Question

Internal energy of $n_1$ moles of $H_2$ at temperature $T$ is equal to the internal energy of $n_2$ moles of He at temperature $2T$. Then the ratio $\frac{n_1}{n_2}$ is

• A. 3/5
• B. 2/3
• C. 6/5
• D. 3/7

Answer 1

The correct option is C 6/5

Internal energy per degree of freedom: $E=\frac{1}{2}kT$ where k is the Boltzman's constant and T is the temp in Kelvin.
For n_1 moles of H_2 $E_1=n_1\ast f_1\ast \frac{1}{2}k{T}_1$
For n_2 moles of He $E_2=n_2\ast f_2\ast \frac{1}{2}k{T}_2$
Given $E_1=E_2$ and $T_2=2T_1$
$f_1$=5 since $H_2$ is a diatomic molecule and has 5 degrees of freedom where as He is mono atomic and has 3 degrees of freedom.
$\therefore n_1\ast f_1\ast \frac{1}{2}k{T}_1=n_2\ast f_2\ast \frac{1}{2}k{T}_2$
$\therefore \frac{n_1}{n_2}=\frac{f_2{T}_2}{f_1{T}_1}=\frac{3\ast 2T_1}{5T_1}=\frac{6}{5}$