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GATE 2025 Chemical Engineering, Q33 · Thermodynamics

The correct answer is 18.5 to 18.9.

Question

Ideal nonreacting gases A and B are contained inside a perfectly insulated chamber, separated by a thin partition, as shown in the figure. TA=TB=273T_{A} = T_{B} = 273 K, PA=PB=1P_{A} = P_{B} = 1 atm, VB=22.4V_{B} = 22.4 L, and VA=3VB.V_{A} = 3V_{B}. The partition is removed, and the two gases mix until final equilibrium is reached. The change in total entropy for the process is ____ J/K (rounded off to 1 decimal place). Given: universal gas constant R = 8.314 J/(mol·K).

Figure for Q33 (Thermodynamics, Entropy of Mixing): Ideal nonreacting gases A and B are contained inside a perfectly insulated chamber, separated by a t…

Official answer: 18.5 to 18.9

Why

GATE 2025 official key: 18.5 to 18.9 J/K. VB=22.4V_{B} = 22.4 L at 273 K, 1 atm ⇒nB=1\Rightarrow n_{B} = 1 mol; VA=3VB⇒nA=3V_{A} = 3V_{B} \Rightarrow n_{A} = 3 mol. Total n = 4 mol, yA=0.75,yB=0.25.ΔSmix=−R(nAlnyA+nBlnyB)=−8.314×(3ln0.75+1×ln0.25)≈18.7y_{A} = 0.75, y_{B} = 0.25. \Delta S_{mix} = −R(n_{Aln} y_{A} + n_{Bln} y_{B}) = −8.314 \times (3ln0.75 + 1 \times ln0.25) \approx 18.7 J/K.

Explanation cross-checked for consistency. Final answer matches the official IIT answer key.

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