Question

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

• A. $1.5$
• B. $2.5$
• C. $2$
• D. $4$

Answer 1

Answer: (C)

Ideal gas equation is given by,
$pV=nRT$ ...(i)
For oxygen, $p=1atm,V=1L,n=n_{O_{2}}$
Therefore Eq. (i) becomes
$1\times 1=n_{O_{2}}RT$
$\Rightarrow n_{O_{2}}=\frac{1}{R T}$
For nitrogen $p=0.5atm,V=2L,n=n_{N_{2}}$
$\therefore 0.5\times 2=n_{N_{2}}RT$
$\Rightarrow n_{N_{2}}=\frac{1}{R T}$
For mixture of gasses,
$P_{m i x }V_{m i x}=n_{m i x}RT$
Here $n_{m i x}=n_{O_{2}}+n_{N_{2}}$
$\therefore \frac{P_{m i x } V_{m i x}}{R T}=\frac{1}{R T}+\frac{1}{R T}$
$\Rightarrow P_{m i x}V_{m i x}=2$
$P_{mix}=2$