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 $\frac{10 n_{1}}{n_{2}}$ is

Answer 1

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, $\frac{10 n_{1}}{n_{2}}=12$