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Standard IX

Chemistry

Boyle's Law

A 100 ml vess...

Question

A 100 ml vessel containing $O_{2} (g)$ at 1.0 at and 400 K is connected to a 300 ml vessel containing NO(g) at 1.5 atm and 400 K by means of a narrow tube of the negligible volume where gases react to form $N O_{2}$ . Final pressure of mixture will be:

1.125 atm

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0.125 atm

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1 atm

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1.5 atm

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Solution

The correct option is B 1.125 atm
Volume of $O_{2} (g) V = 100 m l$
$T = 400 k$ pressure $P = 1$ atm
$n = \frac{P V}{R T}$ millimole
$= \frac{1 \times 100}{0.0821 \times 400} = 3 m m o l e$
volume of No = $300 m l$
$T = 400 k P = 1.5 a t m$
$n = \frac{1.5 \times 300}{0.0821 \times 400}$ millimole
$= 13.7 m m o l$
$2 N O + O_{2} \to 2 N O_{2}$
3 millimole react with 6 millimole No
Number of mole = $13.7 m o l$
Final pressure $P = \frac{n R T}{V} = \frac{13.7 \times 0.0821 \times 400}{400}$
$= 1.125 a t m$

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A 100 ml vessel containing O2(g) at 1.0 at and 400 K is connected to a 300 ml vessel containing NO(g) at 1.5 atm and 400 K by means of a narrow tube of the negligible volume where gases react to form NO2. Final pressure of mixture will be:

A 100 ml vessel containing $O_{2} (g)$ at 1.0 at and 400 K is connected to a 300 ml vessel containing NO(g) at 1.5 atm and 400 K by means of a narrow tube of the negligible volume where gases react to form $N O_{2}$ . Final pressure of mixture will be: