1. Set up pressures in the two gas compartments

Let the cross-sectional area of the cylinder be $A$.

• Volume of gas above piston: $V_1 = A\ell_1$
• Volume of gas below piston: $V_2 = A\ell_2$

Each side contains $n$ moles of ideal gas at temperature $T$, so by ideal gas law:

$P_1 V_1 = nRT \quad \Rightarrow \quad P_1 = \frac{nRT}{A\ell_1}$

$P_2 V_2 = nRT \quad \Rightarrow \quad P_2 = \frac{nRT}{A\ell_2}$

Here:

• $P_1$ = pressure of gas above piston
• $P_2$ = pressure of gas below piston

Since $\ell_1 > \ell_2$, we have $P_2 > P_1$ which makes sense because the lower compartment has smaller volume.

2. Apply force balance on the stationary piston

For the piston to remain stationary, net force on it must be zero.

• Upward force due to lower gas: $F_{\text{up}} = P_2 A$
• Downward forces:
  • due to upper gas: $P_1 A$
  • weight of piston: $mg$

So equilibrium gives:

$P_2 A = P_1 A + mg$

Hence,

$mg = A(P_2 - P_1)$

Substitute $P_1$ and $P_2$:

$mg = A\left(\frac{nRT}{A\ell_2} - \frac{nRT}{A\ell_1}\right)$

$mg = nRT\left(\frac{1}{\ell_2} - \frac{1}{\ell_1}\right)$

$mg = nRT\left(\frac{\ell_1 - \ell_2}{\ell_1\ell_2}\right)$

Therefore,

$m = \frac{nRT}{g}\left(\frac{\ell_1 - \ell_2}{\ell_1\ell_2}\right)$

3. Match with the options

This is exactly Option A:

$\boxed{\frac{nRT}{g}\left(\frac{\ell_1 - \ell_2}{\ell_1\ell_2}\right)}$

4. Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

So they agree.