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NCERT Solutions For Class 11 Chemistry Chapter 5

The Class 11 Chemistry Chapter 5 – States of Matter discusses the concepts including gaseous state, thermal energy, different laws, and various types of intermolecular force. To help students get a better understanding of the topics, the NCERT textbook has related questions at the end of Chapter 5.

Students can refer to NCERT Solutions for Class 11 Chemistry Chapter 5 – States of Matter to get accurate answers to the textbook questions. Subject-matter experts at Extramarks have prepared these solutions ensuring that they are easy to comprehend and have authentic answers to all the questions.

NCERT Solutions for Class 11 Chemistry Chapter 5 – States of Matter

Access NCERT Solutions for Class 11 Chemistry Chapter 5 – States of matter

Let us quickly give a glance at the Properties of the Four States of Matter

Solids: In solids, particles can vibrate at their own position, but they are not free to move as the molecules, atoms, and ions are all packed together closely. The application of an external force on solids can change their volume and shape.

Liquid: In liquids, the molecules, atoms, and ions are incompressible. When the pressure and temperature are constant, liquids have a fixed volume. Solids being exposed to high temperatures can become liquids, depending upon the properties of pressure.

Gas: In gases, the interacting particles have an intermolecular force of attraction with enough kinetic energy which changes the molecular forces to zero. The molecular space between the molecules in gases is large and hence liquids can be transformed into gaseous states.

Plasma: In this state of matter the ions are highly electrically conductive. The behaviour of matter is dominated by them due to the magnetic fields and currents they produce.

Factors like pressure, temperature, mass, and volume impact the behaviour of matter in different states of matter. Chemical reactions depend on the physical state of the matter, however, the chemical properties of a substance remain the same despite the change in state.

Intermolecular Forces

Intermolecular forces are forces that interact between the particles of matter, however, these are not to be confused with electrostatic forces. The force that exists between two oppositely charged ions is called electrostatic force. The types of Intermolecular forces in existence are:

Dispersion Forces or London Forces
Dipole-Dipole Forces
Dipole-Induced Dipole Forces
Hydrogen Bond

Thermal Energy

Thermal energy is the energy that arises when there is a motion of atoms and molecules in a body. It is the quantity of the particles’ average kinetic energy. The thermal energy increases with the increase in temperature.

Intermolecular Forces Vs Thermal Interactions

The thermal energy of particles keeps the molecules apart while the intermolecular forces of a substance keep them together. Hence, a balance between thermal energy and intermolecular forces of molecules results in the three states of matter.

Gaseous State

The physical properties of the molecules in the gaseous state are:

Pressure is exerted by gases in all directions.
Gases have a lower density as compared to solids and liquids.
Gas molecules are highly compressible.
Gases are completely miscible,i.e. they mix completely in all proportions.
Gases do not have volume or a definite shape and can take the shape of the container.
Gases have negligible intermolecular forces of attraction. Some experiments conducted led to the discovery of certain laws that regulate the behaviour of the molecules of gases.

Boyle’s Law

Under Boyle’s law, the pressure of the molecules of gas and its volume are inversely proportional under isothermal conditions.

P1V1=P2V2 (Under constant T)

Here V= Volume, T= Temperature, and P= Pressure.

If there is an expansion in a fixed amount of gas occupying volume V1 at pressure P1 under constant temperature T, it leads volume to become V2 and pressure to P2.

Charles’s Law

The direct relation between volume and absolute temperature under isobaric conditions is covered in Charles’ Law. This law discusses how gases expand with a rise in temperature and decrease in volume with a decrease in temperature.

V1=T2

V2= T1

Where V= Volume, P= Pressure, and T= Temperature

Gay-Lussac’s Law

Gay Lussac’s Law says that pressure exerted by a fixed amount of gas changes with the change in the absolute temperature of gas.

P ∝ T (at constant V)

Here, P= Pressure, V= Volume and T= Temperature

Avogadro Law

This law says that equal volumes of all gases contain equal numbers of molecules if they are under the same circumstances of pressure and temperature.

V ∝ n (at constant P and T)

Here n = number of molecules, V = Volume, P = Pressure, and T = Temperature.

Ideal Gas Equation

The ideal gas equation is an equation that was formed after combining all the three gas laws – Gay Lussac’s Law, Boyle’s Law, and Avogadro Law.

pV = nRT

Here V = Volume, p = Pressure, R = gas constant, n = number of molecules, and T = Temperature.

Dalton’s Law of Partial Pressure

Dalton’s Law of Partial Pressure says that the sum of partial pressures exerted by a mixture of non-reacting gases is equal to the total pressure exerted by them.

It should be noted here that in a mixture of gases, partial pressure is the pressure exerted by the individual gas.

Ptotal = p1 + p2 + p3 + ……..

Here, p1, p2, ……. are partial pressures.

Kinetic Theory of Gases

A large number of identical molecules are found in gases.
At ordinary temperature, molecules of gas do not have any force of attraction between them which is why gases expand and take up space.
The actual volume occupied by molecules of gases is negligible because the distance between the molecules in gases is pretty large.
The particles of gas travel in all directions in a straight line and collide against each other during their random motion.
There is no loss of kinetic energy as the collision of the particles of gas is elastic in nature.
Every molecule in the gases has a different speed which keeps on changing and so the energy of these molecules also keeps on changing.

Related Questions

Q1. What does matter mean?

A1. Everything that has mass and occupies space is called Matter. A Matter contains minute particles and can exist in multiple states like solid, gas, and liquid. An example of matter can be a stone or fruit juice or a pen or a newspaper. All of these things have mass and occupy space. The behaviour of matter is impacted by several factors like temperature, volume, etc.

Q2. What are the different states of matter?

A2. Matter has four different states;

Solids: This state of matter has compact particles (molecules, atoms, and ions) closely packed with each other.
Liquids: In the liquid state of matter, ions, atoms, and molecules are immiscible.
Gas: In this state, the molecules of gases have large intermolecular spaces between them. Gases also have no fixed volume or shape.
Plasma: In the plasma state of matter, ions are highly electrically conductive.

Students can study further about this chapter by referring to NCERT Solutions Class 11 Chemistry Chapter 5 by Extramarks.

Q3. What are the factors that affect the states of matter?

A3. States of matter are affected by factors like volume, pressure, temperature, and mass. Pressure on an object can cause it to condense, temperature causes an ice cube to melt while mass and volume play their own roles in impacting matter.

Q.1 What will be the minimum pressure required to compress 500 dm³ of air at 1 bar to 200dm³ at 30°C?

Ans.

Given,

Initial pressure, p₁ = 1 bar

Initial volume, V₁ = 500 dm³

Final volume, V₂ = 200 dm³

Since the temperature remains constant, the final pressure (p₂) can be calculated using Boyle’s law.

According to Boyle’s law,

$\begin{array}{l} p_{1} v_{1} = p_{2} v_{2} \\ O r, p_{2} = \frac{p_{1} v_{1}}{v_{2}} \\ = \frac{1 \times 500}{200} b a r \\ = 2.5 bar \end{array}$

Therefore, the minimum pressure required is 2.5 bar.

Q.2 A vessel of 120 mL capacity contains a certain amount of gas at 35 °C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35 °C. What would be its pressure?

Ans.

Given,

Initial pressure, p₁ = 1.2 bar

Initial volume, V₁ = 120 mL

Final volume, V₂ = 180 mL

Since the temperature remains constant, the final pressure (p₂) can be calculated using Boyle’s law.

According to Boyle’s law,

$\begin{array}{l} p_{1} v_{1} = p_{2} v_{2} \\ p_{2} = \frac{p_{1} v_{1}}{v_{2}} \\ = \frac{1.2 \times 120}{180} b a r \\ = 0.8 bar \end{array}$

Therefore, the pressure would be 0.8 bar.

Q.3 Using the equation of state pV = nRT; show that at a given temperature density of a gas is proportional to gas pressure p.

Ans.

The equation of state is given by,

pV = nRT ……….. (i)

Where,

p →Pressure of gas

V →Volume of gas

n→ Number of moles of gas

R→ Gas constant

T →Temperature of gas

From equation (i) we have,

$\begin{array}{l} \frac{n}{V} = \frac{p}{R T} ………… (i i) \\ Re p l a c i n g n with \frac{m}{M} in equation (ii), we have \\ \frac{m}{M V} = \frac{p}{R T} ………… (i i i) \end{array}$

Where,

m →Mass of gas

M →Molar mass of gas

$\begin{array}{l} B u t \frac{m}{M V} = d (d = d e n s i t y of gas) \\ from equation (iii) \\ \frac{d}{M} = \frac{p}{R T} \\ \Rightarrow d = (\frac{M}{R T}) p \end{array}$

Molar mass (M) of a gas is always constant so, at constant temperature (T)

$\begin{array}{l} \frac{M}{R T} = c o n s \tan t \\ d = (c o n s \tan t) p \\ \Rightarrow d \propto p \end{array}$

So, at a given temperature, the density (d) of gas is proportional to its pressure (p)

Q.4 At 0°C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?

Ans.

Density (d) of the substance at temperature (T) can be given as,

$\begin{array}{l} d = \frac{M}{R T} \\ Density of oxide (d_{1}) is given by, \\ d_{1} = \frac{M_{1} p_{1}}{R T} \end{array}$

Here, M₁ and p₁ are the mass and pressure of the oxide respectively.
Density of dinitrogen gas (d₂) can be given as:

$d_{2} = \frac{M_{2} p_{2}}{R T}$

In above equation, M₂ and p₂ are the mass and pressure of the oxide respectively.
According to the given question,

$\begin{array}{l} d_{1} = d_{2} \\ ∴ M_{1} p_{1} = M_{2} p_{2} \end{array}$

Molecular mass of nitrogen, M = 28 g/mol

$\begin{array}{l} N o w, M_{1} = \frac{M_{2} P_{2}}{p_{1}} \\ \frac{28 \times 5}{2} \end{array}$

=70 g/mol

Therefore, the molecular mass of the oxide is 70 g/mol.

Q.5 Pressure of 1 g of an ideal gas A at 27 °C is found to be 2bar. When 2 g of another ideal gas B is introduced in the same flask at same temperature the pressure becomes 3bar.Find a relationship between their molecular masses.

Ans.

The ideal gas equation for ideal gas A is given by,

pAV = n_ART

Here, p_A and n_A is the pressure and number of moles of gas A.

For ideal gas B, the ideal gas equation is:

$p_{B} V = n_{B} R T …………. (i i)$

Where, p and n represent the pressure and number of moles of gas B.

[V and T are constants for gases A and B]

From equation (i),

$\begin{array}{l} p_{A} V = \frac{m_{A}}{M_{A}} R T \\ \Rightarrow \frac{p_{A} M_{A}}{m_{A}} = \frac{R T}{V} ……… (i i i) \end{array}$

From equation (ii),

$\begin{array}{l} p_{B} V = \frac{m_{B}}{M_{B}} R T \\ \Rightarrow \frac{p_{B} M_{B}}{m_{B}} = \frac{R T}{V} ……… (i v) \end{array}$

Here, M_Aand M_B are the molecular masses of gases A and B respectively.

From equations (iii) and (iv),

$\begin{array}{l} p_{B} V = \frac{m_{B}}{M_{B}} R T \\ \Rightarrow \frac{p_{B} M_{B}}{m_{B}} = \frac{R T}{V} ……… (i v) \\ G i v e n, \\ m_{A} = 1 g \\ p_{A} = 2 bar \\ m_{B} = 2 g \\ p_{B} = (3 - 2) = 1 bar \end{array}$

(Since total pressure is 3 bar)

Substituting the above values in equation (v)

$\begin{array}{l} \frac{2 \times M_{A}}{1} = \frac{1 \times M_{B}}{2} \\ \Rightarrow 4 M_{A} = M_{B} \\ Thus, the relationship of molecular masses of A and B can be given as : \\ 4 M_{A} = M_{B} \end{array}$

Q.6 The drain cleaner, Drainex contains small bits of aluminum which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20 °C and one bar will be released when 0.15g of aluminum reacts?

Ans.

The reaction of aluminium with caustic soda (NaOH)can be represented as:

$\underset{\begin{array}{l} 2 \times 27 g \end{array}}{2 A l} + 2 N a O H + 2 H_{2} O \to 2 N a A I O_{2} + \underset{\begin{array}{l} 3 \times 22400 m L \end{array}}{3 H_{2}}$

At STP (273.15 K and 1 atm),

Hence, 54 g (2 × 27 g) of aluminium gives 3 × 22400 mL of H₂.

$∴ 0.15 g A l gives \frac{3 \times 22400 \times 0.15}{54} m L {of H}_{2} i . e ., 186.67 {mL of H}_{2}$

At STP,

$\begin{array}{l} p_{1} = 1 a t m \\ V_{1} = 186.67 mL \\ T_{1} = 273.15 K \end{array}$

Let the volume of dihydrogen be V₂ at p₂ = 0.987 atm (since 1 bar = 0.987 atm) and T₂

= 20°C = (273.15 + 20) K = 293.15 K

$\begin{array}{l} \frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}} \\ \Rightarrow V_{2} = \frac{p_{1} V_{1} T_{2}}{p_{2} T_{1}} \\ = \frac{1 \times 186.67 \times 293.15}{0.987 \times 273.15} \\ = 202.98 mL \\ =203 mL \end{array}$

Therefore, 203 mL of dihydrogen will be released.

Q.7 What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 dm flask at 27 °C?

Ans.

Since,

$\begin{array}{l} p = \frac{m}{M} \frac{R T}{V} \\ For methane ({CH}_{4}), \\ p_{C H_{4}} = \frac{3.2}{16} \times \frac{8.314 \times 300}{9 \times 10^{- 3}} [\frac{S i n c e 9 d m^{3} = 9 \times 10^{- 3} m^{3}}{27 º C = 300 K}] \\ = 5.543 \times 10^{4} p a \\ For carbon dioxide ({CO}_{2}), \\ p_{C O_{2}} = \frac{4.4}{44} \times \frac{8.314 \times 300}{9 \times 10^{- 3}} \\ = 2.771 \times 10^{4} p a \\ Total pressure exerted by the mixture can be obtained as : \\ p = p_{C H_{4}} + p_{C O_{2}} \\ = (5.543 \times 10^{4} + 2.771 \times 10^{4}) p a \\ = 8.314 \times 10^{4} p a \end{array}$

Therefore, total pressure exerted by the mixture is 8.314 × 10⁴ Pa.

Q.8 What will be the pressure of the gaseous mixture when 0.5 L of H₂ at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in a 1L vessel at 27°C?

Ans.

Let the partial pressure of H₂ in the vessel be P_H2

Given:

Alternatively,

$\begin{array}{l} P a r t i a l pressure of Hydrogen in 1 L container is calculated according to Boyle’s law \\ P_{1} V_{1} = P_{2} V_{2} \Rightarrow 0.8 \times 0.5 = P_{2} \times 1 \\ \Rightarrow ∴ P_{2} = p_{H_{2}} = 0.4 bar \\ Similarly … Partial pressure of dioxygen in 1 L container \\ P_{1} V_{1} = P_{2} V_{2} \Rightarrow 0.7 \times 2.0 = P_{2} \times 1 \\ \Rightarrow ∴ P_{2} = p_{O_{2}} = 1.4 bar \\ Therefore … P_{t o t a l} = p_{H_{2}} + p_{O_{2}} = 0.4 + 1.4 = 1.8 bar \end{array}$

Total pressure of the gaseous mixture in the vessel is 1.8 bar

Q.9 Density of a gas is found to be 5.46 g/dm at 27 °C at 2 bar pressure. What will be its density at STP?

Ans.

Given,

$\begin{array}{l} d_{1} = 5.46 g / d m^{3} \\ p_{1} = 2 bar \\ T_{1} = 27 º C = (27 + 273) K = 300 K \\ p_{2} = 1 bar \\ T_{2} = 273 K \\ d_{2} = ? \end{array}$

The density (d₂) of the gas at STP can be calculated using the equation,

$\begin{array}{l} d = \frac{M p}{R T} \\ ∴ \frac{d_{1}}{d_{2}} = \frac{\frac{M p_{1}}{R T_{1}}}{\frac{M p_{2}}{R T_{2}}} \\ \Rightarrow \frac{d_{1}}{d_{2}} = \frac{p_{1} T_{2}}{p_{2} T_{1}} \\ \Rightarrow d_{2} = \frac{p_{2} T_{1} d_{1}}{p_{1} T_{2}} \\ = \frac{1 \times 300 \times 5.46}{2 \times 273} \\ = 3 g d m^{- 3} \end{array}$

Density of the gas at STP will be 3 g dm^–3

Q.10 34.05 mL of phosphorus vapour weighs 0.0625 g at 546 °C and 0.1 bar pressure. What is the molar mass of phosphorus?

Ans.

p = 0.1 bar

V = 34.05 mL = 34.05 × 10^–3 L = 34.05 × 10^–3 dm³

R= 0.083 bar dm³ K^–1 mol^–1

T = 546°C = (546 + 273) K = 819 K

According to the ideal gas equation:

$\begin{array}{l} p V = n R T \\ \Rightarrow n = \frac{p V}{R T} \\ = \frac{0.1 \times 34.05 \times 10^{- 3}}{0.083 \times 819} \\ = 5.01 \times 10^{- 5} mol \\ Molar mass of phosphorus = \frac{0.0625}{5.01 \times 10^{- 5}} = 124.75 {g mol}^{- 1} \end{array}$

The molar mass of phosphorus is 124.75 g mol^–1

Q.11 A student forgot to add the reaction mixture to the round bottomed flask at 27 °C but instead he/she placed the flask on the flame. After a lapse of time, he realized his mistake, and using a pyrometer he found the temperature of the flask was 477 °C. What fraction of air would have been expelled out?

Ans.

If the volume of the round bottomed flask be V.

Then, the volume of air inside the flask at 27° C is V.

Let volume of air inside flask will be equal to volume of flask and let that volume be V

T = 27°C = 300 K

V₂ =?

T₂ = 477° C = 750 K

According to Charles’s law,

$\begin{array}{l} \frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}} \\ \Rightarrow V_{2} = \frac{V_{1} T_{2}}{T_{1}} \\ = \frac{750 V}{300} \\ = 2.5 V \end{array}$

The volume of air expelled out = 2.5 V – V = 1.5 V

$F r a c t i o n of air expelled out = \frac{1.5 V}{2.5 V} = \frac{3}{5}$

Q.12 Calculate the temperature of 4.0 mol of a gas occupying 5 dm³at 3.32 bar.

(R= 0.083 bar dm³ K^–1 mol^–1)

Ans.

Given,

n = 4.0 mol

V = 5 dm³

p = 3.32 bar

R= 0.083 bar dm³ K^–1 mol^–1

Using the ideal gas equation:

pV = nRT

$\begin{array}{l} \Rightarrow T = \frac{p V}{n R} \\ = \frac{3.32 \times 5}{4 \times 0.083} \\ = 50 K \end{array}$

Temperature is 50 K.

Q.13 Calculate the total number of electrons present in 1.4 g of dinitrogen gas.

Ans.

Molar mass of dinitrogen (N₂) = 28 g mol^–1

$\begin{array}{l} 1.4 {g of N}_{2} = \frac{1.4}{28} = 0.05 mol \\ = 0 .05×6 {.02×10}^{23} number of molecules \\ 3 {.01×10}^{22} number of molecules \end{array}$

1 molecule of N₂ contains 14 electrons.
3.01 × 10²² molecules of N₂ contains = 14 × 3.01 × 10²²
= 4.214 × 10²³ electrons

Q.14 How much time would it take to distribute one Avogadro number of wheat grains, if 10¹⁰grains are distributed each second?

Ans.

Avogadro number = 6.02 × 10²³

10 grains are distributed in = 1sec,

Therefore, 6.02 × 10²³ will be calculated as:Time required

$\begin{array}{l} \frac{6.02 \times 10^{23}}{10^{10}} S \\ = 6.02 \times 10^{23} S \\ = \frac{6.02 \times 10^{23}}{60 \times 60 \times 24 \times 365} y e a r s \\ = 1.909 \times 10^{6} y e a r s \\ T i m e taken = 1.909 \times 10^{6} y e a r s \end{array}$

Q.15 Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm³ at 27°C. R = 0.083 bar dm³ K^–1 mol^–1

Ans.

Mass of dioxygen (O₂) = 8 g (Given)

$\begin{array}{l} N o . {of moles of O}_{2} = \frac{M a s s}{M . w t} \\ O_{2} = \frac{8}{32} = 0.25 mole \\ Mass of dihydrogen (H_{2}) = 4 g (Given) \\ N o . {of moles of H}_{2} = \frac{M a s s}{M . w t} \\ N u m b e r {of moles of H}_{2} = \frac{4}{2} = 2 mole \end{array}$

Total number of moles in the mixture = 0.25 + 2 = 2.25 mole

V = 1 dm³

n = 2.25 mol

R= 0.083 bar dm³ K^–1 mol^–1

T = 27°C = 300 K

Total pressure (p) is:

pV = nRT

$\begin{array}{l} O r, p = \frac{n R T}{V} \\ = 2.25 \times 0.083 \times 3001 \\ = 56.025 bar \end{array}$

The total pressure of the mixture is 56.025 bar.

Q.16 Pay load is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the pay load when a balloon of radius 10 m, mass 100 kg is filled with helium at 1.66 bar at 27°C. (Density of air = 1.2 kg m^–3 and R = 0.083 bar dm³ K^–1Mol^–1).

Ans.

Radius of the balloon, r = 10 m

Volume of a sphere is given as:

$\begin{array}{l} V = \frac{4}{3} π r^{3} \\ ∴ Volume of ballon = \frac{4}{3} π r^{3} \\ = \frac{4}{3} \times \frac{22}{7} \times 10^{3} \\ = 4190.5 m^{3} (a p p r o x) \end{array}$

∴ The volume of the displaced air is 4190.5 m³.

Density of air = 1.2 kg m^–3

Mass = Volume × density
Mass of displaced air = 4190.5 × 1.2 kg = 5028.6 kg

Mass of helium (m) inside the balloon is given by,

$\begin{array}{l} m = \frac{M p V}{R T} \\ H e r e, \\ M = 4 \times 10^{- 3} k g {mol}^{- 1} \\ p = 1.66 bar \end{array}$

V=Volume of the balloon
=4190.5 m³

$\begin{array}{l} R = 0.083 {bar dm}^{3} K^{- 1} m o l^{- 1} \\ T = 27 º C = 300 K \\ m = \frac{4 \times 10^{- 3} \times 1.66 \times 4190.5 \times 10^{3}}{0.083 \times 300} \end{array}$

=1117.5 kg (approx)

Now, total mass of the balloon filled with helium = (100 + 1117.5) kg

= 1217.5 kg

Hence, pay load = (5028.6 – 1217.5) kg

= 3811.1 kg

Hence, the pay load of the balloon is 3811.1 kg.

Q.17 Calculate the volume occupied by 8.8 g of CO₂ at 31.1°C and 1 bar pressure.

R= 0.083 bar L K^–1 mol^–1.

Ans.

Since,

$\begin{array}{l} p V = \frac{m}{M} R T \\ \Rightarrow V = \frac{m R T}{M p} \end{array}$

Here,
m = 8.8 g
R = 0.083 bar LK^–1 mol^–1
T = 31.1°C = 304.1 K
M = 44 g
p = 1 bar

$V o l u m e (V) = \frac{8.8 \times 0.083 \times 304.1}{44 \times 1}$

=5.04806L
=5.05L
The volume occupied is 5.05 L.

Q.18 2.9 g of a gas at 95 °C occupied the same volume as 0.184 g of dihydrogen at 17 °C, at the same pressure. What is the molar mass of the gas?

Ans.

Volume (V) occupied by dihydrogen is:

$\begin{array}{l} V = \frac{m}{M} \frac{R T}{p} \\ = \frac{0.184}{2} \times \frac{R \times 290}{p} \end{array}$

Let, the molar mass of the unknown gas be M then the volume (V) occupied by it, can be given as:

$\begin{array}{l} V = \frac{m}{M} \frac{R T}{p} \\ = \frac{2.9}{M} \frac{R \times 368}{p} \\ According to the question, \\ \frac{0.184}{2} \times \frac{R \times 290}{p} = \frac{2.9}{M} \frac{R \times 368}{M} \\ \Rightarrow \frac{0.184 \times 290}{2} = \frac{2.9 \times 368}{M} \\ \Rightarrow M = \frac{2.9 \times 368 \times 2}{0.184 \times 290} \\ = 40 g {mol}^{- 1} \end{array}$

The molar mass of the gas is 40 g mol^–1

Q.19 A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.

Ans.

If the weight of dihydrogen be 20 g and the weight of dioxygen be 80 g.

$\begin{array}{l} No {. of moles of H}_{2} = \frac{M a s s}{M . wt} \\ T h e {number of moles of dihydrogen, n}_{H_{2}} = \frac{20}{2} \\ = 10 moles and the number moles \\ {The number of moles of dioxygen, n}_{O_{2}} = \frac{80}{32} = 2.5 moles \end{array}$

Total pressure of the mixture, p_total= 1 bar
Then, partial pressure of dihydrogen,

$\begin{array}{l} p_{H_{2}} = \frac{n_{H_{2}}}{n_{H_{2}} + n_{O_{2}}} \times p_{t o t a l} \\ = \frac{10}{10 + 2.5} \times 1 \\ = 0.8 bar \end{array}$

The partial pressure of dihydrogen is 0.8 bar.

Q.20 What would be the SI unit for the quantity pV²T² /n?

Ans.

The SI unit for pressure, p is Nm^–2.

The SI unit for volume, V is m³

The SI unit for temperature, T is K.

The SI unit for the number of moles, n is mol.

$\begin{array}{l} Therefore, the SI unit of quantity \frac{p V^{2} T^{2}}{n} is given by \\ = \frac{(N m^{- 2}) {(m^{3})}^{2} {(K)}^{2}}{m o l} \\ = {Nm}^{4} K^{2} m o l^{- 1} \end{array}$

Q.21 In terms of Charles’ law explain why –273°C is the lowest possible temperature.

Ans.

Charles’ law states that at constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature.

At a given pressure the plot of volume vs. temperature is always a straight line for all gasses. If this line is extended to zero then it intersects at –273°C which is the lowest possible temperature.

Q.22 Critical temperature for carbon dioxide and methane are 31.1 °C and –81.9 °C respectively. Which of these has stronger intermolecular forces and why?

Ans.

The intermolecular forces of attraction are stronger in the case of CO₂ because higher is the critical temperature of a gas, easier is its liquefaction. So, the intermolecular forces of attraction between the molecules of a gas are directly proportional to its critical temperature.

Q.23 Explain the physical significance of Van der Waals parameters.

Ans.

Physical significance of ‘a’:

Value of ‘a’ is measure of magnitude of intermolecular attractive forces within the gas and is independent of temperature and pressure.

Physical significance of ‘b’:

‘b’ is measure of the volume of a gas molecule.

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FAQs (Frequently Asked Questions)

Surface tension is caused because of the intermolecular attractive forces. The rise in temperature decreases the surface tension.

There are different types of intermolecular forces like;

Dispersion Forces or London Forces
Dipole-Dipole Forces
Dipole-Induced Dipole Forces
Hydrogen Bond

You can learn more about them through our NCERT Solutions Class 11.

Heat or thermal energy arises when there is a rise in temperature that leads to the collision of atoms and molecules. The thermal energy of a substance and the temperature of the substance are directly proportional. Heat is the flow of thermal energy.

Grades can be improved by referring to the Extramarks’ NCERT Solutions for Class 11 Chemistry Chapter 5. The solutions have answers to all the questions that are asked in Chapter 5 of Class 11 Chemistry. The answers are framed concisely while keeping the language simple.

Extramarks also offers past years’ question papers, sample papers, mock tests, and other learning material that will help students in preparing for exams.

Atoms, ions, and molecules in a liquid are incompressible and not impacted by pressure. They have a set volume if temperature and pressure stay constant. Solids exposed to extreme temperatures turn into liquids that are affected by pressure characteristics.

Liquids have a higher rate of diffusion than solids because the former has more interparticle space than the latter. This in turn leads to higher fluidity than solids as particles are not as closely packed in liquids as they are in solids.

Here are the physical properties of gases;

The density of gases as compared to solids and liquids is much lower.
Gas molecules are very highly compressible.
Pressure is exerted in all directions by gases.
Gases are completely miscible.
They do not have a shape and can conform to the shape of the container.

Class 11 Chemistry Chapter 5 NCERT Solutions has answers to all the questions asked in the back exercise of Chapter 5. The solutions are prepared by subject-matter experts at Extramarks, which ensures that they are easy to understand, detailed, and accurate.

NCERT Solutions For Class 11 Chemistry Chapter 5