Question

The kinetic energy of 3 moles of a diatomic gas molecules in a container at a temperature \( T \) is same as that of kinetic energy of \( n \) moles of monoatomic gas molecules in another container at the same temperature \( T \). The value of \( n \) is:

• A. 3
• B. 4
• C. 2.5
• D. 5
• E. 3.5

Hint:

The total kinetic energy in a gas is distributed among its degrees of freedom, as per the equipartition theorem.

Answer 1

The Correct Option is D

For diatomic gas, the degrees of freedom \( f = 5 \) (at room temperature), and for monoatomic gas \( f = 3 \). 
Using the equipartition theorem: \[ {Total KE for diatomic} = 3 { moles} \times \frac{5}{2} R T = \frac{15}{2} R T \] \[ {Total KE for monoatomic} = n { moles} \times \frac{3}{2} R T \] Equating the two energies: \[ \frac{15}{2} R T = \frac{3}{2} R T \times n \implies n = 5 \]