40 g f a gas occupies `20 dm^(3)` at 300 K and 100 kPa pressure. If the pressure is changed to 50 lPa without changing to temperature, what would be its volume ?

Explanation

`P_(1)v_(1)=P_(2)v_(2)` at constant temperature.
`P_(1)=100 kPa, v_(1)=20 dm^(3), P_(2) =50 kPa, V_(2) ?`
`V_(2)=(P_(1)V_(1))/(P_(2)),v_(2)=(100xx20)/(50)=40 dm^(3)`
`:. V_(2)=40 dm^(3)`