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

C_{p . m}

ln 2

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C_{v . m}

ln 2

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R In 2

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(C_{v, m} - R)

ln 2

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Solution

The correct option is D $(C_{v, m} - R)$ ln 2
n = 1 mole
Entropy of gas,
$Δ S = n C_{v . m} l n \frac{T_{2}}{T_{1}} + R l n \frac{V_{2}}{V_{1}}$
$= C_{v, m} l n 2 + R l n (\frac{1}{2}) = C_{v, m} l n 2 - R l n (2) = (C_{v . m} - R) l n 2$

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Entropy for a Reversible Process

Standard XI Chemistry

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

The correct option is D (Cv,m−R) ln 2n = 1 mole ​​​​​​​Entropy of gas, ΔS=nCv.m lnT2T1+R lnV2V1 =Cv,m ln 2+R ln(12)=Cv,m ln 2−R ln(2)=(Cv.m−R) ln 2

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

The correct option is D $(C_{v, m} - R)$ ln 2
n = 1 mole
Entropy of gas,
$Δ S = n C_{v . m} l n \frac{T_{2}}{T_{1}} + R l n \frac{V_{2}}{V_{1}}$
$= C_{v, m} l n 2 + R l n (\frac{1}{2}) = C_{v, m} l n 2 - R l n (2) = (C_{v . m} - R) l n 2$