Question

The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of $10\,{\text{d}}{{\text{m}}^3}$ to a volume of $100\,{\text{d}}{{\text{m}}^3}$ at $27{\,^ \circ }{\text{C}}$ is:
A. $42.3\,{\text{J}}\,\,{{\text{K}}^{ - 1}}\,{\text{mo}}{{\text{l}}^{ - 1}}$
B. $38.3\,{\text{J}}\,\,{{\text{K}}^{ - 1}}\,{\text{mo}}{{\text{l}}^{ - 1}}$
C. $35.8\,{\text{J}}\,\,{{\text{K}}^{ - 1}}\,{\text{mo}}{{\text{l}}^{ - 1}}$
D. $32.3\,{\text{J}}\,\,{{\text{K}}^{ - 1}}\,{\text{mo}}{{\text{l}}^{ - 1}}$

Answer 1

The entropy change for an isothermal reversible expansion is determined in terms of temperature and volume change. The formula will be reduced in case of isothermal process.

Formula used: $\Delta S = \,n{C_v}\ln \left[ {\dfrac{{{T_f}}}{{{T_i}}}} \right] + \,\,nR\ln \left[ {\dfrac{{{V_f}}}{{{V_i}}}} \right]$

Complete step by step answer:
Suppose an ideal gas is taken in a cylinder having a moving piston. When a gas expands at a constant temperature the volume of the gas increases. The piston moves upward. The gas does some work for this expansion and the randomness of the gas increases hence the entropy increases.
The formula of energy change during the isothermal reversible expansion is as follows:
$\Delta S = \,n{C_v}\ln \left[ {\dfrac{{{T_f}}}{{{T_i}}}} \right] + \,\,nR\ln \left[ {\dfrac{{{V_f}}}{{{V_i}}}} \right]$
Where,
$\Delta S$ is the entropy change.
n is the number of molecules of gas.
${C_v}$ is the heat capacity at constant volume.
R is gas constant.
${T_i}$ is the initial temperature.
${T_f}$ is the final temperature.
${V_f}$ is the final volume.
${V_i}$ is the initial volume.
The entropy has to be calculated for isothermal process so, the entropy formula will be reduced as follows:
$\Delta S = \,nR\ln \left[ {\dfrac{{{V_f}}}{{{V_i}}}} \right]$
Multiply the above formula with 2.303 to convert the ln into log.
$\Delta S = \,2.303\,nR\log \left[ {\dfrac{{{V_f}}}{{{V_i}}}} \right]$
Substitute 2 moles for number of moles of gas, $8.314\,{\text{J}}\,\,{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}$ for gas constant,$10\,{\text{d}}{{\text{m}}^3}$ for initial volume and $100\,{\text{d}}{{\text{m}}^3}$ for final volume.
$\Delta S = 2.303\,\, \times 2\,{\text{mol}} \times 8.314\,{\text{J}}\,\,{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}} \times \log \left[ {\dfrac{{100\,{\text{d}}{{\text{m}}^3}}}{{10\,{\text{d}}{{\text{m}}^3}}}} \right]$
$\Delta S = 38.3\,{\text{J}}\,\,{{\text{K}}^{ - 1}}\,{\text{mo}}{{\text{l}}^{ - 1}}$
So, the entropy change involved in the isothermal reversible expansion of the ideal gas is $38.3\,{\text{J}}\,\,{{\text{K}}^{ - 1}}\,{\text{mo}}{{\text{l}}^{ - 1}}$.

Therefore, option (B) $38.3\,{\text{J}}\,\,{{\text{K}}^{ - 1}}\,{\text{mo}}{{\text{l}}^{ - 1}}$ is correct.

Note: According to ideal gas equation, at constant temperature the relation between volume and pressure is as follows:
$\dfrac{{{V_2}}}{{{V_1}}}\, = \dfrac{{{P_1}}}{{{P_2}}}$ So, $\dfrac{{{V_f}}}{{{V_i}}}$ can be replaced with $\dfrac{{{P_i}}}{{{P_f}}}$so, the expression of entropy change will be,
$\Delta S = \,n{C_P}\ln \left[ {\dfrac{{{T_f}}}{{{T_i}}}} \right] + \,\,nR\ln \left[ {\dfrac{{{P_i}}}{{{P_f}}}} \right]$.