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

1 Answer
Jul 4, 2017

Answer:

Roughly #8.9"L"#

Explanation:

Assuming there is no temperature change then #"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 #"P"_1"V"_1="P"_2"V"_2-=((125*1000)*(2.5*0.001))=((35*1000)*"V"_2=(125000*0.0025=35000"V"_2#.

#"V"_2=(125000*0.0025)/35000~~0.0089"m"^3=8.9"L"#