First slide

laws

Question

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

Moderate

A

450 K
B

228 K
C

420 K
D

570 K

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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