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 :
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 \[\Delta U=\Delta Q-W\Rightarrow 0=\Delta Q-W\Rightarrow \Delta Q=W\] i.e., \[\Delta Q\]= work done by gas in isothermal process which went through from \[({{P}_{1}},{{V}_{1}},T)\] to \[({{P}_{2}},{{V}_{2}},T)\] \[\Rightarrow \,\,\Delta Q=\mu RT\log {{ & }_{e}}\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)\] ?..(ii) For 1 mole of an ideal gas, \[\mu =1,\] so from eqs. (i) and (ii) \[\Delta S=R\,{{\log }_{e}}\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)=R\,In\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)\]
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