A thermally isolated cylindrical closed vesse | vedclass

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(D) Let the cross-sectional area of the cylinder be $A$. Initially,the partition is at $4 \ m$ from the top,so the volume of each part is $V_0 = A 	i

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$6$

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(D) Let the cross-sectional area of the cylinder be $A$. Initially,the partition is at $4 \ m$ from the top,so the volume of each part is $V_0 = A 	imes 4$. Since the partition is diathermic,the temperature $T = 300 \ K$ remains constant for both parts. Let the partition move down by a distance $x$ from the initial position. The new volumes are $V_1' = A(4 + x)$ and $V_2' = A(4 - x)$. Using Boyle's Law $(PV = nRT)$,the pressures in the two parts at equilibrium are: $P_1' = \frac{nRT}{V_1'} = \frac{nRT}{A(4+x)}$ and $P_2' = \frac{nRT}{V_2'} = \frac{nRT}{A(4-x)}$. At equilibrium,the force balance on the partition is: $(P_2' - P_1')A = mg$ $\left[ \frac{nRT}{A(4-x)} - \frac{nRT}{A(4+x)} ight] A = mg$ $nRT \left[ \frac{1}{4-x} - \frac{1}{4+x} ight] = mg$ Substituting the values $n = 0.1 \ mol$,$R = 8.3 \ J \ mol^{-1} \ K^{-1}$,$T = 300 \ K$,$m = 8.3 \ kg$,and $g = 10 \ m/s^2$: $(0.1)(8.3)(300) \left[ \frac{(4+x) - (4-x)}{16-x^2} ight] = (8.3)(10)$ $249 \left[ \frac{2x}{16-x^2} ight] = 83$ $3 \left( \frac{2x}{16-x^2} ight) = 1$ $6x = 16 - x^2 \implies x^2 + 6x - 16 = 0$ $(x+8)(x-2) = 0$. Since $x > 0$,we have $x = 2 \ m$. The distance of the partition from the top is $4 + x = 4 + 2 = 6 \ m$.

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