Question

The change in the entropy of a $1$ mole of an ideal gas which went through an isothermal process from an initial stat $ ({{P}_{1}},{{V}_{1}},T) $ to the final state $ ({{P}_{2}},{{V}_{2}},T) $ is equal to:

• A. zero
• B. $ R\,ln\,T $
• C. $ R\,ln\,\frac{{{V}_{1}}}{{{V}_{2}}} $
• D. $ R\,ln\,\frac{{{V}_{2}}}{{{V}_{1}}} $

Answer 1

Answer: (D)

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) $