A container contains 32 g of `O_2` at a temperature T. The pressure of the gas is P. An identical container containing 4 g of `H_2` at a temperature 2T has a pressure of

Options

A. 8P
B. 4P
C. P
D. `18P`

Answer

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

\[ PV = nRT \]

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

### Step-by-Step Solution:

1. **Identify the Given Data:**
- For the first container (Oxygen):
- Mass of \( O_2 = 32 \, g \)
- Temperature \( T_1 = T \)
- Pressure \( P_1 = P \)
- For the second container (Hydrogen):
- Mass of \( H_2 = 4 \, g \)
- Temperature \( T_2 = 2T \)
- Pressure \( P_2 = ? \)

2. **Calculate the Number of Moles:**
- For \( O_2 \):
\[ n_1 = \frac{\text{mass}}{\text{molar mass}} = \frac{32 \, g}{32 \, g/mol} = 1 \, mol \]
- For \( H_2 \):
\[ n_2 = \frac{4 \, g}{2 \, g/mol} = 2 \, mol \]

3. **Use the Ideal Gas Law:**
- For the first container:
\[ P_1 V = n_1 R T_1 \implies P V = 1 \cdot R \cdot T \implies P = \frac{R T}{V} \]
- For the second container:
\[ P_2 V = n_2 R T_2 \implies P_2 V = 2 \cdot R \cdot (2T) \implies P_2 = \frac{4R T}{V} \]

4. **Relate the Pressures:**
- Since the volumes and the gas constant \( R \) are the same for both containers, we can write:
\[ \frac{P_1}{P_2} = \frac{n_1 T_1}{n_2 T_2} \]
- Substituting the values:
\[ \frac{P}{P_2} = \frac{1 \cdot T}{2 \cdot 2T} = \frac{1}{4} \]
- Therefore, we can express \( P_2 \):
\[ P_2 = 4P \]

5. **Conclusion:**
- The pressure of the hydrogen container is \( P_2 = 4P \).

### Final Answer:
The pressure of the container containing hydrogen is \( 4P \).