A thermally isolated cylindrical closed vessel of height 8 m is kept vertically. It is divided into two equal parts by a diathermic (perfect thermal conductor) frictionless partition of mass 8.3 kg. [2020]

Question

A thermally isolated cylindrical closed vessel of height 8 m is kept vertically. It is divided into two equal parts by a diathermic (perfect thermal conductor) frictionless partition of mass 8.3 kg. Thus the partition is held initially at a distance of 4 m from the top, as shown in the schematic figure below. Each of the two parts of the vessel contains 0.1 mole of an ideal gas at temperature 300 K. The partition is now released and moves without any gas leaking from one part of the vessel to the other. When equilibrium is reached, the distance of the partition from the top (in m) will be ________ (take the acceleration due to gravity = 10 ${ms}^{- 2}$ and the universal gas constant = 8.3 J ${mol}^{- 1} K^{- 1}$ ). [2020]

Smart Achievers · Accepted Answer

(6)

$Initially partition is held at a distance of 4 m from the top.$

$Temperature, T = 300 K$

$P V = P A h = n R T$

$P_{0} 4 A = 0.1 R \times 300$ $(For both)$

$Let piston shifts by x meter$ $and final temperature is T, then$

$P_{2} A (4 - x) = 0.1 R T$

$∴$ $P_{2} A = \frac{0.1 R T}{4 - x} = \frac{R T}{10 (4 - x)}$ $and P_{1} A = \frac{R T}{10 (4 + x)}$

$Finally P_{2} A - P_{1} A = m g = 83 = \frac{R T}{10} (\frac{1}{4 - x} - \frac{1}{4 + x})$

$\Rightarrow \frac{R T}{10} (\frac{2 x}{16 - x^{2}}) = 83$

$0.2 C_{v} \times 300 + m g \cdot x = 0.2 C_{v} T$

$\Rightarrow \frac{3}{2} R \times \frac{2}{10} \times 300 + 83 x = \frac{2}{10} \times \frac{3}{2} R T$

$\Rightarrow (\frac{900 R}{10} + 83 x) = 3 (\frac{R T}{10})$

$\Rightarrow (\frac{300 R}{10} + \frac{83}{3} x) = \frac{R T}{10}$ $\Rightarrow (30 R + \frac{83 x}{3}) (\frac{2 x}{16 - x^{2}}) = 83$

$\Rightarrow 83 (3 + \frac{x}{3}) (\frac{2 x}{16 - x^{2}}) = 83$ $\Rightarrow \frac{(9 + x)}{3} (\frac{2 x}{16 - x^{2}}) = 1$

$\Rightarrow 18 x + 2 x^{2} = 48 - 3 x^{2}$ $\Rightarrow 5 x^{2} + 18 x - 48 = 0$

$∴$ $x = 1.78 \approx 2$

$Hence distance from top when reached equilibrium = 4 + 2 = 6 m$

Solution

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