Question

What is the pressure of 2 moles of ammonia at \[{\rm{27^\circ C}}\] when its volume is 5 litres in the Van der Waals equation? (a=4.17,b=0.03711)
A. 10.33 atm
B. 9.33 atm
C. 9.74 atm
D. 9.2 arm

Answer 1

Hint: All the gases do not follow the gas laws. These gases do not have ideal behaviour. These gases are called real gases. The ideal gas equation, \[{\rm{PV = nRT}}\] does not work on these gases. They follow an equation called Van der Waals equation.

Formula Used:
\[\left( {{\rm{P + }}\frac{{{\rm{a}}{{\rm{n}}^{\rm{2}}}}}{{{{\rm{V}}^{\rm{2}}}}}} \right)\left( {{\rm{V - nb}}} \right){\rm{ = nRT}}\]
where
P=pressure of the gas
V=volume of the gas
a and b=van der Waals constants
R=universal gas constant
T=temperature
n=no.of moles

Complete Step by Step Solution:
The kinetic theory of gases makes two assumptions.
A. There is no force of attraction between the molecules of a gas
B. Volume of the molecules of a gas is negligibly small in comparison to the space occupied by the gas.
These assumptions are not correct for all gases at all temperatures and pressure.

Gases show divergence from ideal gas as molecules are not far apart from each other and interact with each other. At high pressure, these molecules are near to each other and don't hit the walls of the container. But instead, they are pulled back by other molecules due to the attractive force. So, the pressure of real gas is less than that of an ideal gas. So, a correction term an2V2 is added to the pressure in the ideal gas equation. "a" is a constant here.

The volume in the case of real gases acts as small spheres because of the repulsive forces that act on the gases. These forces arise when gas molecules are in contact with each other.
The volume decreases so, the correction term nb is subtracted from the total volume.

The equation becomes as follows:-
\[\left( {{\rm{P + }}\frac{{{\rm{a}}{{\rm{n}}^{\rm{2}}}}}{{{{\rm{V}}^{\rm{2}}}}}} \right)\left( {{\rm{V - nb}}} \right){\rm{ = nRT}}\]
V=volume of the gas
a and b=van der Waals constants
R = Universal gas constant
T = Temperature
n = No.of moles

Here we are given,
T = \[{\rm{27^\circ C = }}\left( {{\rm{27 + 273}}} \right){\rm{K = 300K}}\]
n = 2 moles
Volume = 5 litres
a = 4.17
b = 0.03711
R = \[{\rm{0}}{\rm{.0821atm L mol}}{{\rm{l}}^{{\rm{ - 1}}}}{{\rm{K}}^{{\rm{ - 1}}}}\]

We have to find the pressure using the van der Waals equation.
Putting the above values on the equation, we get
\[\left( {{\rm{P + }}\frac{{{\rm{4}}{\rm{.17}}{{\left( {\rm{2}} \right)}^{\rm{2}}}}}{{{{\left( {\rm{5}} \right)}^{\rm{2}}}}}} \right)\left( {{\rm{5 - 2 \times 0}}{\rm{.03711}}} \right){\rm{ = 2 \times 0}}{\rm{.0821 \times 300}}\]
\[ \Rightarrow \left( {{\rm{P + }}\frac{{{\rm{16}}{\rm{.68}}}}{{{\rm{25}}}}} \right)\left( {{\rm{5 - 0}}{\rm{.074}}} \right){\rm{ = 49}}{\rm{.26}}\]
\[ \Rightarrow \left( {{\rm{P + 0}}{\rm{.667}}} \right)\left( {{\rm{4}}{\rm{.926}}} \right){\rm{ = 49}}{\rm{.26}}\]
\[ \Rightarrow {\rm{P + 0}}{\rm{.667 = 10}}\]
So, P=9.33atm
So, the pressure of ammonia gas is 9.33 atm.
So, option B is correct.

Note: Real gases exhibit ideal behaviour when conditions of temperature and pressure are such that the intermolecular force of attraction is insignificant. The real gases show ideal behaviour when pressure approaches zero. At very low-temperature intermolecular forces become considered as the molecule travels with low average speed due to which the attractive forces become significant. So, real gases at low pressure and high temperature follow ideal behaviour.