Step-by-step Solution

1. Define the Initial and Final States

• Initial State: The cylindrical vessel of total height $H = 8$ m is divided into two equal parts by a partition.

  • Height of the upper part, $h_1 = 4$ m.
  • Height of the lower part, $h_2 = 4$ m.
  • Number of moles of ideal gas in each part, $n_1 = n_2 = n = 0.1$ mole.
  • Initial temperature of the gas in both parts, $T_i = 300$ K.
  • The partition has mass $m = 8.3$ kg.

• Final State (at Equilibrium): The partition is released and settles at a new equilibrium position.

  • Let the final distance of the partition from the top be $x$.
  • Height of the upper part, $h'_1 = x$.
  • Height of the lower part, $h'_2 = H - x = 8 - x$.
  • Since the partition is diathermic (a perfect thermal conductor), the final temperature in both parts will be the same. Let this be $T_f$.
  • Let the final pressures be $P'_1$ and $P'_2$.

2. Mechanical Equilibrium Condition

At equilibrium, the forces acting on the partition must balance. The forces are:

• Force from the upper gas pressure: $P'_1 A$ (downwards)
• Weight of the partition: $mg$ (downwards)
• Force from the lower gas pressure: $P'_2 A$ (upwards)

where $A$ is the cross-sectional area of the cylinder.

The force balance equation is: $P'_2 A = P'_1 A + mg$ $P'_2 - P'_1 = \frac{mg}{A}$

Using the ideal gas law, $PV = nRT$, we can express the pressures as: $P'_1 = \frac{nRT_f}{V'_1} = \frac{nRT_f}{Ax}$ $P'_2 = \frac{nRT_f}{V'_2} = \frac{nRT_f}{A(8-x)}$

Substituting these into the equilibrium equation: $\frac{nRT_f}{A(8-x)} - \frac{nRT_f}{Ax} = \frac{mg}{A}$ Multiplying by $A$, we get: $nRT_f \left( \frac{1}{8-x} - \frac{1}{x} \right) = mg$ $nRT_f \left( \frac{x - (8-x)}{x(8-x)} \right) = mg$ $nRT_f \left( \frac{2x - 8}{x(8-x)} \right) = mg \quad (*)$

3. Energy Conservation and the Isothermal Assumption

The vessel is thermally isolated, so the total energy of the system (gas in both parts + partition) is conserved. The decrease in the partition's potential energy is converted into internal energy of the gas, causing the temperature to rise ($T_f > T_i$). The exact value of $T_f$ would depend on the specific heat (and thus degrees of freedom, $f$) of the ideal gas, which is not provided.

However, in problems of this type where a key parameter ($f$) is missing and an integer answer is expected, it often implies a simplifying assumption. The standard simplification is to assume the process is isothermal, i.e., the temperature of the gas does not change. So we assume $T_f = T_i = 300$ K.

4. Calculation with the Isothermal Assumption

Setting $T_f = T_i = 300$ K in equation (*): $nRT_i \left( \frac{2x - 8}{x(8-x)} \right) = mg$

Now, we substitute the given values:

• $n = 0.1$ mol
• $R = 8.3$ J mol⁻¹K⁻¹
• $T_i = 300$ K
• $m = 8.3$ kg
• $g = 10$ ms⁻²

$(0.1 \text{ mol}) (8.3 \text{ J mol}^{-1}\text{K}^{-1}) (300 \text{ K}) \left( \frac{2(x-4)}{x(8-x)} \right) = (8.3 \text{ kg})(10 \text{ ms}^{-2})$ $249 \left( \frac{2(x-4)}{x(8-x)} \right) = 83$

Divide both sides by 83: $3 \left( \frac{2(x-4)}{x(8-x)} \right) = 1$ $6(x-4) = x(8-x)$ $6x - 24 = 8x - x^2$

Rearranging into a standard quadratic form: $x^2 - 2x - 24 = 0$

Factoring the quadratic equation: $(x-6)(x+4) = 0$

This gives two possible solutions for $x$: $x=6$ or $x=-4$.

5. Final Answer

Since $x$ represents the distance from the top of the vessel, it must be a positive value. Therefore, the physically meaningful solution is: $x = 6 \text{ m}$

When equilibrium is reached, the distance of the partition from the top will be 6 m.