Question

One litre of oxygen at a pressure of $1\text{atm}$ and two litres of nitrogen at a pressure of $0.5\text{atm}$ at the same temperature are introduced into a vessel of one litre volume. If there is no change in the temperature, then final pressure of the mixture of gases is

• A. $1\text{atm}$
• B. $3\text{atm}$
• C. $2\text{atm}$
• D. $4\text{atm}$

Answer 1

The correct option is C $2\text{atm}$

Given:
Volume of oxygen, $V_1=1\text{L}$
Pressure of oxygen, $P_1=1\text{atm}$
Volume of nitrogen, $V_2=2\text{L}$
Pressure of nitrogen, $P_2=0.5\text{atm}$
Volume of vessel, $V=1\text{L}$
To find:
Final pressure of the mixture, $P=?$

We know that ideal gas equation is given by,
$PV=nRT$
So, number of moles of oxygen, $n_1=\frac{P_1{V}_1}{R{T}_1}$............(1)
Similarly, number of moles of nitrogen, $n_2=\frac{P_2{V}_2}{R{T}_2}$............(2)
Also, number of moles of mixture, $n=\frac{PV}{RT}$ .............(3)
Number of moles of the mixture will be equal to the sum of number of moles of oxygen and nitrogen.
From (1), (2) and (3),
$\frac{PV}{RT}=\frac{P_1{V}_1}{R{T}_1}+\frac{P_2{V}_2}{R{T}_2}$
$\Rightarrow P=\frac{P_1{V}_1+P_2{V}_2}{V}$
[ given temperature is same for both gases and does not change on mixing i.e $T_1=T_2=T$ ]
$\Rightarrow P=\frac{1\times 1+0.5\times 2}{1}$
$\Rightarrow P=2\text{atm}$