V} - R) l n 4 \times C_{p}

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Solution

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

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When 1 mol of an ideal gas is compressed to half of its volume,its temperature becomes double. Then the change in entropy (△S) would be:

The correct option is B Cp ln2△S=nCvln(T2T1)+nRln(V2V1)=Cvln(2)+Rln(12) ∵Cp−Cv=R=(Cv−R)ln2 Cv−R−Cp=Cpln2

When 1 mol of an ideal gas is compressed to half of its volume,its temperature becomes double. Then the change in entropy $(△ S)$ would be:

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