A 2.24,L oxygen cylinder shows a leakage at N | vedclass

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(C) At $NTP$,the initial pressure $P_1 = 1\,atm$ and temperature $T = 273\,K$.Initial moles of $O_2$ gas: $n_1 = \frac{P_1 V}{RT} = \frac{1 	imes 2.2

Vedclass · Accepted Answer

$0.075$

Vedclass · Answer

(C) At $NTP$,the initial pressure $P_1 = 1\,atm$ and temperature $T = 273\,K$.Initial moles of $O_2$ gas: $n_1 = \frac{P_1 V}{RT} = \frac{1 	imes 2.24}{0.0821 	imes 273} \approx 0.1\,mol$.Pressure drop is $570\,mm\,Hg$,so the final pressure $P_2 = 760 - 570 = 190\,mm\,Hg = \frac{190}{760}\,atm = 0.25\,atm$.Final moles of $O_2$ gas: $n_2 = \frac{P_2 V}{RT} = \frac{0.25 	imes 2.24}{0.0821 	imes 273} \approx 0.025\,mol$.Moles of gas escaped = $n_1 - n_2 = 0.1 - 0.025 = 0.075\,mol$.

$P \ (bar)$	$\frac{PV}{RT}$
$1$	$1$
$2$	$1$
$3$	$1$
$4$	$1.5$
$5$	$2.0$

$A$ $2.24\,L$ oxygen cylinder shows a leakage at $NTP$. When the leakage is plugged,the pressure drops by $570\,mm\,Hg$. What is the number of moles of gas that escaped?

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