A container contains a certain gas of mass m at high pressure. A little amount of the gas has been allowed to escape from the container and after some time, the pressure of the gas becomes half and its absolute temperature two-third. The mass of the gas escaped is

Options

A. `(2m)/3`
B. `m/2`
C. `m/4`
D. `m/6`

Answer

To solve the problem, we will use the ideal gas law and the information given about the changes in pressure and temperature after some gas has escaped from the container.

### Step-by-Step Solution:

1. **Initial Conditions**:
- Let the initial pressure be \( P_1 \).
- Let the initial temperature be \( T_1 \).
- Let the initial mass of the gas be \( m \).
- The number of moles can be expressed as \( n_1 = \frac{m}{M} \), where \( M \) is the molar mass of the gas.

2. **Final Conditions After Gas Escapes**:
- The pressure after some gas has escaped is \( P_2 = \frac{P_1}{2} \).
- The temperature after some gas has escaped is \( T_2 = \frac{2}{3} T_1 \).
- Let the mass of the gas that has escaped be \( x \). Therefore, the remaining mass of the gas is \( m - x \).
- The number of moles after the gas escapes is \( n_2 = \frac{m - x}{M} \).

3. **Applying the Ideal Gas Law**:
- For the initial state, the ideal gas law gives us:
\[ P_1 V = n_1 R T_1 \quad \text{(1)} \]
- For the final state, the ideal gas law gives us:
\[ P_2 V = n_2 R T_2 \quad \text{(2)} \]

4. **Setting Up the Equations**:
- From equation (1):
\[ P_1 = \frac{n_1 R T_1}{V} = \frac{\frac{m}{M} R T_1}{V} \quad \text{(3)} \]
- From equation (2):
\[ P_2 = \frac{n_2 R T_2}{V} = \frac{\frac{m - x}{M} R \left(\frac{2}{3} T_1\right)}{V} \quad \text{(4)} \]

5. **Substituting Known Values**:
- Substitute \( P_2 = \frac{P_1}{2} \) into equation (4):
\[ \frac{P_1}{2} = \frac{\frac{m - x}{M} R \left(\frac{2}{3} T_1\right)}{V} \]

6. **Equating the Two Expressions**:
- From equations (3) and (4), we can equate:
\[ \frac{P_1}{2} = \frac{\frac{m - x}{M} R \left(\frac{2}{3} T_1\right)}{V} \]
- Rearranging gives:
\[ P_1 = \frac{2(m - x) R \left(\frac{2}{3} T_1\right)}{2MV} \quad \text{(5)} \]

7. **Canceling Common Terms**:
- From equations (3) and (5), we can cancel \( R \) and \( V \):
\[ 1 = \frac{2(m - x) \left(\frac{2}{3} T_1\right)}{2M T_1} \]
- Simplifying gives:
\[ 1 = \frac{(m - x) \cdot \frac{2}{3}}{M} \]

8. **Solving for \( x \)**:
- Rearranging gives:
\[ m - x = \frac{3M}{2} \]
- Therefore:
\[ x = m - \frac{3M}{2} \]
- Since we need to express \( x \) in terms of \( m \), we can rearrange:
\[ x = \frac{m}{2} \]

### Conclusion:
The mass of the gas that has escaped is \( \frac{m}{2} \).