NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter: Gases and Liquids is the best study material you can utilise for effective exam preparation of class 11 and entrance examinations.
NCERT Solutions for Class 11 chemistry chapter 5 is an important chapter in a student’s life. The topics which are taught in class 11 are the basics to the topics to be taught in class 12. Students who want to go for further studies after class 12 should try to understand the topics of class 11 as they are often asked in competitive examinations.
Class 11 NCERT Solutions for chemistry chapter 5 States of Matter: Gases and Liquids
Students are advised to solve the NCERT questions given at the end of each chapter. Solving the questions provided in the NCERT class 11 Chemistry textbook help to examine yourself about the chapter you have just read. After solving all the questions check in which topic or subtopic you have written an incorrect answer or you did not attend the question. If you think you need some more time to study then plan your time table accordingly.
This solution has exemplar problems in MCQ, Long and short answering format along with questions from previous year question papers and sample papers. Practising this solution will give a lot of confidence as they will help you determine your strengths as well as weaknesses.
Concepts involved in Class 11 Chemistry Chapter 5 States of Matter: Gases and Liquids

Intermolecular Forces
 Dispersion Forces Or London Forces
 Dipoledipole Forces
 Dipoleinduced Dipole Forces
 Hydrogen Bond
 Thermal Energy
 Intermolecular Forces Vs Thermal Interactions
 The Gaseous State Ex

The Gas Laws Ex
 Boyle’s Law (Pressurevolume Relationship)
 Charles’ Law (Temperaturevolume Relationship)
 Gay Lussac’s Law (Pressuretemperature Relationship)
 Avogadro Law (Volume – Amount Relationship)

Ideal Gas Equation
 Density And Molar Mass Of A Gaseous Substance
 Dalton’s Law Of Partial Pressures
 Kinetic Molecular Theory Of Gases
 Behaviour Of Real Gases: Deviation From Ideal Gas Behaviour
 Liquefaction Of Gases

Liquid State
 Vapour Pressure
 Surface Tension Ex 5.10.3 – Viscosity.
Class 11 Chemistry Chapter 5 States of Matter: Gases and Liquids Important questions
Q1. Calculate the minimum pressure required to compress 500 \(dm^{ 3 }\)of air at 1 bar to 200 \(dm^{ 3 }\) at \(30^{\circ} C \) ?
Answer:
Initial pressure, P_{1} = 1 bar
Initial volume, V_{1} = 500 \(dm^{ 3 }\)
Final volume, V_{2} = 200 \(dm^{ 3 }\)
As the temperature remains same,the final pressure (P_{2}) can be calculated with the help of Boyle’s law.
Acc. Boyle’s law,
P_{1}V_{1} = P_{2}V_{2}
P_{2 }= \(\frac{P_{1}V_{1}}{V_{2}}\)
= \(\frac{1 \; \times \; 500}{200}\)
= 2.5 bar
∴ the minimum pressure required to compress is 2.5 bar.
Q2. A container with a capacity of 120 mL contains some amount of gas at \(35^{\circ}\) C and 1.2 bar pressure. The gas is transferred to another container of volume 180 mL at \(35^{\circ} C\). Calculate what will be the pressure of the gas?
Answer:
Initial pressure, P_{1} = 1.2 bar
Initial volume, V_{1} = 120 mL^{ }
Final volume, V_{2} = 180 mL
As the temperature remains same, final pressure (P_{2}) can be calculated with the help of Boyle’s law.
According to the Boyle’s law,
P_{1}V_{1} = P_{2}V_{2}
P_{2 }= \(\frac{P_{1}V_{1}}{V_{2}}\)
= \(\frac{1.2 \; \times \; 120}{180}\)
= 0.8 bar
Therefore, the min pressure required is 0.8 bar.
Q3. Prove that at a given temp density of a gas is proportional to the gas pressure by using the equation of state pV = nRT.
Answer:
The equation of state is given by,
pV = nRT ……..(1)
Where, p = pressure
V = volume
N = number of moles
R = Gas constant
T = temp
\(\frac{n}{V}\) = \(\frac{p}{RT}\)Replace n with \(\frac{m}{M}\), therefore,
\(\frac{m}{MV}\) = \(\frac{p}{RT}\)……..(2)Where, m = mass
M = molar mass
But, \(\frac{m}{V}\) = d
Where, d = density
Therefore, from equation (2), we get
\(\frac{d}{M}\) = \(\frac{p}{RT}\)d = (\(\frac{M}{RT}\)) p
d \(\propto\) p
Therefore, at a given temp, the density of gas (d) is proportional to its pressure (p).
Q4. At \(0^{\circ}\) C, the density of a certain oxide of a gas at 2 bars is equal to that of dinitrogen at 5 bars. Calculate the molecular mass of the oxide.
Answer:
Density (d) of the substance at temp (T) can be given by,
d = \(\frac{Mp}{RT}\)
Now, density of oxide (d_{1}) is as given,
\(d_{1}\) = \(\frac{M_{1}p_{1}}{RT}\)Where, M_{1 }= mass of the oxide
p_{1 }= pressure of the oxide
Density of dinitrogen gas (d_{2}) is as given,
\(d_{2}\) = \(\frac{M_{1}p_{2}}{RT}\)Where, M_{2 }= mass of the oxide
p_{2 }= pressure of the oxide
Acc to the question,
d_{1 }= d_{2}
Therefore, \(M_{1}p_{1} = M_{2}p_{2}\)
Given:
\(p_{1}\) = 2 bar \(p_{2}\) = 5 barMolecular mass of nitrogen, \(M_{2}\) = 28 g/mol
Now, \(M_{1}\)
= \(\frac{M_{ 2 }p_{2}}{p_{ 1 }}\)
= \(\frac{ 28 × 5 }{ 2 }\)
= 70 g/mol
Therefore, the molecular mass of the oxide is 70 g/mol.
Q5. A pressure of 1 g of an ideal gas X at \(27^{\circ}\) C is found to be 2 bars. When 2 g of another ideal gas is added in the same container at same temp the pressure becomes 3 bars. Find the relation between their molecular masses.
Answer:
For ideal gas A, the ideal gas equation is given by,
\(p_{X}V = n_{X}RT\)……(1)Where \(p_{X}\) and \(n_{X}\) represent the pressure and number of moles of gas X.
For ideal gas Y, the ideal gas equation is given by,
\(p_{Y}V = n_{Y}RT\)……(2)Where, \(p_{Y}\) and \(n_{Y}\) represent the pressure and number of moles of gas Y.
[V and T are constants for gases X and Y]From equation (1),
\(p_{ X }V = \frac{m_{ X }}{M_{ X }}\) RT \(\frac{p_{ X }M_{ X }}{m_{ X }}\) = \(\frac{ R T}{ V }\) ……(3)From equation (2),
\(p_{ Y }V =\frac{m_{ Y }}{M_{ Y }}\) RT \(\frac{p_{ Y }M_{ Y }}{m_{ Y }}\) = \(\frac{ R T}{V}\) …… (4)Where, \(M_{ X }\) and \(M_{ Y }\) are the molecular masses of gases X and Y respectively.
Now, from equation (3) and (4),
\(\frac{p_{ X }M_{ X }}{m_{ X }}\) = \(\frac{p_{ Y }M_{ Y }}{m_{ Y }}\) ….. (5)Given,
\(m_{ X }\) = 1 g \(p_{ X }\) = 2 bar \(m_{ Y }\) = 2 g \(p_{ Y }\) = (3 – 2) = 1 bar (Since total pressure is 3 bar)Substituting these values in equation (5),
\(\frac{2 \; \times \; M_{X} }{1}\) = \(\frac{1 \; \times \; M_{Y} }{2}\)4 \(M_{ X }\) = \(M_{ Y }\)
Therefore, the relationship between the molecular masses of X and Y is,
4 \(M_{ X }\) = \(M_{ Y }\)
Q6. The drain cleaner has small bits of aluminum, which react with caustic soda to produce dihydrogen. What volume of dihydrogen at \(20^{\circ}\) C and 1 bar will be released when 0.15 g of aluminum reacts?
Answer:
The reaction of aluminum with caustic soda is as given below:
2Al + 2NaOH + 2H_{2}O \(\rightarrow\) 2NaAlO_{2} + 3H_{2}
At Standard Temperature Pressure (273.15 K and 1 atm), 54 g ( 2 × 27 g) of Al gives 3 ×22400 mL of H_{2}.
Therefore, 0.15 g Al gives:
= \(\frac{3 \; \times \; 22400 \; \times \; 0.15}{54}\) mL of H_{2}
= 186.67 mL of H_{2}
At Standard Temperature Pressure,
\(p_{ 1 }\) = 1 atm \(V_{ 1 }\) = 186.67 mL \(T_{ 1 }\) = 273.15 KLet the volume of dihydrogen be \(V_{ 2 }\) at \(p_{ 2 }\) = 0.987 atm (since 1 bar = 0.987 atm) and \(T_{ 2 }\) = \(20^{\circ}\) C = (273.15 + 20) K = 293.15 K.
Now,
\(\frac{p_{ 1 }V_{ 1 }}{T_{ 1 }}\) = \(\frac{p_{ 2 }V_{ 2 }}{T_{ 2 }}\) \(V_{ 2 } = \frac{p_{ 1 }V_{ 1 }T_{ 2 }}{p_{ 2 }T_{ 1 }}\)= \(\frac{1 \; \times \; 186.67 \; \times \; 293.15}{0.987 \; \times \; 273.15}\)
= 202.98 mL
= 203 mL
Hence, 203 mL of dihydrogen will be released.
Q7. Calculate the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 \(dm^{ 3 }\) at flask at 27°.
Answer:
It is known that,
p = \(\frac{m}{M}\) \(\frac{RT}{V}\)
For methane (CH_{4}),
\(p_{CH_{ 4 }}\)= \(\frac{ 3.2 }{ 16 }\) × \(\frac{8.314 \; \times \;300 }{9 \; \times \; 10^{ – 3 }}\) [Since 9 dm^{3 }= \(9 \; \times \; 10^{ 3 }\) m^{3} \(27^{\circ}\)C = 300 K]
= 5.543 × \(10^{ 4 }\) Pa
For carbon dioxide (CO_{2}),
\(p_{CO_{ 2 }}\)= \(\frac{ 4.4 }{ 44 }\) × \(\frac{8.314 \; \times \;300 }{9 \; \times \; 10^{ – 3 }}\)
= 2.771 × \(10^{4}\) Pa
Total pressure exerted by the mixture can be calculated as:
p = \(p_{CH_{ 4 }}\) + \(p_{CO_{ 2 }}\)
= (5.543 × \(10^{ 4 }\) + 2.771 × \(10^{4}\)) Pa
= 8.314 × \(10^{ 4 }\) Pa
Q8. Calculate the pressure of the gaseous mixture when 0.5 L of H_{2} at 0.8 bars and 2.0 L of dioxygen at 0.7 bars are introduced in a 1L container at \(27^{\circ}\).
Answer:
Let the partial pressure of \({H_{ 2 }}\) in the container be \(p_{H_{ 2 }}\).
Now,
\({p_{ 1 }}\) = 0.8 bar \({p_{ 2 }}\) = \(p_{H_{ 2 }}\) \({V_{ 1 }}\) = 0.5 L \({V_{ 2 }}\) = 1 LIt is known that,
\({p_{ 1 }}\) \({V_{ 1 }}\) = \({p_{ 2 }}\) \({V_{ 2 }}\) \({p_{ 2 }}\) = \(\frac{p_{ 1 } \; \times \; V_{ 1 }}{V_{ 2 }}\) \(p_{H_{ 2 }}\) = \(\frac{ 0.8 \; \times \; 0.5 }{ 1 }\)= 0.4 bar
Now, let the partial pressure of O_{2} in the container be \(p_{O_{ 2 }}\).
Now,
\({p_{ 1 }}\) = 0.7 bar \({p_{ 2 }}\) = \(p_{O_{ 2 }}\) \({V_{ 1 }}\) = 2.0 L \({V_{ 2 }}\) = 1 L \({p_{ 1 }}\) \({V_{ 1 }}\) = \({p_{ 2 }}\) \({V_{ 2 }}\) \({p_{ 2 }}\) = \(\frac{p_{ 1 } \; \times \; V_{ 1 }}{V_{ 2 }}\) \(p_{O_{ 2 }}\) = \(\frac{ 0.7 \; \times \; 20 }{ 1 }\)= 1.4 bar
Total pressure of the gas mixture in the container can be obtained as:
\(p_{total}\) = \(p_{H_{ 2 }}\) + \(p_{O_{ 2 }}\)= 0.4 + 1.4
= 1.8 bar
Q9. A density of a gas is 5.46 g/dm^{3} at \(27^{\circ}\)C at 2 bar pressure. Calculate its density at Standard Temperature Pressure.
Answer:
Given,
d_{1} = 5.46 g/dm^{3}
p_{1 }= 2 bar
T_{1} = \(27^{\circ}\)C = (27 + 273) K = 300 K
p_{2 }= 1 bar
T_{2 }= 273 K
d_{2 }= ?
The density ( d_{2 }) of the gas at STP can be calculated using the equation,
d = \(\frac{Mp}{RT}\) \(\frac{d_{1}}{d_{2}}\) = \(\frac{\frac{M \;p_{ 1 }}{R \; T_{ 1 }}}{\frac{M \; p_{ 2 }}{R \; T_{ 2 }}}\) \(\frac{d_{1}}{d_{2}}\) = \(\frac{p_{1} \; T_{2}}{p_{2} \; T_{1}}\)
d_{2} = \(\frac{p_{2} \; T_{1} \; d_{1}}{p_{1} \; T_{2}}\)
= \(\frac{1 \; \times 300 \; \times 5.46}{2 \; \times 273}\)
= 3 g dm^{3}
Hence, the density of the gas at STP will be 3 g dm^{3}
Q10. 34.05 mL of phosphorus vapour has weight 0.0625 g at \(546^{\circ} C\) and 0.1 bar pressure. Calculate the molar mass of phosphorus.
Answer:
Given,
p = 0.1 bar
V = 34.05 mL = 34.05 × 10^{3} \(dm^{ 3 }\)
R = 0.083 bar \(dm^{ 3 }\) at^{ }K^{1} mol^{1}
T = \(546^{\circ} C \) = (546 + 273) K = 819 K
The no of moles (n) can be calculated using the ideal gas equation as:
pV = nRT
n = \(\frac{ pV }{ RT }\)
= \(\frac{ 0.1 \times 34.05 × 10^{3} }{ 0.083 \times 819 }\)
= 5.01 × 10^{5} mol
Therefore, molar mass of phosphorus = \(\frac{ 0.0625 }{ 5.01 \times 10^{5} }\)
= 1247.5 g mol^{1}
Q11. A student forgot to add the reaction mixture to the container at \(27^{\circ}\) C but instead, he placed the container on the flame. After a lapse of time, he came to know about his mistake, and using a pyrometer he found the temp of the container \(477^{\circ}\) C. What fraction of air would have been expelled out?
Answer:
Let the volume of the container be V.
The volume of the air inside the container at \(27^{\circ}\) C is V.
Now,
V_{1} = V
T_{1} = \(27^{\circ}\) C = 300 K V_{2} = ?
T_{2} = \(477^{\circ}\) C = 750 K
Acc to Charles’s law,
\(\frac{V_{1}}{T_{1}}\) = \(\frac{V_{2}}{T_{2}}\) \(V_{ 1 }\) = \(\frac{V_{ 1 }T_{ 2 }}{T_{ 1 }}\)= \(\frac{750V }{300 }\)
= 2.5 V
Therefore, volume of air expelled out
= 2.5 V – V = 1.5 V
Hence, fraction of air expelled out
= \(\frac{1.5V }{ 2.5V }\)
= \(\frac{3 }{ 5 }\)
Q12. What is the temp of 4.0 mol of gas occupying 5 dm^{3} at 3.32 bar? ( R = 0.083 bar \(dm^{ 3 }\) at K^{1} mol^{1}).
Answer:
Given,
N= 4.0 mol
V = 5 \(dm^{ 3 }\)
p = 3.32 bar
R = 0.083 bar \(dm^{ 3 }\) at^{ }K^{1} mol^{1}
The temp (T) can be calculated using the ideal gas equation as:
pV = nRT
T = \(\frac{ pV }{ nR }\)
= \(\frac{3.32 \; \times \; 5 }{ 4 \;\times \; 0.083 }\)
= 50 K
Therefore, the required temp is 50 K.
Q13. What is the total no of electrons present in 1.4 g of dinitrogen gas?
Answer:
Molar mass of dinitrogen (N_{2}) = 28 g mol^{1}
Thus, 1.4 g of N_{2 }
= \(\frac{ 1.4 }{ 28 }\)
= 0.05 mol
= 0.05 × 6.02 × 10^{23} no of molecules
= 3.01 × 10^{23 }no. of molecules
Now, 1 molecule of N_{2 }has 14 electrons.
Therefore, 3.01 × 10^{23 }molecules of N_{2} contains,
= 14 × 3.01 × 1023
= 4.214 × 10^{23} electrons
Q14. How much time would it take to distribute 1 Avogadro no. of wheat grains, if 10^{10} grains are distracted each second?
Answer:
Avogadro no. = 6.02 × 10^{23}
Therefore, time taken
= \(\frac{6.02 \; \times \; 10^{23}}{10^{10}} s\)
= 6.02 × 10^{13} s
= \(\frac{6.02 \; \times \; 10^{23}}{60 \; \times \; 60 \; \times \; 24 \; \times \; 365 } years\)
= 1.909 × 10^{6} years
Therefore, the time taken would be 1.909 × 10^{6} years.
Q15. What is the total pressure in the mixture of 4 g of dihydrogen and 8 g of dioxygen in a container of 1 \(dm^{ 3 }\) at K^{1} mol^{1}?
Answer:
Given:
Mass of O_{2 }= 8 g
No. of moles
= \(\frac{ 8 }{ 32 }\)
= 0.25 mole
Mass of H_{2} = 4 g
No. of moles
= \(\frac{ 4 }{ 2 }\)
= 2 mole
Hence, total no of moles in the mixture
= 0.25 + 2
= 2.25 mole
Given:
V = 1 \(dm^{ 3 }\)
n = 2.25 mol
R = 0.083 bar \(dm^{ 3 }\) at K^{1} mol^{1}
T = \(27^{\circ}\) C = 300 K
Total pressure :
pV = nRT
p = \(\frac{ nRT }{ V }\)
= \(\frac{225 \; \times \; 0.083 \; \times \; 300 }{ 1 }\)
= 56.025 bar
Therefore, the total pressure of the mixture is 56.025 bar.
Q16. The difference between the mass of displaced air and the mass of the balloon is known as pay load. What is the pay load when a balloon of radius is 10 m, mass 100 kg is filled with helium at 1.66 bar at \(27^{\circ}\) C.
(Density of air = 1.2 kg m^{3} and R = 0.083 bar \(dm^{ 3 }\) at K^{1} mol^{1})
Answer:
Given:
r = 10 m
Therefore, volume of the balloon
= \(\frac{4}{3}\) πr^{3}
= \(\frac{ 4 }{ 3 }\; \times \; \frac{ 22 }{ 7 } \; \times \; 10^{3}\)
= 4190.5 m^{3} (approx.)
Therefore, the volume of the displaced air
= 4190.5 × 1.2 kg
= 5028.6 kg
Mass of helium,
= \(\frac{ MpV }{ RT }\)
Where, M = 4 × 10^{3} kg mol^{1}
p = 1.66 bar
V = volume of the balloon
= 4190.5 m^{3}
R = 0.083 0.083 bar \(dm^{ 3 }\) at K^{1} mol^{1}
T = 27 °C = 300 K
Then,
m = \(\frac{ 4 \; \times \; 10^{3} \; \times \; 1.66 \; \times \; 4190.5 \; \times \; 10^{3}}{0.083 \; \times \; 300}\)
= 1117.5 kg (approx.)
Now, total mass with helium,
= (100 + 1117.5) kg
= 1217.5 kg
Therefore, pay load,
= (5028.6 – 1217.5)
= 3811.1 kg
Therefore, the pay load of the balloon is 3811.1 kg.
Q17. What is the volume occupied by 8.8 g of CO_{2} at \(31.1^{\circ}\) C and 1 bar pressure? Given that R = 0.083 bar \(dm^{ 3 }\) at K^{1} mol^{1.}
Answer:
pVM = mRT
V = \(\frac{mRT}{Mp}\)
Given:
m = 8.8 g
R = 0.083 bar \(dm^{ 3 }\) at K^{1} mol^{1.}
T = 31.1 °C = 304.1 K
M = 44 g
p = 1 bar
Thus, Volume (V),
= \(\frac{8.8 \; \times \; 0.083 \; \times \; 304.1}{ 44 \; \times \; 1}\)
= 5.04806 L
= 5.05 L
Therefore, the volume occupied is 5.05 L.
Q18. 2.9 g of a gas at \(95^{\circ}\) C occupied the same volume as 0.184 g of dihydrogen at \(17^{\circ}\) C, at the same pressure. Calculate the molar mass of the gas.
Answer:
Volume,
V = \(\frac{mRT}{Mp}\)
= \(\frac{0.184 \; \times \; R \; \times \; 290}{ 2 \; \times \; p}\)
Let M be the molar mass of the unknown gas.
Volume occupied by the unknown gas is,
= \(\frac{mRT}{Mp}\)
= \(\frac{2.9 \; \times \; R \; \times \; 368}{ M \; \times \; p}\)
According to the ques,
\(\frac{0.184 \; \times \; R \; \times \; 290}{ 2 \; \times \; p}\) = \(\frac{2.9 \; \times \; R \; \times \; 368}{ M \; \times \; p}\) \(\frac{0.184 \; \times \; 290}{ 2 }\) = \(\frac{2.9 \; \times \; 368}{ M }\)M = \(\frac{2.9 \;\times \; 368 \; \times \; 2}{0.184 \; \times \; 290}\)
= 40 g mol^{1}
Therefore, the molar mass of the gas is 40 g mol^{1}
Q19. A mixture of dioxygen and dihydrogen at 1 bar pressure has 20% by weight of dihydrogen. What is the partial pressure of dihydrogen?
Answer:
Let the weight of dihydrogen be 20 g.
Let the weight of dioxygen be 80 g.
No. of moles of dihydrogen (n_{H2}),
= \(\frac{20}{2}\)
= 10 moles
No. of moles of dioxygen (n_{O2}),
= \(\frac{80}{32}\)
= 2.5 moles
Given:
p_{total} = 1 bar
Therefore, partial pressure of dihydrogen (p_{H2}),
= \(\frac{ n_{H_{2}} }{ n_{H_{2}} \; + \; n_{O_{2}} }\) × p_{total}
= \(\frac{ 10 }{ 10 \; + \; 2.5 } \; \times \; 1\)
= 0.8 bar
Therefore, the partial pressure of dihydrogen is 0.8 bar.
Q20. What will be the SI unit for the quantity \(\frac{ pV^{ 2 }T^{ 2 } }{ n }\)?
Answer:
SI unit of pressure, p = \(Nm^{ 2 }\)
SI unit of volume, V = \(m^{ 3 }\)
SI unit of temp, T = K
SI unit of number of moles, n = mol
Hence, SI unit of \(\frac{ pV^{ 2 }T^{ 2 } }{ n }\) is,
= \(\frac{(Nm^{ 2 }) \; (m^{3})^{2} \; (K)^{2}}{mol}\)
= \(Nm^{ 4 }K^{ 2 }mol^{ 1 }\)
Q21. According to Charles’ law explain why \(273^{\circ}\) C is the lowest possible temp.
Answer:
According to Charles’ law
At constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temp.
It was found that for all gasses (at any given pressure), the plot of volume vs. temp. (in \(^{\circ}\)C) is a straight line.
If we extend the line to zero volume, then it intersects the tempaxis at \(273^{\circ}\) C. That is the volume of any gas at\(273^{\circ}\) C is 0. This happens because all gasses get transferred into liquid form before reaching \(273^{\circ}\) C.
Therefore, it can be said that\(273^{\circ}\) C is the lowest possible temp.
Q22. Critical temp of methane and carbon dioxide are \(81.9^{\circ}\) C and \(31.1^{\circ} C\) respectively. Which of the following have stronger intermolecular forces? Why?
Answer:
If the critical temp of a gas is higher then it is easier to liquefy. That is the intermolecular forces of attraction among the molecules of gas are directly proportional to its critical temp.
Therefore, in CO_{2} intermolecular forces of attraction are stronger.
Q23. What is the physical significance of Van der Waals parameters?
Answer:
The physical significance of ‘a’:
The magnitude of intermolecular attractive forces within gas is represented by ‘a’.
The physical significance of ‘b’:
The volume of a gas molecule is represented by ‘b’.
Also Access 
NCERT Exemplar for class 11 chemistry Chapter 5 
CBSE Notes for class 11 chemistry Chapter 5 
The NCERT Class 11 chemistry States of Matter: Gases and Liquids has many important topics which are very much important from the examination point of view like: Intermolecular forces, van der Waals forces, Dispersion Forces or London Forces, DipoleDipole Forces, Boyle’s Law (PressureVolume Relationship), the gas laws:Charles’ Law, Gay Lussac’s Law, Avogadro Law and Ideal gas Equation.
These chapters are also important from the competitive examination point of view. As chemistry is a common subject for Engineering and medical entrance examinations so students are advised to study chemistry in a better way.