JEE Advance - Chemistry (1994 - No. 8)
An LPG (liquefied petroleum gas) cylinder weighs 14.8 kg when empty. When full, it weighs 29.0 kg and shows a pressure of 2.5 atm. In the course of use at 27o C, the weight of the full cylinder reduces to 23.2 kg. Find out the volume of the gas in cubic meters used up at the normal usage conditions find the final pressure inside the cylinder. Assume LPG to be n-butane with normal boiling point of 0o C.
Volume of gas used up = 2.46 m3, Final pressure = 1.48 atm
Volume of gas used up = 2.40 m3, Final pressure = 1.50 atm
Volume of gas used up = 2.50 m3, Final pressure = 1.45 atm
Volume of gas used up = 2.46 m3, Final pressure = 1.478 atm
Volume of gas used up = 2.463 m3, Final pressure = 1.48 atm
Explanation
Weight of butane gas in filled cylinder = 29 $$-$$ 14.8 kg = 14.2 kg
$$\Rightarrow$$ During the course of use, weight of cylinder reduces to 23.2 kg
$$\Rightarrow$$ Weight of butane gas remaining now = 23.2 $$-$$ 14.8 = 8.4 kg
Also, during use, V (cylinder) and T remains same.
Therefore, $${{{p_1}} \over {{p_2}}} = {{{n_1}} \over {{n_2}}}$$
$$ \Rightarrow {p_2} = \left( {{{{n_2}} \over {{n_1}}}} \right){p_1} = \left( {{{8.4} \over {14.2}}} \right) \times 2.5$$ [Here, $${{{n_2}} \over {{n_1}}} = {{{w_2}} \over {{w_1}}}$$]
= 1.48 atm
Also, pressure of gas outside the cylinder is 1.0 atm.
$$ \Rightarrow pV = nRT$$
$$ \Rightarrow V = {{nRT} \over p} = {{(14.2 - 8.4) \times {{10}^3}} \over {58}} \times {{0.082 \times 30} \over 1}L$$
= 2460 L = 2.46 m3
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