A diatomic gas (γ=1.4) does 200 J of work when it is expanded isobarically. The heat given to the gas in the process is :

Solution

Physics sir · 4 steps

At constant pressure, heat and work share the same nΔ T, so their ratio is just Cₚ/R.

1. Work at constant pressure. In an isobaric process W = PΔ V = nRΔ T. So nRΔ T = 200 J.
2. Heat at constant pressure. The heat supplied is Q = nCₚΔ T. With γ = 1.4, Cₚ = (γ R)/(γ - 1) = 1.4/0.4R = 7/2R.

   Q/W = (nCₚΔ T)/(nRΔ T) = γ/(γ - 1) = 7/2

3. First-law check. Δ U = nCᵥΔ T = 5/2 × 200 = 500 J. Then Q = Δ U + W = 500 + 200 = 700 J, the same result.
4. So. Q = 7/2 × 200 = 700 J, which is option (c).

The idea to keep. For an isobaric process remember Q : Δ U : W = Cₚ : Cᵥ : R. For a diatomic gas that is 7 : 5 : 2, so one known quantity gives you the other two.

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