In an adiabatic expansion, the temperature of one mole of an ideal monatomic gas (γ=5/3) decreases from 60 K to 50 K. The work done by the gas in the process is: [Take R=8.3 J mol⁻¹K⁻¹]

Solution & Explanation

Principle: in an adiabatic process no heat is exchanged (Q = 0), so by the first law the work done by the gas equals the drop in internal energy: W = −ΔU. For an ideal gas the work done by the gas in an adiabatic change is W = n·R·(T₁ − T₂)/(γ − 1), or equivalently W = n·R·ΔT/(1 − γ). Here n = 1, R = 8.3 J mol⁻¹K⁻¹, T₁ = 60 K, T₂ = 50 K, γ = 5/3. Use W = n·R·(T₁ − T₂)/(γ − 1): numerator = (1)·(8.3)·(60 − 50) = 8.3·10 = 83 J. Denominator = γ − 1 = 5/3 − 1 = 2/3. So W = 83/(2/3) = 83·(3/2) = 124.5 J. The gas expands and cools, so it does positive work: W = +124.5 J, matching option C. Check of sign: temperature falls (expansion), internal energy decreases, and that energy leaves as work done by the gas — positive, as found.