A vessel has 6g of hydrogen at pressure P and temperature 500K. A small hole is made in it so that hydrogen leaks out. How much hydrogen leaks out if the final pressure is P/2 and temperature falls to 300 K ?

Options

A. 2 g
B. 3 g
C. 4 g
D. 1 g

Explanation

To solve the problem of how much hydrogen leaks out of the vessel, we will use the Ideal Gas Law and the information provided in the question.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Initial mass of hydrogen, \( m_1 = 6 \, \text{g} \)
- Initial pressure, \( P_1 = P \)
- Initial temperature, \( T_1 = 500 \, \text{K} \)
- Final pressure, \( P_2 = \frac{P}{2} \)
- Final temperature, \( T_2 = 300 \, \text{K} \)

2. **Convert Mass to Moles:**
- The molar mass of hydrogen (H₂) is approximately \( 2 \, \text{g/mol} \).
- Calculate the initial number of moles, \( n_1 \):
\[ n_1 = \frac{m_1}{\text{Molar mass}} = \frac{6 \, \text{g}}{2 \, \text{g/mol}} = 3 \, \text{mol} \]

3. **Use the Ideal Gas Law:**
- The Ideal Gas Law is given by \( PV = nRT \).
- Since the volume of the vessel is constant, we can set up the relationship between the initial and final states:
\[ \frac{P_1 n_1}{T_1} = \frac{P_2 n_2}{T_2} \]

4. **Substitute Known Values:**
- Substitute \( P_1 = P \), \( P_2 = \frac{P}{2} \), \( T_1 = 500 \, \text{K} \), and \( T_2 = 300 \, \text{K} \):
\[ \frac{P \cdot 3}{500} = \frac{\frac{P}{2} \cdot n_2}{300} \]

5. **Simplify the Equation:**
- Cross-multiply to solve for \( n_2 \):
\[ 3P \cdot 300 = \frac{P}{2} \cdot n_2 \cdot 500 \]
\[ 900P = \frac{500P}{2} n_2 \]
\[ 900P = 250P n_2 \]
- Divide both sides by \( P \) (assuming \( P \neq 0 \)):
\[ 900 = 250 n_2 \]
- Solve for \( n_2 \):
\[ n_2 = \frac{900}{250} = 3.6 \, \text{mol} \]

6. **Calculate the Change in Moles:**
- The change in the number of moles, \( \Delta n \):
\[ \Delta n = n_1 - n_2 = 3 - 3.6 = -0.6 \, \text{mol} \]
- Since we are looking for the amount that leaked out, we take the absolute value:
\[ \Delta n = 0.6 \, \text{mol} \]

7. **Convert Moles Back to Mass:**
- Calculate the mass of hydrogen that leaked out:
\[ \Delta m = \Delta n \times \text{Molar mass} = 0.6 \, \text{mol} \times 2 \, \text{g/mol} = 1.2 \, \text{g} \]

### Final Answer:
The amount of hydrogen that leaked out is **1.2 grams**.