
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
161.4k+ views
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.
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.
Recently Updated Pages
Two pi and half sigma bonds are present in A N2 + B class 11 chemistry JEE_Main

Which of the following is most stable A Sn2+ B Ge2+ class 11 chemistry JEE_Main

The enolic form of acetone contains a 10sigma bonds class 11 chemistry JEE_Main

The specific heat of metal is 067 Jg Its equivalent class 11 chemistry JEE_Main

The increasing order of a specific charge to mass ratio class 11 chemistry JEE_Main

Which one of the following is used for making shoe class 11 chemistry JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

NIT Cutoff Percentile for 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts of Chemistry

NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Degree of Dissociation and Its Formula With Solved Example for JEE
