A) zero
B) \[\text{R}\,\text{In}\,\text{T}\]
C) \[\text{R}\,\text{In}\,\frac{{{\text{V}}_{1}}}{{{\text{V}}_{2}}}\]
D) \[\text{R}\,\text{In}\,\frac{{{\text{V}}_{\text{2}}}}{{{\text{V}}_{\text{1}}}}\]
Correct Answer: D
Solution :
Key Idea: In an isothermal process, there is no change in internal energy of gas, i.e., \[\Delta U=0.\] The change in entropy of an ideal gas \[\Delta S=\frac{\Delta Q}{T}\] ?(i) In isothermal process, there is no change in internal energy of gas, i.e., \[\Delta U=0\] \[\therefore \] \[\Delta U=\Delta Q-W\] \[\Rightarrow \] \[0=\Delta Q-W\] \[\Rightarrow \] \[\Delta Q=W\] ie., \[\text{Q =}\]work done by gas in isothermal process which went through from \[\text{(}{{\text{P}}_{1}}\text{,}{{\text{V}}_{1}}\text{,T)}\] to \[\text{(}{{\text{P}}_{2}}\text{,}{{\text{V}}_{2}}\text{,T)}\] or \[\Delta Q=\mu RT{{\log }_{e}}\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)\] ?(ii) For 1 mole of an ideal gas, \[\mu =1\] So, from Eqs. (i) and (ii), we get or \[\Delta S=R{{\log }_{e}}\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)\] \[=R\ln \left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)\]
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