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

To solve this problem, we need to understand the concept of the degrees of freedom in a diatomic gas molecule. The degrees of freedom determine the energy distribution among translation, rotation, and vibration in gas molecules. For a diatomic molecule at room temperature or NTP (Normal Temperature and Pressure), the vibrational modes are not usually active due to lower energy levels. Hence, they exhibit five degrees of freedom: three translational and two rotational.

The formula for average kinetic energy per molecule due to degrees of freedom is given by:
\( KE = \frac{f}{2}kT \)
where \( f \) is the degrees of freedom, \( k \) is the Boltzmann constant, and \( T \) is the temperature.

1. At NTP, for a diatomic gas, \( f = 5 \). Therefore, \( KE_{NTP} = \frac{5}{2}kT \).
2. At a high temperature, vibrational modes are excited, adding 2 more degrees of freedom, making \( f = 7 \). Hence, \( KE_{high} = \frac{7}{2}kT \).
3. The ratio of kinetic energies is:
\[ \text{Ratio} = \frac{KE_{high}}{KE_{NTP}} = \frac{\frac{7}{2}kT}{\frac{5}{2}kT} = \frac{7}{5} \]

This shows that the ratio of kinetic energy of a diatomic gas molecule at high temperature to that at NTP is \( \frac{7}{5} \).

Answer 2

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: \[ \boxed{\frac{7}{5}}. \]