Question

Internal energy of $n_1$ moles of hydrogen at temperature $T$ is equal to internal energy of $n_2$ moles of helium at temperature $2T$, then the ratio $n_1 : n_2$ is [Degree of freedom of $\text{He} = 3$, Degree of freedom of $\text{H}_2 = 5$]

• A. 5 : 3
• B. 6 : 5
• C. 2 : 3
• D. 3 : 5

Hint:

To avoid carrying fractional factors through algebraic lines, write out the product of degrees of freedom, moles, and temperature directly for each side: $5 \times n_1 \times 1 = 3 \times n_2 \times 2$. This simplifies instantly to $5n_1 = 6n_2$, yielding the ratio $\frac{6}{5}$ in a single mental step.

Answer 1

The Correct Option is B

Step 1: Internal energy formula.
For an ideal gas with $f$ degrees of freedom, the internal energy of $n$ moles at temperature $T$ is \[ U = \frac{f}{2}nRT. \]
Step 2: Energy of the hydrogen.
Hydrogen is diatomic, $f_1 = 5$, at temperature $T$: \[ U_1 = \frac{5}{2}n_1RT. \]
Step 3: Energy of the helium.
Helium is monatomic, $f_2 = 3$, at temperature $2T$: \[ U_2 = \frac{3}{2}n_2R(2T) = 3n_2RT. \]
Step 4: Equate the energies.
\[ \frac{5}{2}n_1RT = 3n_2RT. \]
Step 5: Cancel $RT$.
\[ \frac{5}{2}n_1 = 3n_2. \]
Step 6: Form the ratio.
\[ \frac{n_1}{n_2} = \frac{3\times2}{5} = \frac{6}{5}, \] so $n_1 : n_2 = 6 : 5$, option (B). \[ \boxed{n_1 : n_2 = 6 : 5} \]