Equilibrium Position of Diathermic Partition in Isolated Gas Cylinder

To find the distance of the partition from the top at equilibrium, we analyze the forces acting on the diathermic partition when the system reaches mechanical and thermal equilibrium.

1. Given Data

• Total height of the vessel, $H = 8\text{ m}$
• Initial position of the partition from the top, $x_0 = 4\text{ m}$
• Mass of the partition, $m = 8.3\text{ kg}$
• Number of moles of gas in each compartment, $n = 0.1\text{ mole}$
• Temperature at equilibrium, $T = 300\text{ K}$
• Acceleration due to gravity, $g = 10\text{ ms}^{-2}$
• Universal gas constant, $R = 8.3\text{ J mol}^{-1}\text{K}^{-1}$

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2. Equilibrium Equations

Let $x$ be the distance of the partition from the top in meters at equilibrium, and let $A$ be the cross-sectional area of the cylinder.

• Volume of the top compartment: $V_1 = A \cdot x$

• Volume of the bottom compartment: $V_2 = A \cdot (8 - x)$

Since the partition is diathermic, thermal equilibrium ensures that both compartments have the same temperature $T = 300\text{ K}$.

Using the ideal gas equation $P V = n R T$:

• Pressure in the top compartment ($P_1$): $P_1 = \frac{n R T}{V_1} = \frac{n R T}{A x}$

• Pressure in the bottom compartment ($P_2$): $P_2 = \frac{n R T}{V_2} = \frac{n R T}{A (8 - x)}$

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3. Mechanical Equilibrium

At equilibrium, the upward force exerted by the bottom gas balances the downward force exerted by the top gas and the weight of the partition: $P_2 A = P_1 A + mg$

Dividing both sides by $A$: $P_2 - P_1 = \frac{mg}{A}$

Substitute the expressions for $P_1$ and $P_2$: $\frac{n R T}{A(8 - x)} - \frac{n R T}{A x} = \frac{mg}{A}$

Canceling $A$ from both sides: $\frac{n R T}{8 - x} - \frac{n R T}{x} = mg$

Factoring out $n R T$: $n R T \left( \frac{1}{8 - x} - \frac{1}{x} \right) = mg$

$n R T \left( \frac{x - (8 - x)}{x(8 - x)} \right) = mg$

$n R T \left( \frac{2x - 8}{8x - x^2} \right) = mg$

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4. Numerical Calculation

First, evaluate the terms $n R T$ and $mg$: $n R T = 0.1 \times 8.3 \times 300 = 249\text{ J}$ $mg = 8.3 \times 10 = 83\text{ N}$

Substitute these values into the equilibrium equation: $249 \left( \frac{2x - 8}{8x - x^2} \right) = 83$

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

$3(2x - 8) = 8x - x^2$

$6x - 24 = 8x - x^2$

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

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

Since $x$ represents a physical distance inside the $8\text{ m}$ cylinder, $x > 0$: $x = 6\text{ m}$

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Final Answer

The distance of the partition measured from the top when equilibrium is attained will be 6.