Equal molar amount of `H _(2),` He having molecular weight of 2 and 4 respectively are filled at same temperature in two containers of equal volumes. If `H_(2)` gas has a pressure of 4 atmospheres, then He gas will have pressure as :-

Answer

To solve the problem, we will use the Ideal Gas Law, which states that:

\[ PV = nRT \]

where:
- \( P \) = pressure of the gas,
- \( V \) = volume of the gas,
- \( n \) = number of moles of the gas,
- \( R \) = universal gas constant,
- \( T \) = temperature of the gas.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- For \( H_2 \) (Hydrogen):
- Pressure, \( P_{H_2} = 4 \) atm
- Molecular weight = 2
- For \( He \) (Helium):
- Molecular weight = 4
- Both gases are at the same temperature and are in containers of equal volume.

2. **Apply the Ideal Gas Law for \( H_2 \):**
- Using the equation \( PV = nRT \):
\[ P_{H_2} V = n_{H_2} RT \]
- Rearranging gives:
\[ P_{H_2} = \frac{n_{H_2} RT}{V} \]

3. **Substitute the Known Values for \( H_2 \):**
- We know \( P_{H_2} = 4 \) atm, so:
\[ 4 = \frac{n_{H_2} RT}{V} \]

4. **Apply the Ideal Gas Law for \( He \):**
- Similarly, for \( He \):
\[ P_{He} V = n_{He} RT \]
- Rearranging gives:
\[ P_{He} = \frac{n_{He} RT}{V} \]

5. **Since Both Gases are Equimolar:**
- Given that equal molar amounts are filled, we have:
\[ n_{H_2} = n_{He} = n \]
- Therefore, we can write:
\[ P_{He} = \frac{n RT}{V} \]

6. **Relate the Pressures:**
- Since both equations for pressure have the same form and the same \( n \), \( R \), \( T \), and \( V \):
\[ P_{He} = P_{H_2} \]
- Thus:
\[ P_{He} = 4 \text{ atm} \]

### Final Answer:
The pressure of the Helium gas \( P_{He} \) is **4 atm**.