If the absolute temperature of a gas having volume `V cm^(3)` is doubled and the pressure is reduced to half, the final volume will be

Options

A. `0.25 V`
B. `0.50 V`
C. `2V^(2)`
D. `4V`

Answer

D

Explanation

Here, `T_(2) = 2T_(1)` and `P_(2) = (P_(1))/(2) V_(1) = V, V_(2) = ?` Final volume,
`:. V_(2) = (P_(1)V_(1))/(T_(1)) xx (T_(2))/(P_(2)) = (P_(1)V)/(T_(1)) xx (2T_(1))/(P_(1)//2) = 4V`