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laws

Question

One litre of helium gas at a pressure 76 cm. of Hg and temperature $27^{o} C$ is heated till its pressure and volume are double. The final ternperature attained by the gas is:

Easy

A

$327^{0} C$
B

$927^{0} C$
C

$1027^{0} C$
D

$827^{0} C$

Solution

Initial volume of the gas $V_{1} = 1 litre initial pressure of the gas P_{1} = 70 cm of Hg$
Final pressure of the gas $P_{2} = 2 \times 76 = 152 cm of Hg$
Final volume of the gas $V_{2} = 2 litre$
According the Gay Lussac law we have
$\frac{P_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}} or \frac{76 \times 1}{300} = \frac{152 \times 2}{T_{2}}$
$\frac{1}{300} = \frac{4}{T_{2}} or T_{2} = 4 \times 300 = 1200 K = 927^{0} C$

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