Question

A $100ml$ vessel containing $O_2\left(g\right)$ at $1.0atm$ and $400K$ is connected to a $300ml$ vessel containing $NO\left(g\right)$ at $1.5atm$ and $400K$ by means of a narrow tube of negligible volume where gases react to form $N{O}_3$. Final pressure of mixture will be:

• A. $1.125atm$
• B. $0.125atm$
• C. $1atm$
• D. $1.5atm$

Answer 1

The correct option is A $1.125atm$

$10ml$ vessel of $O_2$ at temp $T=400k$ at pressure 1 atm

$n=\frac{Pv}{RT}=\frac{1\times 100ml}{0.0821\times 400k}=3.045mmol$

$300ml$ vessel of No at temperature $T=300K$ at pressure $1.5atm$

$n=\frac{Pv}{RT}=\frac{1.5\times 300}{0.0821\times 300k}=18.27mmol$

$2NO+O_2\to 2N{O}_2$

$2mol$ No require $1mol{O}_2$

$18.27$ milli Mol No require $\frac{1}{2}\times 18.27=9.135mMol$

$O_2$ is limiting reagent

$1mol{O}_2$ require 2 mol No

$3.045$ milli mol $O_2$ requiree $6.045mol$ No

total number of m mole of NO after reaction $=18.27-6.045=12.225$

Mole of $N{O}_2=6.045$

Total number of mole $=12.25+6.045$

$=18.27$

$P=\frac{nRT}{v}=\frac{18.27\times 0.0821\times 400}{400}$