A) zero
B) \[R\ln T\]
C) \[R\ln \frac{{{V}_{1}}}{{{V}_{2}}}\]
D) \[R\ln \frac{{{V}_{2}}}{{{V}_{1}}}\]
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., \[\Delta U=0\]. \[\therefore \] \[\Delta U=\Delta Q-W\] \[\Rightarrow \] \[0=\Delta Q-W\] \[\Rightarrow \] \[\Delta Q=W\] i.e., AQ = work done by gas in isothermal process which went through from \[({{P}_{1}},{{V}_{1}},T)\] to \[({{P}_{2}},{{V}_{2}},T)\] or \[\Delta Q=\mu RT\,{{\log }_{e}}\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)\] ... (ii) For 1 mole of an ideal gas, u == 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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