Question

One litre of oxygen at a pressure of $1 \,atm$ and two litres of nitrogen at a pressure of $0.5\, atm$, are introduced into a vessel of volume $1\, L$. If there is no change in temperature, the final pressure of the mixture of gas (in atm) is

• A. 1.5
• B. 2.5
• C. 2
• D. 4

Answer 1

Answer: (C)

2

Answer 2

Ideal gas equation is given by
$pV = nRT ...(i)$
For oxygen, $p=1 atm , V =1 L , n = n _{0}$
Therefore equation (i) becomes
$\therefore 1 \times 1= n _{0,} RT$
$\Rightarrow n _{0_{2}}=\frac{1}{ RT }$
For nitrogen $p=0.5 atm , V =2 L , n = nN$
$\therefore 0.5 \times 2= n _{ N _{2}} RT$
$\Rightarrow n _{ N _{2}}=\frac{1}{ RT }$
For mixture of gas
$p _{mix} V _{mix}= n _{mix} RT$
Here, $n _{mix}= n O + n _{ N _{2}}$
$\therefore \frac{ p _{mix} V _{mix}}{ RT }=\frac{1}{ RT }+\frac{1}{ RT }$
$\Rightarrow p _{mix} V _{mix}=2$