GATE Chemical 2025 | Question: 38

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. The spring constant is $10 \mathrm{kN} / \mathrm{m}$. At the initial equilibrium state (shown in the figure), the spring is unstretched. The gas is expanded reversibly by adding $362.5 \:J$ of heat. At the final equilibrium state, the piston presses against the stoppers. Neglecting the heat loss to the surroundings, the final equilibrium temperature of the gas is $\_\_\_\_\_ \mathrm{K}$ (rounded off to the nearest integer).

Answer 1

• 1. Moles of gas (n):

  • P1​=1bar=1×10⁵Pa (Correct conversion)
  • V1​=2L=2×10−3m3 (Correct conversion)
  • T1​=300K
  • $R = 8.314 \, \text{J/(mol·K)}$
  • $n = \frac{P_1 V_1}{R T_1} = \frac{(1 \times 10^5 \, \text{Pa})(2 \times 10^{-3} \, \text{m}^3)}{(8.314 \, \text{J/(mol·K)})(300 \, \text{K})} \approx 0.0802 \, \text{mol}$.

• 2. Volume change (ΔV):

  • A=100cm2=100×(10−2m)2=100×10−4m2=1.0×10−2m2 (Correct conversion)
  • x=5cm=0.05m (Correct conversion)
  • ΔV=A⋅x=(1.0×10−2m2)(0.05m)=5.0×10−4m3.

• 3. Work against atmosphere (Watm​):

  • Pamb​=1bar=1×105Pa (This assumes the ambient pressure is the same as the initial gas pressure, which is a reasonable assumption for equilibrium with an unstretched spring).
  • Watm​=Pamb​ΔV=(1×105Pa)(5.0×10−4m3)=50J.

• 4. Spring work (Wspring​):

  • k=10kN/m=10000N/m
  • Wspring​=21​kx2=21​(10000N/m)(0.05m)2=21​(10000)(0.0025)=12.5J.

• 5. Total work (Wtotal​):

  • Wtotal​=Watm​+Wspring​=50J+12.5J=62.5J.

• 6. Change in internal energy (ΔU):

  • ΔU=Q−Wtotal​=362.5J−62.5J=300J.

• 7. Temperature rise (ΔT):

  • For a monatomic ideal gas, ΔU=nCv​ΔT=n(23​R)ΔT.
  • $\Delta T = \frac{\Delta U}{n (\frac{3}{2}R)} = \frac{300 \, \text{J}}{(0.0802 \, \text{mol})(\frac{3}{2} \times 8.314 \, \text{J/(mol·K)})} = \frac{300}{0.0802 \times 12.471} = \frac{300}{1.000} \approx 299.99 \, \text{K} \approx 300 \, \text{K}$.

• 8. Final temperature (T2​):

  • T2​=T1​+ΔT=300K+300K=600K.