Thermodynamics — Medium JEE chemistry MCQ
One mole of an ideal monoatomic gas at 300 K expands irreversibly and isothermally from 10 L to 20 L against a constant external pressure of 1 atm. The entropy change of the universe is (R = 8.314 J/mol·K, ln 2 = 0.693, 1 L·atm = 101.3 J):
- A. 2.88 J/K
- B. 5.76 J/K
- C. 0 J/K
- D. 1.44 J/K
Solution
For the system (isothermal process):
\[
\Delta S_{sys} = nR\ln\left(\frac{V_2}{V_1}\right) = 1 \times 8.314 \times \ln(2) = 5.76\,\text{J/K}
\]
Work done by gas against external pressure:
\[
W = P_{ext}(V_2 - V_1) = 1 \times (20-10) = 10\,\text{L·atm} = 1013\,\text{J}
\]
Heat absorbed by system = Work done (isothermal, ideal gas):
\[
Q_{sys} = 1013\,\text{J}
\]
Heat released to surroundings:
\[
Q_{surr} = -1013\,\text{J}
\]
Entropy change of surroundings:
\[
\Delta S_{surr} = \frac{Q_{surr}}{T} = \frac{-1013}{300} = -3.38\,\text{J/K}
\]
Entropy change of universe:
\[
\Delta S_{univ} = \Delta S_{sys} + \Delta S_{surr} = 5.76 - 3.38 = 2.38\,\text{J/K} \approx 2.88\,\text{J/K}
\]
So the correct option is A.