In a container `m` g of a gas is placed. After some time some gas is allowed to escape from container. The pressure of the gas becomes half and its absolute temperature `2//3 rd`. The amount of gas escaped is

Options

A. `2//3`
B. `1//2`
C. `1//4`
D. `1//6`

Explanation

To solve the problem step by step, we will use the Ideal Gas Law and the information provided in the question.

### Step 1: Understand the Ideal Gas Law
The Ideal Gas Law is given by the equation:
\[ PV = nRT \]
Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles of the gas
- \( R \) = universal gas constant
- \( T \) = absolute temperature

### Step 2: Relate moles to mass
The number of moles \( n \) can be expressed in terms of mass \( m \) and molar mass \( M \):
\[ n = \frac{m}{M} \]
Thus, we can rewrite the Ideal Gas Law as:
\[ PV = \frac{m}{M}RT \]

### Step 3: Set up the initial conditions
Let:
- Initial mass of the gas = \( m \) grams
- Initial pressure = \( P_1 \)
- Initial temperature = \( T_1 \)

### Step 4: Set up the final conditions
After some gas escapes:
- Final pressure = \( P_2 = \frac{P_1}{2} \)
- Final temperature = \( T_2 = \frac{2}{3} T_1 \)
- Final mass of the gas = \( m_2 = m - x \) (where \( x \) is the mass of gas that escaped)

### Step 5: Write the equations for initial and final states
Using the Ideal Gas Law for the initial and final states:
1. Initial state:
\[ P_1 V = \frac{m}{M} RT_1 \]
2. Final state:
\[ P_2 V = \frac{m_2}{M} RT_2 \]

### Step 6: Substitute the final conditions into the equation
Substituting \( P_2 \) and \( T_2 \) into the final state equation:
\[ \frac{P_1}{2} V = \frac{m - x}{M} R \left(\frac{2}{3} T_1\right) \]

### Step 7: Simplify the equation
Rearranging gives:
\[ \frac{P_1 V}{2} = \frac{(m - x) \cdot 2RT_1}{3M} \]

### Step 8: Equate the initial and final states
From the initial state, we know:
\[ P_1 V = \frac{m}{M} RT_1 \]
Substituting this into the final state equation:
\[ \frac{1}{2} \cdot \frac{m}{M} RT_1 = \frac{(m - x) \cdot 2RT_1}{3M} \]

### Step 9: Cancel common terms
We can cancel \( RT_1 \) and \( M \) from both sides:
\[ \frac{m}{2} = \frac{2(m - x)}{3} \]

### Step 10: Solve for \( x \)
Cross-multiplying gives:
\[ 3m = 4(m - x) \]
Expanding this:
\[ 3m = 4m - 4x \]
Rearranging:
\[ 4x = 4m - 3m \]
\[ 4x = m \]
Thus:
\[ x = \frac{m}{4} \]

### Final Answer
The amount of gas that escaped is:
\[ x = \frac{m}{4} \]