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Calculate the weight of $CH_4$ in a 9 litre cylinder at 16 atm and $27^{\large\circ}$ (R = 0.08 litre atm $K^{-1}$)

$(a)\;96g\qquad(b)\;98g\qquad(c)\;99g\qquad(d)\;95g$

Can you answer this question?
 
 

1 Answer

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Given
P = 16 atm
V = 9 litre
T = 300K
m = 16
$\therefore PV = \large\frac{w}{m}RT$
$w = \large\frac{mPV}{RT}$
$w = \large\frac{16\times16\times9}{0.08\times300}$
$ = \large\frac{2304}{24}$
= 96 g
Hence answer is (a)
answered Feb 11, 2014 by sharmaaparna1
 

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