A given mass of ideal gas has volume `V` at pressure `P` and room temperature `T`. Its pressure is first increased by `50%` and then decreased by `50%` (both at constant temperature). The volume becomes

Explanation

Case 1: `P_(1)V_(1)=P_(2)V_(2),P_(2)=(150xxP_(1))/(100)=3/2P_(1)`
`P_(1)V_(1)=3/2P_(1)V_(2),thereforeV_(2)2/3V_(1)`
Case 2: `P_(2)V_(2)=P_(3)=(50)/(100)P_(2)=1/2P_(2)`
`P_(2)2/3V_(1)=1/2P_(2)V_(3),thereforev_(3)=4/5V_(1)`