A) zero
B) \[R\,In\,T\]
C) \[R\,In\,\frac{{{V}_{1}}}{{{V}_{2}}}\]
D) \[R\,In\,\frac{{{V}_{2}}}{{{V}_{1}}}\]
Correct Answer: D
Solution :
Key Idea: In an isothermal process, internal energy remains same as temperature is constant. The change in entropy of an ideal gas \[\Delta S=\frac{\Delta Q}{T}\] ...(i) In isothermal process, temperature does not change, that is, internal energy which is a function of temperature will remain same, i.e., \[\Delta U=0.\] First law of thermodynamics gives \[\Delta U=\Delta Q-W\] or \[0=\Delta Q-W\] or \[\Delta Q=W\] i.e.,\[\Delta Q=\] 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,\[\mu =1,\] so from Eqs. (i) and (ii) \[\Delta S=R{{\log }_{e}}\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)=R\,ln\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)\]
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