A gas is contained in a 5.0–liter container at a pressure of 100 kPa. What size container would be required to contain the gas at a pressure of 50 kPa, assuming the temperature remains constant?
P1 V1 = P2 V2
100 5 = 50 V2
500 = 50 V2
V2 = 500 / 50
V2 = 10.0 L
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Answer : The volume of gas is, 10 L
Explanation:
According top the Boyle's Law, the pressure of a gas is inversely proportional to the volume of the gas.
\(P\propto \frac{1}{V}\)
or,
\(\frac{P_1}{P_2}=\frac{V_2}{V_1}\)
where,
\(V_1\) = initial volume of gas = 5 L
\(V_2\) = final volume of gas = ?
\(P_1\) = initial pressure of gas = 100 kPa
\(P_2\) = final pressure of gas = 50 kPa
Now put all the given values in the above formula, we get the final volume of the gas.
\(\frac{100kPa}{50kPa}=\frac{V_2}{5L}\)
\(V_2=10L\)
Therefore, the volume of gas is, 10 L