A diatomic gas (γ=1.4) does 100 J of work in an isobaric expansion. The heat given to the gas is :

Solution

Physics sir · 4 steps

At constant pressure the work is nRΔ T, so write the heat in terms of that same nRΔ T.

1. Work at constant pressure. For an ideal gas at constant pressure, W = PΔ V = nRΔ T. So here nRΔ T = 100 J.
2. Heat at constant pressure. The heat supplied is Q = nCₚΔ T. Using Cₚ = (γ R)/(γ - 1), you can write it directly in terms of the work.

   Q = γ/(γ - 1) nRΔ T = 1.4/0.4 × 100 J

3. Check with first law. The internal energy rises by Δ U = (nRΔ T)/(γ - 1) = 250 J. Then Q = Δ U + W = 250 + 100 = 350 J, which matches.
4. So. Q = 3.5 × 100 = 350 J, option (c).

The idea to keep. In an isobaric process the heat, internal energy change and work are in the ratio Cₚ : Cᵥ : R. For a diatomic gas that is 7 : 5 : 2.

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