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laws

Question

Two flasks of capacity 4 litres and 8 litres connected by a narrow tube contain a gas at a pressure of 60cm of Hg and at a temperature of 27^oC. When the larger flask alone is dipped in a bath at 127^oC, keeping the temperature of the smaller flask constant at 27^oC, then the resultant pressure is

Moderate

A

72 cm of Hg
B

36 cm of Hg
C

144 cm of Hg
D

60 cm of Hg

Solution

no. of moles remain constant
$\begin{array}{l} \frac{P_{0} \times 4}{R 300} + \frac{P_{0} \times 8}{R (300)} = \frac{P (4)}{300 R} + \frac{P (8)}{400 R} \\ \frac{12 P_{0}}{300 R} = \frac{P}{100 R} (\frac{4}{3} + 2) \\ 4 P_{0} = (\frac{10}{3}) P \\ P = \frac{12 P_{0}}{10} = \frac{12 \times 60}{10} = 72 cm of Hg \end{array}$

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