Inside a cylinder having insulating walls and dosed at ends is a movable piston, which divides the cylinder into two compartments. On one side of the piston is a mass m of a gas and on the other side a mass 2 m of the same gas. What fraction of volume of the cylinder will be occupied by the Larger mass of the gas when the piston is in equilibrium Consider that the movable piston is conducting so that the temperature is the same throughout

Options

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

Explanation

To solve the problem, we need to analyze the situation using the principles of the kinetic theory of gases and the ideal gas law.

1. **Understanding the Setup**: We have a cylinder divided into two compartments by a movable piston. On one side, there is a mass \( m \) of gas, and on the other side, there is a mass \( 2m \) of the same gas. The walls of the cylinder are insulating, and the piston is conducting, meaning the temperature is uniform throughout.

2. **Using the Ideal Gas Law**: The ideal gas law states that \( PV = nRT \), where:
- \( P \) is the pressure,
- \( V \) is the volume,
- \( n \) is the number of moles of gas,
- \( R \) is the universal gas constant,
- \( T \) is the temperature.

3. **Defining Variables**: Let:
- \( V_1 \) be the volume occupied by mass \( m \),
- \( V_2 \) be the volume occupied by mass \( 2m \).

4. **Pressure Equilibrium**: Since the piston is in equilibrium, the pressure on both sides must be equal:
\[ P_1 = P_2 \]

5. **Applying the Ideal Gas Law**: For the gas with mass \( m \):
\[ P V_1 = n_1 R T \]
For the gas with mass \( 2m \):
\[ P V_2 = n_2 R T \]
Here, \( n_1 \) and \( n_2 \) are the number of moles of gas on each side. Since the gas is the same, we can relate the number of moles to the mass:
\[ n_1 = \frac{m}{M} \quad \text{and} \quad n_2 = \frac{2m}{M} \]
where \( M \) is the molar mass of the gas.

6. **Setting Up the Equations**: Substituting \( n_1 \) and \( n_2 \) into the ideal gas equations:
\[ P V_1 = \frac{m}{M} R T \quad \text{(1)} \]
\[ P V_2 = \frac{2m}{M} R T \quad \text{(2)} \]

7. **Dividing the Equations**: Since \( P \) and \( R T \) are the same for both sides, we can divide equation (1) by equation (2):
\[ \frac{V_1}{V_2} = \frac{m}{2m} = \frac{1}{2} \]

8. **Expressing Volumes**: This means:
\[ V_1 = \frac{1}{2} V_2 \]

9. **Total Volume**: The total volume \( V \) of the cylinder is:
\[ V = V_1 + V_2 = \frac{1}{2} V_2 + V_2 = \frac{3}{2} V_2 \]

10. **Finding the Fraction**: We want to find the fraction of the total volume occupied by the larger mass of gas (which occupies volume \( V_2 \)):
\[ \text{Fraction occupied by } 2m = \frac{V_2}{V} = \frac{V_2}{\frac{3}{2} V_2} = \frac{2}{3} \]

Thus, the fraction of the volume of the cylinder occupied by the larger mass of gas is \( \frac{2}{3} \).

### Final Answer:
The fraction of volume occupied by the larger mass of gas is \( \frac{2}{3} \).