In the reported figure, there is a cyclic process ABCDA on a sample of 1 mol of a diatomic gas. The temperature of the gas during the process A → B and C → D are T₁ and T₂ (T₁ > T₂) respectively.
Choose the correct option out of the following for work done if processes BC and DA are adiabatic.

Solution

Physics sir · 4 steps

On an adiabat the work depends only on the temperature drop, so A to D does the same work as B to C.

1. Read the cycle. AB is an isotherm at T₁ and CD is an isotherm at T₂. BC and DA are adiabats, so this is a Carnot cycle. Here W_(XY) means the work done by the gas as it goes from X to Y.
2. Adiabatic work. With Q = 0, the first law gives W = -Δ U = nCᵥ(Tᵢₙᵢₜᵢₐₗ - T_(final)). For 1 mol of a diatomic gas, Cᵥ = 5/2R.

   W_(BC) = 5/2R (T₁ - T₂) = W_(AD)

3. Why they match. B to C goes from T₁ down to T₂. A to D also goes from T₁ down to T₂. The same temperature drop gives the same work, even though the two adiabats are different curves.

   Option	Work values	Verdict
   (a)	W_(AB) = RT₁ln r, W_(DC) = RT₂ln r, with the same ratio r = V_B/V_A = V_C/V_D	unequal, since T₁ > T₂
   (b)	W_(AD) = W_(BC) = 5/2R(T₁ - T₂)	true
   (c)	W_(BC) + W_(DA) = 5/2R(T₁ - T₂) + 5/2R(T₂ - T₁) = 0	false, the sum is zero
   (d)	W_(AB) > 0 (expansion), W_(CD) < 0 (compression)	false, W_(AB) > W_(CD)

4. So. Both adiabatic legs run between the same two temperatures in the same direction, so W_(AD) = W_(BC). The answer is (b).

The idea to keep. Adiabatic work is nCᵥΔ T and nothing else. Any two adiabats between the same two isotherms do equal work, opposite in sign when you run one of them backwards.

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