An LPG cylinder containing containing 15 kg butane at `27^(@)C` and 10 atm pressure is leaking. After one day, its pressure decreased to 8 atm. The quantity of the gas leaked is

Options

A. 1 kg
B. 2 kg
C. 3 kg
D. 4 kg

Answer

C

Explanation

PV=nRT. Here, T and V are constant. Hence,
`(P_(1))/(P_(2))=(n_(1))/(n_(2))=(w_(1))/(w_(2))`.
`:. (10)/(8)=(15)/(w_(2))" or " w_(2)=12" kg"`.
Hence, gas leaked=15-12=3 kg