Question

One litre of oxygen at a pressure of $1\,atm\,$ and two litres of nitrogen at a pressure of $0.5\,atm\,$ are added into a vessel of volume $1\,L\,$ . If the change in temperature is zero, then find the final pressure of the mixture of gas (in $atm\,$ ) will be:

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

Answer 1

Answer: (C)

The Ideal gas equation is given by
$PV=nRT$ ... (i)
For oxygen, $P=1atm$ , $V=1L$ , $n=n_{\text{o}_{2}}$
Therefore, Eq. (i) becomes
$\therefore 1\times 1=n_{O_{2}}RT$
$\therefore n_{O_{2}}=\frac{1}{R T}$
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 the mixture of gas
$P_{mix}V_{mix}=n_{mix}RT$
​
Here, $n_{\text{mix}}=n_{\text{o}_{2}}+n_{\text{N}_{2}}$
$\frac{P_{\text{mix}} V_{\text{mix}}}{R T}=\frac{1}{R T}+\frac{1}{R T}$
$\Rightarrow P_{mix}V_{mix}=2$
$\Rightarrow P_{mix}=2\,atm\,$