Question

One mole of an ideal gas at 300K in thermal contact with surroundings expands isothermally from $1.0\,L$ to $2.0\,L$ against a constant pressure of $3.0\,atm$. In this process, the change in entropy of surrounding $(\Delta {S_{surr}})$ in $J{K^{ - 1}}$ is $(1\,Latm = 103\,J)$.
A. $ + 5.673$
B. $ + 1.013$
C. $ - 1.013$
D. $ - 5.673$

Answer 1

Hint: In an isothermal process, the change in internal energy is due to the transfer of energy as work and heat. But considering as ideal gas there is no change in internal energy of the system. The change in entropy is the change in heat transferred during the work done under a certain temperature.

Complete answer:
Isothermal process is a type of thermodynamic process in which the temperature of the system remains constant. This typically occurs when a system is in contact with an outside thermal reservoir, and the change in the system will occur slowly enough to allow the system to continue to adjust to the temperature of the reservoir through heat exchange. In an isothermal process the internal energy of an ideal gas is constant. To know the change in entropy is:
$\Delta {S_{surr}} = \dfrac{{\Delta q}}{T}$
Where
Q is heat
T is temperature
$\Delta q = \Delta u + \Delta w$
$\Delta q = $ Change in heat
$\Delta u = $ Change in internal energy
$\Delta w = $ Change in work done
In an isothermal process there is no change in internal energy therefore$\Delta u = 0$. Hence $\Delta q = \Delta w$
Now in an isothermal process heat gained by a surrounding is heat lost by a system.
Hence
$\Delta {q_{system}} = - \Delta {q_{surr}}$
Therefore:
$\Delta {S_{surr}} = \dfrac{{\Delta {q_{surr}}}}{T}$
$ = - \dfrac{{\Delta {q_{sys}}}}{T} = \dfrac{{ - \Delta w}}{T}$
In thermodynamics, work performed by a system is energy transferred by the system to its surroundings,
$\Delta w = - p(dv)$
Therefore,
$\Delta {S_{surr}} = - \dfrac{{P(dV)}}{T}$
$ = \dfrac{{ - 3(2 - 1)}}{{300}}\times 103\,J$
$ = - 1.03\,J$
Therefore the change in entropy is -1.03 J

Therefore the correct option is (C). $ - 1.013$

Note:
The difference between the reversible and free expansions is found in the entropy of the surroundings. In both cases, the surroundings are at a constant temperature, T. the minus sign is used since the heat transferred to the surroundings is equal in magnitude and opposite in sign to the heat, Q, transferred to the system.