Question

The ratio of kinetic energy of a diatomic gas molecule at a high temperature to that of NTP is:

• A. \( \frac{3}{2} \)
• B. \( \frac{5}{3} \)
• C. \( \frac{5}{7} \)
• D. \( \frac{7}{5} \)

Hint:

For diatomic gases, the degrees of freedom vary with temperature. At NTP, \( f = 5 \) (translational + rotational), while at high temperatures, \( f = 7 \) (including vibrational modes).

Answer 1

The Correct Option is D

Step 1: Understanding Kinetic Energy of a Gas Molecule - The average kinetic energy of a gas molecule is given by: \[ KE = \frac{f}{2} k_B T, \] where: - \( f \) is the degrees of freedom of the gas. - \( k_B \) is Boltzmann's constant. - \( T \) is the temperature. - A diatomic gas has different degrees of freedom at different temperatures: - At Normal Temperature and Pressure (NTP), only translational and rotational motion contribute, so: \[ f_{\text{NTP}} = 5. \] - At higher temperatures, vibrational modes get activated, so: \[ f_{\text{high}} = 7. \]
Step 2: Finding the Ratio of Kinetic Energies - The ratio of kinetic energies is: \[ \frac{KE_{\text{high}}}{KE_{\text{NTP}}} = \frac{\frac{7}{2} k_B T}{\frac{5}{2} k_B T}. \] - Cancelling common terms: \[ \frac{KE_{\text{high}}}{KE_{\text{NTP}}} = \frac{7}{5}. \] Thus, the correct answer is: \[ \boxed{\frac{7}{5}}. \]