2 litres of gas is enclosed in a vessel at a pressure of 760 mm Hg. If the temperature remains constant, calculate the pressure when volume changes to 4 $d m^{3}$ .

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Solution

At constant temperature,

$P_{1} V_{1} = P_{2} V_{2}$

$P_{1} = 760 m m H g, V_{1} = 2 L, V_{2} = 4 d m^{3} = 4 L$

Then, $P_{2} = \frac{760 \times 2}{4} = 380 m m H g$

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2 litres of gas is enclosed in a vessel at a pressure of 760 mm Hg. If the temperature remains constant, calculate the pressure when volume changes to 4 dm3.

At constant temperature,P1V1=P2V2P1=760 mm Hg, V1=2L, V2=4dm3=4LThen, P2=760×24=380 mm Hg

2 litres of gas is enclosed in a vessel at a pressure of 760 mm Hg. If the temperature remains constant, calculate the pressure when volume changes to 4 $d m^{3}$ .

At constant temperature,

$P_{1} V_{1} = P_{2} V_{2}$

$P_{1} = 760 m m H g, V_{1} = 2 L, V_{2} = 4 d m^{3} = 4 L$

Then, $P_{2} = \frac{760 \times 2}{4} = 380 m m H g$