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

Options

A. p/8
B. p
C. 2p
D. 8p

Explanation

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

\[ PV = nRT \]

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 Variables for Oxygen Gas**:
- For the oxygen gas (O₂), we have:
- Number of moles, \( n = 1 \) mole
- Temperature, \( T = T \)
- Pressure, \( P = P \)
- Molar mass = 32 (not needed for the calculation)
- Using the ideal gas equation:
\[ PV = nRT \implies P \cdot V = 1 \cdot R \cdot T \]
- Therefore, we can express the pressure of the oxygen gas as:
\[ P = \frac{RT}{V} \]

2. **Identify the Variables for Helium Gas**:
- For the helium gas (He), we have:
- Number of moles, \( n = 1 \) mole
- Temperature, \( T = 2T \)
- Pressure, \( P_{He} \) (unknown)
- Using the ideal gas equation:
\[ P_{He} \cdot V = nRT \implies P_{He} \cdot V = 1 \cdot R \cdot (2T) \]
- Therefore, we can express the pressure of the helium gas as:
\[ P_{He} = \frac{R(2T)}{V} \]

3. **Relate the Pressures**:
- Now we can relate the pressures of the two gases:
\[ \frac{P}{P_{He}} = \frac{\frac{RT}{V}}{\frac{R(2T)}{V}} \]
- The \( R \) and \( V \) cancel out:
\[ \frac{P}{P_{He}} = \frac{T}{2T} = \frac{1}{2} \]
- Rearranging gives us:
\[ P_{He} = 2P \]

4. **Conclusion**:
- The pressure of the helium gas is:
\[ P_{He} = 2P \]

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

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