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Question Number 55308 by Tawa1 last updated on 21/Feb/19

	Answered by Rio Mike last updated on 21/Feb/19

	${P C l}_{5} \underset{△}{\to} {P C l}_{3} + {C l}_{2}$ $f r o m e q u a t i o n 1 m o l \to 24 {d m}^{3}$ $a t r o o m t e m p e r a t u r e a n d p r e s s u r e$ $(760 M m H g)$ $\Rightarrow 0.01 m o l l i b e r a e s {x d m}^{3}$ $s o {x d}^{} m^{3} = (24 \times 0.01) {d m}^{3}$ ${x c m}^{3} = (24 \times 0.01) \times 10^{3} {c m}^{3}$ $u s i n g {b o y l e}^{'} s l a w$ $a t c o n s t a n t t e m p e r a t u r e$ $p_{1} v_{1} = p_{2} v_{2}$ $760 \times 24000 = p_{2} (24 \times 0.01) \times 10^{3}$ $\Rightarrow p_{2} = \frac{760 \times 24000}{(24 \times 0.01) \times 10^{3}} M m H g .$
	Commented by Tawa1 last updated on 23/Feb/19

	$God bless you sir$
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