A vertical closed cylinder is separated into two parts by a frictionless piston of mass m and of negligible thickness. The piston is free to move along the length of the cylinder. The length of the cylinder above the piston is `l_(1)`, and that below the piston is `l_(2)`, such that `l_(1)gtl_(2)`. Each part of the cylinder contains n moles of an ideal gas at equal temeprature T. If the pistion is stationary, its mass, m, will be given by : (R is universal gas constant and g is the acceleration due to gravitey)

Options

A. `(RT)/(ng)[(l_(1)-3l_(2))/(l_(1)l_(2))]`
B. `(RT)/(g)[(2l_(1)+l_(2))/(l_(1)l_(2))]`
C. `(nRT)/(g)[(1)/(l_(2))+(1)/(l_(1))]`
D. `(nRT)/(g)[(l_(1)-l_(2))/(l_(1)l_(2))]`

Explanation

To solve the problem, we need to analyze the forces acting on the piston in the stationary condition and apply the ideal gas law. Here’s a step-by-step solution:

### Step 1: Understand the System
We have a vertical closed cylinder with a frictionless piston separating two parts. The upper part has length \( l_1 \) and the lower part has length \( l_2 \). Each part contains \( n \) moles of an ideal gas at the same temperature \( T \).

### Step 2: Apply the Ideal Gas Law
For an ideal gas, the pressure \( P \) can be expressed using the ideal gas law:
\[ PV = nRT \]
Where:
- \( P \) is the pressure,
- \( V \) is the volume,
- \( n \) is the number of moles,
- \( R \) is the universal gas constant,
- \( T \) is the temperature.

### Step 3: Calculate the Pressures
The volume of gas above the piston is:
\[ V_1 = A \cdot l_1 \]
The pressure \( P_1 \) in the upper part of the cylinder can be expressed as:
\[ P_1 = \frac{nRT}{V_1} = \frac{nRT}{A \cdot l_1} \]

Similarly, for the lower part of the cylinder:
\[ V_2 = A \cdot l_2 \]
The pressure \( P_2 \) in the lower part of the cylinder is:
\[ P_2 = \frac{nRT}{V_2} = \frac{nRT}{A \cdot l_2} \]

### Step 4: Set Up the Force Balance
The forces acting on the piston must balance for it to be stationary. The upward force due to the pressure from the lower part is:
\[ F_{\text{up}} = P_2 \cdot A \]
The downward forces are the pressure from the upper part and the weight of the piston:
\[ F_{\text{down}} = P_1 \cdot A + mg \]

Setting these forces equal gives us:
\[ P_2 \cdot A = P_1 \cdot A + mg \]

### Step 5: Substitute Pressures
Substituting the expressions for \( P_1 \) and \( P_2 \):
\[ \frac{nRT}{A \cdot l_2} \cdot A = \frac{nRT}{A \cdot l_1} \cdot A + mg \]

### Step 6: Simplify the Equation
Canceling \( A \) from both sides:
\[ \frac{nRT}{l_2} = \frac{nRT}{l_1} + mg \]

### Step 7: Rearranging for \( mg \)
Rearranging gives:
\[ mg = \frac{nRT}{l_2} - \frac{nRT}{l_1} \]
\[ mg = nRT \left( \frac{1}{l_2} - \frac{1}{l_1} \right) \]

### Step 8: Solve for Mass \( m \)
Dividing both sides by \( g \):
\[ m = \frac{nRT}{g} \left( \frac{1}{l_2} - \frac{1}{l_1} \right) \]

### Step 9: Final Expression
This can be rewritten as:
\[ m = \frac{nRT}{g} \cdot \frac{l_1 - l_2}{l_1 l_2} \]

### Conclusion
Thus, the mass \( m \) of the piston is given by:
\[ m = \frac{nRT (l_1 - l_2)}{g \cdot l_1 \cdot l_2} \]