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(C) The internal energy $U$ of an ideal gas is given by $U = \frac{f}{2} nRT$,where $f$ is the degrees of freedom,$n$ is the number of moles,$R$ is th

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$1$

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(C) The internal energy $U$ of an ideal gas is given by $U = \frac{f}{2} nRT$,where $f$ is the degrees of freedom,$n$ is the number of moles,$R$ is the universal gas constant,and $T$ is the absolute temperature.For $He$ (a monoatomic gas),the degrees of freedom $f_1 = 3$. Given $n_1$ moles at $10T$,the internal energy is $U_1 = \frac{3}{2} n_1 R (10T) = 15 n_1 RT$.For $H_2$ (a diatomic gas),the degrees of freedom $f_2 = 5$. Given $n_2$ moles at $6T$,the internal energy is $U_2 = \frac{5}{2} n_2 R (6T) = 15 n_2 RT$.Equating the internal energies,$U_1 = U_2$,we get $15 n_1 RT = 15 n_2 RT$.Therefore,$n_1 = n_2$,which implies $\frac{n_1}{n_2} = 1$.

If the internal energy of $n_1$ moles of $He$ at temperature $10T$ is equal to the internal energy of $n_2$ moles of hydrogen $(H_2)$ at temperature $6T$,find the ratio $\frac{n_1}{n_2}$.

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