When one mole of an ideal gas is compressed to half its initial volume and simultaneously heated to twice…

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 $(\Delta \mathrm{S})$ is
  1. $\mathrm{C}_{\mathrm{V}} \ln 2$
  2. $\mathrm{C}_{\mathrm{p}} \ln 2$
  3. $\mathrm{R} \ln 2$
  4. $\left(\mathrm{C}_{\mathrm{V}}-\mathrm{R}ight) \ln 2$

Solution

When there is simultaneously change in temperature and volume (or pressure)
$\Delta \mathrm{S}=n C_{v} \ln \left(\frac{T_{2}}{T_{1}}ight)+n R \ln \left(\frac{V_{2}}{V_{1}}ight)$
$=C_{V} \log _{e}\left(\frac{2}{1}ight)+R \log _{e}\left(\frac{1}{2}ight)$
$=C_{V} \log _{e} 2-R \log _{e} 2$
$=\left(C_{V}-Right) \log _{e} 2$
hence (d) .

Asked in: JEE-TOPICTESTS-CHEMISTRY

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