Question

Internal energy of $n_1$ mol of hydrogen of temperature $T$ is equal to the internal energy of $n_2$ mol of helium at temperature $2T$ . The value of $\frac{10n_1}{n_2}$ is

Answer 1

Answer: 12

Internal energy of $n$ moles of an ideal gas at temperature $T$ is given by,
$\text{U}=\frac{f}{2}\text{n}\text{RT}$ , where $f=$ degrees of freedom
$U_1=U_2$
$f_1{n}_1{T}_1=f_2{n}_2{T}_2$
$\therefore \frac{n_1}{n_2}=\frac{f_2{T}_2}{f_1{T}_1}=\frac{3\times 2}{5\times 1}=\frac{6}{5}=1.2$
Here, $f_2$ = degrees of freedom of He = 3
and, $f_1$ = degrees of freedom of H₂ = 5
Now, $\frac{10n_1}{n_2}=12$