Chemegate

GATE 2025 Chemical Engineering, Q48 · Thermodynamics

The correct answer is 595 to 605.

Question

An ideal monoatomic gas is contained inside a cylinder-piston assembly connected to a Hookean spring, as shown in the figure. The piston is frictionless and massless, with cross-sectional area 100cm2100 cm² and a stopper 5 cm away. Ambient pressure is 1 bar. Initial gas conditions: Vinitial=2V_{initial} = 2 L, Tinitial=300T_{initial} = 300 K. The spring constant is 10 kN/m and is initially unstretched. The gas is expanded reversibly by adding 362.5 J of heat, until the piston presses against the stoppers at the final equilibrium state. Neglecting heat loss to the surroundings, the final equilibrium temperature of the gas is ____ K (rounded off to the nearest integer). Given: for a monoatomic ideal gas, Cv=C_{v} = (3/2)R, where R = 8.314 J/(mol·K).

Figure for Q48 (Thermodynamics, Ideal Gas Processes): An ideal monoatomic gas is contained inside a cylinder-piston assembly connected to a Hookean spring…

Official answer: 595 to 605

Why

GATE 2025 official key: 595 to 605 K. ΔV=Area×stroke=0.01m2×0.05\Delta V = Area \times stroke = 0.01 m² \times 0.05 m =0.0005m3.= 0.0005 m³. Force balance gives P(x) =Pamb+= P_{amb} + kx/A, so work done by gas W =PambΔV+½kx2=100000×0.0005+0.5×10000×0.052=50+12.5=62.5= P_{amb}\cdot \Delta V + ½k\cdot x² = 100000 \times 0.0005 + 0.5 \times 10000 \times 0.05² = 50 + 12.5 = 62.5 J. Energy balance: ΔU=\Delta U = Q W = 362.5 62.5 = 300 J. Since n =P1V1/(RT1)= P₁V₁/(RT₁) and nR =P1V1/T1=100000×0.002/300=0.667= P₁V₁/T₁ = 100000 \times 0.002/300 = 0.667 J/K, ΔU=n(3/2)RΔT=(3/2)×0.667×ΔT=300ΔT=300\Delta U = n\cdot (3/2)R\cdot \Delta T = (3/2) \times 0.667 \times \Delta T = 300 \Rightarrow \Delta T = 300 K, so Tfinal=300+300=600T_{final} = 300+300 = 600 K.

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