Two moles a monoatomic gas is mixed with six moles of a diatomic gas. The molar specific heat of the mixture at constant volume is :

Solution

Physics sir · 3 steps

Add up the internal energy of both gases, then share it over all 8 moles: a mole-weighted average.

1. Each gas alone. A monoatomic gas has 3 degrees of freedom, so C_V = 3/2R. A diatomic gas has 5, so C_V = 5/2R.
2. Why average. Heating the mixture by Δ T at constant volume needs n₁ C_(V1)Δ T + n₂ C_(V2)Δ T. Divide this by the total moles n₁ + n₂ to get the heat per mole of mixture.

   C_V = (n₁ C_(V1) + n₂ C_(V2))/(n₁ + n₂) = (2 × 3/2R + 6 × 5/2R)/(2 + 6) = (3R + 15R)/8

3. So. C_V = 18R/8 = 9/4R, which is option (d).

The idea to keep. For any gas mixture, the molar specific heat is the mole-weighted average of the individual values. Pair each C_V with its own number of moles.

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