# Question #a77a0

Apr 14, 2015

The volume is 40.6 L.

The temperature remains constant while $P$ and $V$ change.

This is a Boyle's Law problem.

$\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$

${P}_{1} = \text{2.00 atm}$; ${V}_{1} = \text{20.3 L}$
${P}_{2} = \text{1.00 atm}$; ${V}_{2} = \text{?}$

${V}_{2} = \text{20.3 L" × (2.00 cancel("atm"))/(1.00 cancel("atm")) = "40.6 L}$

The volume of the gas is 40.6 L