A vessel contains 1 mole of `O_2` gas (relative molar mass 32) at a temperature T. The pressure of the gas is P. An identical vessel containing one mole of He gas (relative molar mass 4) at temperature 2T has a pressure of

Options

A. `P/8`
B. P
C. 2P
D. 8P

Explanation

To solve the problem step by step, we can use the Ideal Gas Law and the relationship between pressure, volume, and temperature.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- For the \( O_2 \) gas:
- Moles (\( n_1 \)) = 1 mole
- Molar mass = 32 g/mol
- Temperature (\( T_1 \)) = \( T \)
- Pressure (\( P_1 \)) = \( P \)
- For the \( He \) gas:
- Moles (\( n_2 \)) = 1 mole
- Molar mass = 4 g/mol
- Temperature (\( T_2 \)) = \( 2T \)
- Pressure (\( P_2 \)) = ?

2. **Use the Ideal Gas Law:**
The Ideal Gas Law states that:
\[ PV = nRT \]
where \( P \) = pressure, \( V \) = volume, \( n \) = number of moles, \( R \) = universal gas constant, and \( T \) = temperature.

3. **Set Up the Equation for Both Gases:**
For \( O_2 \):
\[ P_1V = n_1RT_1 \implies PV = 1 \cdot R \cdot T \]
For \( He \):
\[ P_2V = n_2RT_2 \implies P_2V = 1 \cdot R \cdot (2T) \]

4. **Relate the Two Equations:**
Since both vessels are identical, the volume \( V \) is the same for both gases. We can divide the two equations:
\[ \frac{P_1V}{T_1} = \frac{P_2V}{T_2} \]
This simplifies to:
\[ \frac{P}{T} = \frac{P_2}{2T} \]

5. **Solve for \( P_2 \):**
Rearranging the equation gives:
\[ P_2 = P \cdot \frac{2T}{T} = 2P \]

6. **Conclusion:**
Therefore, the pressure of the helium gas \( P_2 \) is \( 2P \).

### Final Answer:
The pressure of the helium gas is \( 2P \).