Question

The value of \( C_v \) for 1 mol of \( \text{N}_2 \) gas predicted from the principle of equipartition of energy, ignoring vibrational contribution, is .......... J K\(^{-1}\) mol\(^{-1}\) (rounded up to two decimal places).

Hint:

For diatomic gases, the value of \( C_v \) at constant volume is given by \( C_v = \frac{5}{2} R \), where \( R \) is the gas constant.

Answer 1

Correct Answer: 20 - 21

Step 1: Principle of equipartition of energy.
According to the principle of equipartition of energy, each degree of freedom contributes \( \frac{1}{2} R \) to the heat capacity at constant volume (\( C_v \)), where \( R \) is the gas constant. For a diatomic molecule like \( \text{N}_2 \), there are 5 degrees of freedom: 3 translational and 2 rotational. Thus, the contribution to \( C_v \) is: \[ C_v = \frac{5}{2} R \] Substituting the value of \( R = 8.3 \, \text{J K}^{-1} \text{mol}^{-1} \), we get: \[ C_v = \frac{5}{2} \times 8.3 = 20.75 \, \text{J K}^{-1} \text{mol}^{-1} \]

Step 2: Conclusion.
The value of \( C_v \) for 1 mol of \( \text{N}_2 \) gas is 20.79 J K\(^{-1}\) mol\(^{-1}\) (rounded to two decimal places).