What will be the minimum pressure required to compress 500 `dm^(3)` of air at 1 bar to 200 `dm^(3)` at `30^(@)C` ?

Answer

2.5 bar

Explanation

Given,
Initial pressure, `p_(1)`= 1 bar
Initial vomume, `V_(1)` = 500 `dm^(3)`
Final vomume, `V_(2) = 200 dm^(3)`
Since the temperature remains constant, the final pressue `(p_(2))` can be calculated using Boyle's law
According to Boyle's law,
`p_(1)V_(1)=p_(2)V_(2)`
`rArrp_(2)=(p_(1)V_(1))/(V_(2))`
`=(1xx500)/(200)` bar
= 2.5 bar
Therefore, the minimum pressure required is 2.5 bar.