Internal energy of `n_1` moles of hydrogen at temperature 150 K is equal to the in ternal energy of `n_2` moles of helium at temperature 300 K The ratio of `n_1//n_2` is

Options

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

Answer

To solve the problem, we need to find the ratio of the number of moles of hydrogen (\(n_1\)) to the number of moles of helium (\(n_2\)) given that their internal energies are equal at specified temperatures.

### Step-by-Step Solution:

1. **Understand Internal Energy Formula**:
The internal energy \(U\) of an ideal gas can be expressed as:
\[ U = \frac{F}{2} nRT \]
where \(F\) is the degrees of freedom, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is the temperature.

2. **Identify the Gases**:
- For hydrogen (\(H_2\)), which is a diatomic gas, the degrees of freedom \(F\) is 5.
- For helium (\(He\), a monoatomic gas), the degrees of freedom \(F\) is 3.

3. **Write the Internal Energy for Each Gas**:
- For hydrogen at 150 K:
\[ U_{H_2} = \frac{5}{2} n_1 R (150) \]
- For helium at 300 K:
\[ U_{He} = \frac{3}{2} n_2 R (300) \]

4. **Set the Internal Energies Equal**:
According to the problem, the internal energies are equal:
\[ \frac{5}{2} n_1 R (150) = \frac{3}{2} n_2 R (300) \]

5. **Cancel Common Terms**:
We can cancel \(R\) and \(\frac{1}{2}\) from both sides:
\[ 5 n_1 (150) = 3 n_2 (300) \]

6. **Simplify the Equation**:
This simplifies to:
\[ 5 n_1 \cdot 150 = 3 n_2 \cdot 300 \]
Dividing both sides by 150:
\[ 5 n_1 = 3 n_2 \cdot 2 \]
Which simplifies to:
\[ 5 n_1 = 6 n_2 \]

7. **Find the Ratio \( \frac{n_1}{n_2} \)**:
Rearranging gives:
\[ \frac{n_1}{n_2} = \frac{6}{5} \]

### Final Answer:
The ratio of \( \frac{n_1}{n_2} \) is \( \frac{6}{5} \).