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} l n 2

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C_{v, m} l n 2

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R l n 2

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(C_{v, m} - R) l n 2

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Solution

The correct option is C $(C_{v, m} - R) l n 2$
$Δ S = n C_{v, m} \times l n (\frac{T 2}{T 1}) + R \times l n (\frac{V 2}{V 1})$

$Δ S = C_{v, m} \times l n 1 + R \times l n \frac{1}{2} = (C_{v, m} - R) l n 2$

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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 C (Cv,m−R)ln2ΔS=nCv,m×ln(T2T1)+R×ln(V2V1)ΔS=Cv,m×ln1+R×ln12=(Cv,m−R)ln2

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 C $(C_{v, m} - R) l n 2$
$Δ S = n C_{v, m} \times l n (\frac{T 2}{T 1}) + R \times l n (\frac{V 2}{V 1})$

$Δ S = C_{v, m} \times l n 1 + R \times l n \frac{1}{2} = (C_{v, m} - R) l n 2$