JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Heat & Thermodynamics

5.6 liter of helium gas at STP is adiabatically compressed to 0.7 liter. Taking the initial temperature to be

T_{1}

, the work done in the process is

Answer: (A)

Solution

For an adiabatic process,

T V^{γ - 1} = constant

. For helium (monatomic),

γ = 5/3

, so

γ - 1 = 2/3

T_{1} (5.6)^{2/3} = T_{2} (0.7)^{2/3}

T_{2} = T_{1} (\frac{5.6}{0.7})^{2/3} = T_{1} (8)^{2/3} = T_{1} \cdot 4 = 4 T_{1}

At STP, 5.6 liters = 0.25 mol (since 22.4 L = 1 mol). Work done on the gas in an adiabatic process:

W = \frac{n R Δ T}{γ - 1} = \frac{0.25 R ( 4 T _{1} - T _{1} )}{2/3} = \frac{0.25 R \cdot 3 T _{1} \cdot 3}{2} = \frac{9 R T _{1}}{8}

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Single correct option