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. `P_(18)`

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 1: Calculate the number of moles of \( O_2 \)

Given:
- Mass of \( O_2 = 32 \, \text{g} \)
- Molar mass of \( O_2 = 32 \, \text{g/mol} \)

Using the formula for number of moles:

\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{32 \, \text{g}}{32 \, \text{g/mol}} = 1 \, \text{mol} \]

### Step 2: Apply the Ideal Gas Law for \( O_2 \)

Using the Ideal Gas Law for \( O_2 \):

\[ PV = nRT \]

Substituting the values we have:

\[ P \cdot V = 1 \cdot R \cdot T \]

Thus,

\[ PV = RT \quad \text{(1)} \]

### Step 3: Calculate the number of moles of \( H_2 \)

Given:
- Mass of \( H_2 = 4 \, \text{g} \)
- Molar mass of \( H_2 = 2 \, \text{g/mol} \)

Using the formula for number of moles:

\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{4 \, \text{g}}{2 \, \text{g/mol}} = 2 \, \text{mol} \]

### Step 4: Apply the Ideal Gas Law for \( H_2 \)

For the \( H_2 \) gas, the temperature is \( 2T \):

\[ P'V = nRT' \]

Substituting the values we have:

\[ P' \cdot V = 2 \cdot R \cdot (2T) \]

This simplifies to:

\[ P'V = 4RT \quad \text{(2)} \]

### Step 5: Relate the two equations

From equation (1):

\[ PV = RT \]

From equation (2):

\[ P'V = 4RT \]

### Step 6: Solve for \( P' \)

Now, we can relate \( P \) and \( P' \):

\[ P'V = 4PV \]

Dividing both sides by \( V \) (since \( V \) is the same for both containers):

\[ P' = 4P \]

### Conclusion

The pressure of the \( H_2 \) gas in the second container is:

\[ P' = 4P \]

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