If the volume of a container of gas is reduced, what will happen to the pressure inside the container?

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If the volume of a container of gas is reduced, what will happen to the pressure inside the container?

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Answer 1 · 2016-11-28T13:31:57

At constant $temperature, P_{1} V_{1} = P_{2} V_{2}$ $"temperature, " P_1V_1=P_2V_2$ , which is $Boyle's law$ $"Boyle's law"$ and so.........

Explanation:

Pressure varies as the ratio $\frac{V_{1}}{V_{2}}$ $V_1/V_2$ . If the gas were enclosed in a piston, the pressure should increase as the piston is compressed, and the pressure should decrease as the piston expands. This is intuitively sound.

Of course, we can combine with $Charles' Law$ $"Charles' Law"$ to give $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ $(P_1V_1)/T_1=(P_2V_2)/T_2$ , the combined gas law, and with $Avogadro's law$ $"Avogadro's law"$ , $V \propto n$ $Vpropn$ , to give the Ideal Gas law, $P V = n R T$ $PV=nRT$ , for an ideal gas.

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If the volume of a container of gas is reduced, what will happen to the pressure inside the container?

1 Answer

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