Question

For a diatomic gas molecule the value of \(C_p\) in \(J mol^{-1} K^{-1}\) is \((R = 8.2 J mol^{-1} K^{-1})\)

• A. 20.5
• B. 41
• C. 10.25
• D. 12.3
• E. 28.7

Hint:

For monatomic gas: \(C_v = 3R/2\), \(C_p = 5R/2\). For diatomic: \(C_v = 5R/2\), \(C_p = 7R/2\).

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
The question asks for the molar specific heat at constant volume (\(C_v\)) for a diatomic gas. According to the law of equipartition of energy, the total internal energy of a gas is distributed equally among its degrees of freedom. Each degree of freedom contributes \(\frac{1}{2}kT\) to the average energy per molecule, or \(\frac{1}{2}RT\) to the molar internal energy.
Step 2: Key Formula or Approach:
The molar specific heat at constant volume is defined as the rate of change of molar internal energy (\(U\)) with respect to temperature (\(T\)).
\[ C_v = \frac{dU}{dT} \] For a gas with \(f\) degrees of freedom, the molar internal energy is given by:
\[ U = \frac{f}{2}RT \] Therefore, the formula for \(C_v\) becomes:
\[ C_v = \frac{d}{dT}\left(\frac{f}{2}RT\right) = \frac{f}{2}R \] Step 3: Detailed Explanation:
A diatomic gas molecule (like O\(_2\), N\(_2\)) at normal temperatures has 5 degrees of freedom:
- 3 translational degrees of freedom (movement along x, y, and z axes).
- 2 rotational degrees of freedom (rotation about two axes perpendicular to the line connecting the atoms).
Vibrational modes are generally not active at normal temperatures.
So, for a diatomic gas, \(f = 5\).
Now, we can substitute the values of \(f\) and R into the formula for \(C_v\).
Given:
- Degrees of freedom, \(f = 5\).
- Universal gas constant, \(R = 8.2\) Jmol\(^{-1}\)K\(^{-1}\).
\[ C_v = \frac{5}{2}R \] \[ C_v = \frac{5}{2} \times 8.2 \] \[ C_v = 2.5 \times 8.2 \] \[ C_v = 20.5 \, \text{Jmol}^{-1}\text{K}^{-1} \] Step 4: Final Answer:
The value of \(C_v\) for the diatomic gas is 20.5 Jmol\(^{-1}\)K\(^{-1}\), which corresponds to option (A).