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 find the ratio of `n_1/n_2` where the internal energy of `n_1` moles of hydrogen at temperature 150 K is equal to the internal energy of `n_2` moles of helium at temperature 300 K, we can follow these steps:

### Step 1: Write the formula for internal energy
The internal energy (U) of an ideal gas can be expressed as:
\[ U = \frac{F}{2} nRT \]
where:
- \( F \) = degrees of freedom,
- \( n \) = number of moles,
- \( R \) = universal gas constant,
- \( T \) = temperature in Kelvin.

### Step 2: Identify the degrees of freedom
For hydrogen (H₂), which is a diatomic molecule:
- Degrees of freedom \( F_{H_2} = 5 \) (3 translational + 2 rotational).

For helium (He), which is a monoatomic molecule:
- Degrees of freedom \( F_{He} = 3 \) (3 translational).

### Step 3: Write the internal energy equations
For hydrogen:
\[ U_{H_2} = \frac{5}{2} n_1 R (150) \]

For helium:
\[ U_{He} = \frac{3}{2} n_2 R (300) \]

### Step 4: Set the internal energies equal
Since the internal energies are equal:
\[ \frac{5}{2} n_1 R (150) = \frac{3}{2} n_2 R (300) \]

### Step 5: Simplify the equation
We can cancel \( R \) from both sides and simplify:
\[ \frac{5}{2} n_1 (150) = \frac{3}{2} n_2 (300) \]

Dividing both sides by \( \frac{1}{2} \):
\[ 5 n_1 (150) = 3 n_2 (300) \]

### Step 6: Further simplify
This simplifies to:
\[ 5 n_1 \cdot 150 = 3 n_2 \cdot 300 \]
\[ 750 n_1 = 900 n_2 \]

### Step 7: Solve for the ratio \( n_1/n_2 \)
Rearranging gives:
\[ \frac{n_1}{n_2} = \frac{900}{750} \]
\[ \frac{n_1}{n_2} = \frac{6}{5} \]

### Conclusion
Thus, the ratio of \( n_1/n_2 \) is:
\[ \frac{n_1}{n_2} = \frac{6}{5} \]

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