An `L.P.G` cylinder contains `15kg` of butane gas at `27^(@)C` and 10 atm pressure It was leaking and its pressure fell down to 8 atm pressure after one day Calculate the amount of leaked gas .

Options

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

Answer

C

Explanation

According to ideal gas equation, `PV = nRT`
`:. (P_(1))/(P_(2)) = (n_(1))/(n_(2))` [where V, T are constant]
or `(P_(1))/(P_(2)) = (w_(1))/(w_(2))`
Hence `(10)/(8) = (15)/(w_(2)) " or " w_(2) = 12 kg`
`:.` Gas leaked `= 15 - 12 = 3 kg`