Question

The entropy change in the isothermal reversible expansion of 2 moles of an ideal gas from 10L to 100L at 300K is:
(A) $42.3{\text{ J}}{{\text{K}}^{ - 1}}$
(B) ${\text{35}}{\text{.8 J}}{{\text{K}}^{ - 1}}$
(C) ${\text{38}}{\text{.3 J}}{{\text{K}}^{ - 1}}$
(D) ${\text{32}}{\text{.3 J}}{{\text{K}}^{ - 1}}$

Answer 1

Hint: In an isothermal process, temperature remains constant i.e. $\Delta {\text{T = 0}}$. Joule's law states that if q=0 then there is no change in the internal energy U and hence, change in entropy entirely depends on change in volume of the gas.

Complete step by step answer:
Change in entropy in an isothermal irreversible process is given by
$\Delta {\text{S = 2}}{\text{.303 nR log}}\dfrac{{{{\text{V}}_{{\text{final}}}}}}{{{{\text{V}}_{{\text{initial}}}}}}$
We have ${{\text{V}}_{{\text{initial}}}}{\text{ = 10 d}}{{\text{m}}^3}$ and ${{\text{V}}_{{\text{final}}}}{\text{ = 100 d}}{{\text{m}}^3}$
Substituting these values in the given equation we get,
$\Delta {\text{S = 2}}{\text{.303 }} \times {\text{ 2 }} \times {\text{ 8}}{\text{.314 }} \times {\text{ log }}\dfrac{{100}}{{10}}$
Since, ${\text{log }}\dfrac{{100}}{{10}}{\text{ = log 10 = 1}}$
$\Delta {\text{S = 2}}{\text{.303 }} \times {\text{ 2 }} \times {\text{ 8}}{\text{.314 }}$
${\text{ = 38}}{\text{.29}}$
$\approx {\text{ 38}}{\text{.3 K}}{{\text{J}}^{ - 1}}$
So, the change in entropy is ${\text{38}}{\text{.3 J}}{{\text{K}}^{ - 1}}$ i.e. option C is correct.

Additional information:
Whether the process is reversible or irreversible, in case of free expansion of a gas, the amount of work done is always 0.
We know that internal energy of a system is given by the formula: ${\text{q + ( - W)}}$
Where, $\Delta {\text{U}}$= change in internal energy of a system, q= heat supplied and W= Amount of work done on the system.
However, this equation changes as the type of process changes.
When an ideal gas is subjected to isothermal expansion i.e. $\Delta {\text{T = 0}}$ in vacuum the work done W = 0 as ${{\text{p}}_{{\text{ex}}}}{\text{ = 0}}$. As determined by Joule experimentally q =0, thus $\Delta {\text{U }} = 0$.

Note: The process after which the system as well as the surroundings return to their original states is called a reversible process. Whereas the process in which on completion, permanent changes occur in the system as well as the surrounding is called an irreversible process.