Two moles of an ideal gas with C_P/C_V = 5/3 are mixed with 3 moles of another ideal gas with C_P/C_V = 4/3. The value of C_P/C_V for the mixture is :

Solution

Physics sir · 4 steps

Do not average the γ values; average C_V and C_P by moles, then divide.

1. Find each CV. For an ideal gas C_V = R/(γ - 1). With γ = 5/3, C_V = 3/2R. With γ = 4/3, C_V = 3R.
2. Find each CP. Use C_P = C_V + R. That gives C_P = 5/2R for the first gas and C_P = 4R for the second.

   γₘᵢₓ = (n₁ C_(P1) + n₂ C_(P2))/(n₁ C_(V1) + n₂ C_(V2)) = (2 × 5/2R + 3 × 4R)/(2 × 3/2R + 3 × 3R) = 17R/12R

3. Why this works. The heat needed to warm the mixture by 1 K is the sum of the heats for each gas. So the total heat capacities add, weighted by the number of moles. The ratio γ itself does not add.
4. So. γₘᵢₓ = 17/12 ≈ 1.42, which is option (d).

The idea to keep. For gas mixtures, heat capacities add mole by mole. Find the total C_P and the total C_V first, then take the ratio. Never average γ directly.

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