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laws

Question

A vessel containing 0.1 $m^{3}$ of air at 76 cm of Hg is connected to an evacuated vessel of capacity 0.09 $m^{3}$ . The resultant air pressure is:

Moderate

A

20 cm of Hg
B

30 cm of Hg
C

40 cm of Hg
D

60 cm of Hg

Solution

Due to free expansion , temperature of the gas remains constant .At constant temperature
$P_{1} V_{1} = P_{2} V_{2}$
$P_{2} = (\frac{V_{1}}{V_{2}}) P_{1} = \frac{0.1 m^{3}}{0.1 + 0.09} \times 76 cm of Hg$
= 40 cm of Hg

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