# Calculate the volume occupied by 96g $CH_4$ at 16 atm and $27^{\large\circ}C$.(R = 0.08litre atm $K^{-1}mol^{-1}$.

$(a)\;4\;litre\qquad(b)\;5\; litre\qquad(c)\;7\;litre\qquad(d)\;9\;litre$

Given
P = 16 atm
m = 16
T = 27+273 = 300K
V = ?
$\therefore we\;have\;PV = \large\frac{w}{m}RT$
Or $V = \large\frac{wRT}{Pm}$
$= \large\frac{96\times0.08\times300}{16\times16}$
= 9 litre