An ideal gas, C̅ᵥ=5/2 R, is expanded adiabatically against a constant pressure of 1 atm untill it doubles in volume. If the initial temperature and pressure is 298 K and 5 atm, respectively then the final temperature is _________ K (nearest integer).
[c̅ᵥ is the molar heat capacity at constant volume]

Solution

Chemistry ma’am · 4 steps

Fixed outside pressure makes it irreversible: use Δ U = -PₑₓₜΔ V.

1. First law. Adiabatic means q = 0, so Δ U = w. The gas pushes against a fixed 1 atm, so w = -Pₑₓₜ(V₂ - V₁). For an ideal gas, Δ U = nC̅ᵥ(T₂ - T₁).
2. Volume change. The volume doubles, so V₂ - V₁ = V₁. From the ideal gas law, V₁ = nRT₁/P₁ with P₁ = 5 atm.

   n·5/2R (T₂ - 298) = -(1)·(nR × 298)/5

3. Solve for T2. n and R cancel. That leaves 5/2(T₂ - 298) = -59.6, so T₂ - 298 = -23.84 and T₂ = 274.16 K.
4. So. The final temperature is about 274.2 K, which rounds to 274.

The common slip: 226. Using the reversible relation TV^(γ-1) = constant with γ = 1.4 gives 298 × 2^(-0.4) ≈ 226. That relation does not apply when the gas expands against a fixed 1 atm.

The idea to keep. Against a constant external pressure, an adiabatic expansion is irreversible. Use nC̅ᵥΔ T = -PₑₓₜΔ V. Keep TV^(γ-1) for reversible adiabatic changes only.

Answer from the NTA final answer key. Solution by PrepWiser.