One mole of a monoatomic gas is mixed with three moles of a diatomic gas. The molecular specific heat of mixture at constant volume is α²/4 R J / mol K; then the value of α will be _________. (Assume that the given diatomic gas has no vibrational mode).

Solution

Physics sir · 4 steps

Mix the gases by adding up their total heat capacity, then share it over all 4 moles.

1. Cv of each gas. A monoatomic gas has 3 degrees of freedom, so C_V = 3/2R. A diatomic gas with no vibration has 5 (3 translational, 2 rotational), so C_V = 5/2R.
2. Weight by moles. Heat capacities add, so weight each C_V by its number of moles and divide by the total of 1 + 3 = 4 moles.

   C_(V,mix) = (1 × 3/2R + 3 × 5/2R)/(1 + 3) = (3/2R + 15/2R)/4 = 9/4R

3. Match the form. Compare with α²/4R: this gives α² = 9.
4. So. α = 3, so the answer is 3.

The common slip: 9. Stopping at α² = 9 and reporting it gives 9. The question asks for α, so take the square root to get 3.

The idea to keep. For a gas mixture, C_V is the mole-weighted average: (n₁C_(V1) + n₂C_(V2))/(n₁ + n₂). Never take a plain average when the mole numbers differ.

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