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Standard XII

Chemistry

Heat and Work

The change in...

Question

The change in the entropy of a mole of ideal gas which went through an isothermal process from an initial state ( $P_{1}, V_{1}, T$ ) to the final state ( $P_{2}, V_{2}, T$ ) is equal to

zero

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R ln T

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ln \frac{V_{1}}{V_{2}}

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ln \frac{V_{2}}{V_{1}}

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Solution

The correct option is D $ln \frac{V_{2}}{V_{1}}$
Work done by $n$ moles of a gas when its volume changes from $V_{1}$ to $V_{2}$ is,
$W = n R T {log}_{e} \frac{V_{2}}{V_{1}}$
For $n = 1$
$W = R T ln \frac{V_{2}}{V_{1}}$

For an isothermal process, $Δ U = 0$

$∴ Δ Q = Δ U + W = 0 + R T ln \frac{V_{2}}{V_{1}}$
$∴$ Entropy $= \frac{Δ Q}{T} = R ln \frac{V_{2}}{V_{1}}$

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The change in the entropy of a mole of ideal gas which went through an isothermal process from an initial state (P1,V1,T) to the final state (P2,V2,T) is equal to

The correct option is D lnV2V1Work done by n moles of a gas when its volume changes from V1 to V2 is,W=nRTlogeV2V1For n=1W=RTlnV2V1For an isothermal process, ΔU=0∴ΔQ=ΔU+W=0+RTlnV2V1∴ Entropy=ΔQT=RlnV2V1

The change in the entropy of a mole of ideal gas which went through an isothermal process from an initial state ( $P_{1}, V_{1}, T$ ) to the final state ( $P_{2}, V_{2}, T$ ) is equal to