# A 2.5 L container of xenon had a pressure change from 125 kpa to 35 kpa what is the new volume?

Jul 4, 2017

Roughly $8.9 \text{L}$
Assuming there is no temperature change then ${\text{P"_1"V"_1="P"_2"V}}_{2}$, where the pressure (in Pa) x the volume in (m^3) is a constant.
In this case ${\text{P"_1"V"_1="P"_2"V"_2-=((125*1000)*(2.5*0.001))=((35*1000)*"V"_2=(125000*0.0025=35000"V}}_{2}$.
$\text{V"_2=(125000*0.0025)/35000~~0.0089"m"^3=8.9"L}$