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A cylinder container with the moveable piston initially hold 3.0 mole of a gas at 8.0 atm pressure and a volume of 5.0 litre. If piston is moved to create a volume 10.0 litre , while simultaneously with drawing 1.5 mole of gas. Calculate the final pressure.

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

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PV = nRT (or)
$\large\frac{PV}{n} = RT$
$P_1 = 8 atm$
$V_1 = 5.0 litre$ = 3.0 mole
$P_2 = ?$
$V_2 = 10 litre$
$n_2 = 1.5 mole$
Now $\large\frac{P_1V_1}{n_1} = \large\frac{P_2V_2}{n_2}$
$\large\frac{8\times5}{3} =\large\frac{P_2\times10}{1.5}$
$\large\frac{8\times5\times1.5}{10} = P_2$
$\large\frac{40\times1.5}{10\times3} = P_2$
$\large\frac{6.0}{3} = P_2$
$P_2 = 2 atm$
Hence answer is (b)
answered Feb 23, 2014 by sharmaaparna1

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