A mixture of one mole of monoatomic gas and one mole of a diatomic gas (rigid) are kept at room temperature (27° C). The ratio of specific heat of gases at constant volume respectively is:

Solution

Physics sir · 3 steps

Count the active degrees of freedom of each gas; the mixing and the 27 °C are only there to distract you.

1. Equipartition. Each degree of freedom contributes 1/2RT per mole to the internal energy. So a gas with f degrees of freedom has C_V = f/2R per mole.
2. Count each gas. A monoatomic molecule has only 3 translational degrees of freedom, so C_V = 3/2R. A rigid diatomic molecule has 3 translational and 2 rotational, so f = 5 and C_V = 5/2R. Vibration is frozen because the molecule is rigid.

   (C_V)ₘₒₙₒ/(C_V)_(dia) = (3/2R)/(5/2R) = 3/5

3. So. Taken in the order monoatomic then diatomic, the ratio is 3/5, option (b). Being in a mixture does not change each gas's own C_V, and neither does the temperature.

The idea to keep. C_V = f/2R per mole: f = 3 for monoatomic, 5 for rigid diatomic. Read the word "respectively" carefully so you keep the ratio in the order asked.

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