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An L.P.G Cylinder contains 15kg of butane gas at $27^{\large\circ}C$ 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.

$(a)\;1kg\qquad(b)\;2kg\qquad(c)\;3kg\qquad(d)\;4kg$

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1 Answer

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Volume of cylinder is constant
$P_1V = n_1RT$
$P_2V = n_2RT$
Or $\large\frac{P_1}{P_2} =\large\frac{ n_1}{n_2} = \large\frac{\large\frac{w_1}{m}}{\large\frac{w_2}{m}}$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\large\frac{w_1}{w_2}$
$\large\frac{10}{8} = \large\frac{15}{w_2}$
$w_2 = \large\frac{15\times8}{10}$
$w_2 = \large\frac{120}{10}$
$w_2 = 12kg$
$\therefore$ Gas leaked = 15-12
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=3kg$
Hence answer is (c)
answered Feb 23, 2014 by sharmaaparna1
 

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