Question

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

Answer 1

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"`