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 piston is $$l_1$$, and that below the piston is $$l_2$$, such that $$l_1 > l_2$$. Each part of the cylinder contains n moles of an ideal gas at equal temperature T. If the piston is stationary, its mass m will be given by: (R is universal gas constant and g is the acceleration due to gravity)

We have a vertical cylinder of uniform cross-sectional area $$A$$. A light, frictionless piston of mass $$m$$ separates the cylinder into an upper part of length $$l_1$$ and a lower part of length $$l_2$$, with $$l_1 > l_2$$. Each part contains $$n$$ moles of an ideal gas at the same absolute temperature $$T$$. Because the piston is at rest, the forces acting on it must balance perfectly.

For each gas sample we write the ideal-gas equation. The ideal-gas law states

$$PV = nRT,$$

where $$P$$ is pressure, $$V$$ is volume, $$n$$ is the number of moles, $$R$$ is the universal gas constant and $$T$$ is the absolute temperature.

First for the gas above the piston, whose pressure is $$P_1$$ and volume is $$V_1 = A\,l_1$$, we obtain

$$P_1 V_1 = nRT \;\;\Longrightarrow\;\; P_1 (A\,l_1) = nRT \;\;\Longrightarrow\;\; P_1 = \frac{nRT}{A\,l_1}.$$

Next for the gas below the piston, whose pressure is $$P_2$$ and volume is $$V_2 = A\,l_2$$, we write

$$P_2 V_2 = nRT \;\;\Longrightarrow\;\; P_2 (A\,l_2) = nRT \;\;\Longrightarrow\;\; P_2 = \frac{nRT}{A\,l_2}.$$

Now we look at the forces acting on the piston. The upward force on the piston is the pressure exerted by the lower gas times area $$A$$, namely $$P_2A$$. The downward forces are the pressure exerted by the upper gas, $$P_1A$$, and the weight of the piston, $$mg$$. For mechanical equilibrium (piston stationary) we therefore set

$$\text{Upward force} = \text{Downward force}$$

$$P_2A = P_1A + mg.$$

Solving for the weight term, we write

$$mg = A\,(P_2 - P_1).$$

We now substitute the expressions for $$P_1$$ and $$P_2$$ obtained from the ideal-gas law:

$$mg = A\!\left(\frac{nRT}{A\,l_2} - \frac{nRT}{A\,l_1}\right).$$

The area $$A$$ cancels out:

$$mg = nRT\!\left(\frac{1}{l_2} - \frac{1}{l_1}\right).$$

We combine the fractions inside the parentheses:

$$\frac{1}{l_2} - \frac{1}{l_1} = \frac{l_1 - l_2}{l_1\,l_2}.$$

Substituting this result gives

$$mg = nRT\!\left(\frac{l_1 - l_2}{l_1\,l_2}\right).$$

Finally we isolate the mass $$m$$ by dividing both sides by $$g$$:

$$m = \frac{nRT}{g}\!\left(\frac{l_1 - l_2}{l_1\,l_2}\right).$$

This expression matches option B.

Hence, the correct answer is Option B.