Question

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

Answer 1

The expansion of a gas occurs as reversible or irreversible. In isothermal reversible expansion the pressure is zero and temperature is constant. The entropy S is calculated as heat upon temperature. In isothermal reversible expansion, the internal energy change is 0, so its heat q is the work done in reversible process that is ${{W}_{rev}}=-2.303nRT\log \dfrac{{{V}_{2}}}{{{V}_{1}}}$ .

Complete answer:
We have been given an isothermal reversible expansion of 2 moles of an ideal gas that has initial volume,${{V}_{1}}$ as 10$d{{m}^{3}}$ and final volume ${{V}_{2}}$ as 100$d{{m}^{3}}$ at a temperature T of $27{}^\circ C$. We have to find the entropy change which is$\Delta S$.
As entropy change is calculated as $\Delta S=\dfrac{{{q}_{rev}}}{T}$ , where q is heat and T is temperature. We will take out the value of q in the isothermal reversible expansion. From 1st law of thermodynamics we have, change in internal energy is equal to heat + work done as:
$\Delta U=q+W$, in isothermal expansion the temperature is constant and the pressure is 0, so
$\Delta U=0$, therefore, q = -W
W is the work done in isothermal reversible expansion of an ideal gas that has value ${{W}_{rev}}=-2.303nRT\log \dfrac{{{V}_{2}}}{{{V}_{1}}}$, so putting this in entropy formula we have,
$\Delta S=\dfrac{2.303nRT\log \dfrac{{{V}_{2}}}{{{V}_{1}}}}{T}$
$\Delta S=2.303nR\log \dfrac{{{V}_{2}}}{{{V}_{1}}}$, where R is gas constant = 8.314$J{{K}^{-1}}mo{{l}^{-1}}$, n = 2
So,$\Delta S=2.303\times 2\times 8.314\,\left( \log \dfrac{100}{10} \right)$
$\Delta S=$38.3$J{{K}^{-1}}mo{{l}^{-1}}$

So, option B is correct.

Note:
The work done is equal to pressure and change in volume as $W=P\Delta V$, in isothermal reversible expansion the pressure is 0 because external pressure is equal to internal pressure. From PV = nRT, we will put $P=\dfrac{nRT}{V}$ in a work done formula. Then on integrating the work done in isothermal reversible expansion the value is calculated as ${{W}_{rev}}=-2.303nRT\log \dfrac{{{V}_{2}}}{{{V}_{1}}}$.