One litre of oxygen at a pressure of $1 atm$ and two litres of nitrogen at a pressure of $0.5 atm$ 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

1 atm

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3 atm

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2 atm

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4 atm

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Solution

The correct option is C $2 atm$
Given:
Volume of oxygen, $V_{1} = 1 L$
Pressure of oxygen, $P_{1} = 1 atm$
Volume of nitrogen, $V_{2} = 2 L$
Pressure of nitrogen, $P_{2} = 0.5 atm$
Volume of vessel, $V = 1 L$
To find:
Final pressure of the mixture, $P = ?$

We know that ideal gas equation is given by,
$P V = n R T$
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, $n = \frac{P V}{R T}$ .............(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),
$\frac{P V}{R T} = \frac{P_{1} V_{1}}{R T_{1}} + \frac{P_{2} V_{2}}{R T_{2}}$
$\Rightarrow P = \frac{P_{1} V_{1} + P_{2} V_{2}}{V}$
[ given temperature is same for both gases and does not change on mixing i.e $T_{1} = T_{2} = T$ ]
$\Rightarrow P = \frac{1 \times 1 + 0.5 \times 2}{1}$
$\Rightarrow P = 2 atm$

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Ideal Gas Equation

Standard XII Physics

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One litre of oxygen at a pressure of 1 atm and two litres of nitrogen at a pressure of 0.5 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

One litre of oxygen at a pressure of $1 atm$ and two litres of nitrogen at a pressure of $0.5 atm$ 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