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: Understanding the Question:
The question asks for the mathematical ratio of the number of moles of two different gases ($n_1 / n_2$) when their internal thermodynamic energies are perfectly balanced under different temperature states.
We have $n_1$ moles of hydrogen ($\text{H}_2$) at temperature $T$ and $n_2$ moles of helium (He) at temperature $2T$.

Step 2: Key Formula or Approach:
The internal energy $U$ of an ideal gas system can be calculated from its degrees of freedom $f$, number of moles $n$, and absolute temperature $T$ using the classical equipartition theorem:
$$U = \frac{f}{2}nRT$$

Step 3: Detailed Explanation:
Let's express the individual internal energy equations for both gas samples:
For Hydrogen (diatomic gas, $f_1 = 5$):
$$U_1 = \frac{5}{2}n_1RT$$ For Helium (monoatomic gas, $f_2 = 3$, at temperature $2T$):
$$U_2 = \frac{3}{2}n_2R(2T) = 3n_2RT$$ The problem states that these two internal energies are equal ($U_1 = U_2$):
$$\frac{5}{2}n_1RT = 3n_2RT$$ Cancel out the shared universal constants $R$ and $T$ from both sides:
$$\frac{5}{2}n_1 = 3n_2$$ Rearrange the algebraic terms to isolate the ratio $\frac{n_1}{n_2}$:
$$\frac{n_1}{n_2} = 3 \times \frac{2}{5} = \frac{6}{5}$$ Therefore, the ratio $n_1 : n_2$ is 6 : 5.

Step 4: Final Answer:
The ratio of the number of moles is 6 : 5, which corresponds exactly to option (B).