Question

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?

Answer 1

P1* V1 = P2 * V2

100* 5 =  50 * V2

500 = 50 V2

V2 = 500 / 50

V2 = 10.0 L

hope this helps!

Answer 2

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