Question

In a 20 liter flask, 60 liters of $ {{\text{N}}_{\text{2}}} $ at $ 760{\text{ mm}} $ , 32 liters of $ {{\text{H}}_{\text{2}}} $ at $ 740{\text{ mm}} $ are mixed. The total pressure of the mixture at the same temperature is:
(A) 3740 mm
(B) 1120 mm
(C) 2442 mm
(D) 3464 mm

Answer 1

Hint: To answer this question, you must recall the ideal gas equation and the kinetic molecular gas theory. The ideal gas equation gives a relation between the properties that define any gas, namely, pressure exerted by the gas, volume occupied by the gas, number of moles of the gas and the temperature at which the system is maintained.

Formula used: Ideal Gas equation: $ {\text{PV}} = {\text{nRT}} $
Where, $ {\text{P}} $ denotes the pressure exerted by the gas on the walls of the container
 $ {\text{V}} $ denotes the volume occupied by the gas or the volume of the given container containing the gas
 $ {\text{n}} $ denotes the number of moles of the gas
And, $ {\text{T}} $ denotes the temperature of the gas.

Complete step by step solution
In the given question, we have two different gases, kept at different temperature and pressure conditions. When mixed together, the pressure and the volume of the gases will change. However, the number of moles of the individual gases would remain same. Thus, using the ideal gas equation, we can find the initial and final moles of the gases.
We know that, $ {\text{PV}} = {\text{nRT}} $
We can write the equation as $ {\text{n}} = \dfrac{{{\text{PV}}}}{{{\text{RT}}}} $
So, the number of moles of nitrogen gas in initial conditions $ = {{\text{n}}_{{{\text{N}}_{\text{2}}}}} = \dfrac{{{\text{PV}}}}{{{\text{RT}}}} = \dfrac{{{\text{760 \times 60}}}}{{{\text{RT}}}} = \dfrac{{45600}}{{{\text{RT}}}} $
The number of moles of hydrogen gas in initial conditions $ = {{\text{n}}_{{{\text{H}}_2}}} = \dfrac{{740 \times 32}}{{{\text{RT}}}} = \dfrac{{23680}}{{{\text{RT}}}} $
Thus, the number of moles in the mixture are $ = \dfrac{{69280}}{{{\text{RT}}}} $
Now, we can find the pressure of the mixture using the Ideal gas Equation as $ {\text{P}} = \dfrac{{{\text{nRT}}}}{{\text{V}}} = \dfrac{{69280}}{{{\text{RT}}}} \times \dfrac{{{\text{RT}}}}{{20}} = 3464{\text{ mm}} $
The correct answer is D.

Note
The ideal gas equation does not apply to real gases. A real gas shows ideal behaviour at low pressure and high temperature conditions. The real gas equation is a modified version of the ideal gas equation that takes into consideration the factors that were ignored by the ideal gas law.