A vertical cylinder closed at both ends is fitted with a smooth piston dividing the volume into two parts each containing one mole of air. At the equilibrium temperature of 320 K, the upper and lower parts are in the ratio 4 : 1. The ratio will become 3 : 1 at a temperature of

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450 K

570 K

420 K

228 K

answer is A.

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Detailed Solution

$(p_{2} - p_{1}) A = m g$

or $\frac{m g}{A} = \frac{R T_{i}}{V_{1}} - \frac{R T_{i}}{4 V_{1}} = \frac{3 R T_{i}}{4 V_{1}}$ …(i)

Similarly, in second case,

$\frac{m g}{A} = \frac{R T_{f}}{V_{2}} - \frac{R T_{f}}{3 V_{2}} = \frac{2 R T_{f}}{3 V_{2}}$ …(ii)

Further, 5V₁ = 4V₂
Equating Eqs. (i) and (ii), we get

$\frac{3 T_{i}}{4 V_{1}} = \frac{2 T_{f}}{3 V_{2}}$

$\Rightarrow$ $T_{f} = \frac{9}{8} \times \frac{V_{2}}{V_{1}} \times T_{i}$

$= \frac{9}{8} \times \frac{5}{4} \times 320$

= 450 K

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