Question

The 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 $2T$ . What will be the value of the ratio $\frac{n_1}{n_2}$ ?

Answer 1

Answer: 1.2

The internal energy of $n$ moles of an ideal gas at temperature T is given by
$\text{U}=\frac{f}{2}\text{nRT}$ ( $f=$ 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