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

Explanation

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 parameters for the first vessel (O₂ gas)**:
- Number of moles, \( n_1 = 1 \) mole,
- Temperature, \( T_1 = T \),
- Pressure, \( P_1 = P \),
- Volume, \( V_1 = V \) (assumed identical for both vessels).

Using the Ideal Gas Law for the first vessel:
\[ P \cdot V = n_1 \cdot R \cdot T_1 \]
Substituting the values:
\[ P \cdot V = 1 \cdot R \cdot T \]
This can be rearranged to:
\[ P = \frac{RT}{V} \tag{1} \]

2. **Identify the parameters for the second vessel (He gas)**:
- Number of moles, \( n_2 = 1 \) mole,
- Temperature, \( T_2 = 2T \),
- Pressure, \( P_2 = xP \),
- Volume, \( V_2 = V \) (identical volume).

Using the Ideal Gas Law for the second vessel:
\[ P_2 \cdot V = n_2 \cdot R \cdot T_2 \]
Substituting the values:
\[ xP \cdot V = 1 \cdot R \cdot (2T) \]
Rearranging gives:
\[ xP \cdot V = 2RT \tag{2} \]

3. **Relate equations (1) and (2)**:
From equation (1), we know:
\[ RT = PV \]
Substituting this into equation (2):
\[ xP \cdot V = 2(PV) \]
Simplifying gives:
\[ xP \cdot V = 2PV \]
Dividing both sides by \( PV \) (assuming \( P \) and \( V \) are not zero):
\[ x = 2 \]

### Final Answer:
The value of \( x \) is \( 2 \).

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