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?

Verified Answer :

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