If three moles of monoatomic gas (γ=5/3) is mixed with two moles of a diatomic gas (γ=7/5), the value of adiabatic exponent γ for the mixture is

Solution

Physics sir · 4 steps

Do not average the gammas: add the heat capacities of both gases, then take Cₚ/Cᵥ for the mixture.

1. Heat capacities. A monoatomic gas has Cᵥ = 3/2R and a diatomic gas has Cᵥ = 5/2R. For any ideal gas, Cₚ = Cᵥ + R.
2. Mix the Cv. The internal energies of the two gases add, so the mixture's Cᵥ is the mole-weighted average of the two values.

   C_(v,mix) = (3 × 3/2R + 2 × 5/2R)/(3 + 2) = (4.5R + 5R)/5 = 1.9R

3. Find Cp. C_(p,mix) = C_(v,mix) + R = 2.9R.
4. So. γₘᵢₓ = 2.9R/1.9R = 29/19 ≈ 1.526, closest to 1.52, option (b).

The idea to keep. For a gas mixture, weight Cᵥ by moles, then use Cₚ = Cᵥ + R. Shortcut: (n₁ + n₂)/(γ - 1) = n₁/(γ₁ - 1) + n₂/(γ₂ - 1).

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