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Ncert Solutions for Class 11 Chemistry Chapter 5 Pdf Download

NCERT Solutions Class 11 Chemistry Chapter 5 – Free PDF Download

NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matterare available on this page to help students get a clear idea of the basic concepts. Detailed and step-by-step solutions according to the latest term-II CBSE Syllabus 2021-22 are provided to every question listed in the NCERT textbook can be found here. This chapter mainly provides complete information regarding the different laws, kinds of thermal energy and intermolecular forces. NCERT Solutions for Class 11 Chemistry are designed with the main aim of helping the Class 11 students face the second term exams without fear.

Solving the exercise questions using the solutions will provide a strong grip on the basic concepts. By regular practise, students will be able to analyse their areas of weakness and work on them for a better score in the term-II Class 11 Chemistry exam. The solutions contain authentic information by the experts after conducting a wide research on the important concepts. Also, these NCERT Solutions can be downloaded in a PDF format (for free) by clicking the download button provided above.

*The complete and accurate NCERT Solutions for Class 11 Chemistry Chapter 5 will be updated soon…

Download NCERT Solutions Class 11 Chemistry Chapter 5 PDF

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

"States of Matter" is the fifth chapter in the NCERT Class 11 Chemistry textbook. This chapter contains several fundamental concepts related to intermolecular forces and how they affect the physical state of a substance. It also deals with several other important concepts associated with the liquid and gaseous states of matter. This is the reason why "states of matter" is regarded by many as one of the most important chapters in the NCERT Class 11 Chemistry textbook.

The topics covered in this chapter include intermolecular forces and thermal energy, gas laws, the ideal gas equation, the kinetic theory of gases, the deviation of gases from ideal behaviour, liquefaction of gases and the liquid state. The NCERT Solutions for Class 11 Chemistry (Chapter 5) provided on this page deal with the following types of questions:

  • Numerical problems based on Boyle's law, Charles's law, Gay-Lusscac's law and Avogadro's law.
  • Numerical problems on calculating partial pressure.
  • Questions on critical temperature and pressure.
  • Questions on Van der Waals forces and other types of intermolecular forces.

Subtopics of Class 11 Chemistry Chapter 5: States of Matter

  1. Intermolecular Forces
    • Dispersion Forces Or London Forces
    • Dipole-dipole Forces
    • Dipole-induced Dipole Forces
    • Hydrogen Bond
  2. Thermal Energy
  3. Intermolecular Forces Vs Thermal Interactions
  4. The Gaseous State Ex
  5. The Gas Laws Ex
    • Boyle's Law (Pressure-volume Relationship)
    • Charles' Law (Temperature-volume Relationship)
    • Gay Lussac's Law (Pressure-temperature Relationship)
    • Avogadro Law (Volume – Amount Relationship)
  6. Ideal Gas Equation
    • Density And Molar Mass Of A Gaseous Substance
    • Dalton's Law Of Partial Pressures
  7. Kinetic Molecular Theory Of Gases
  8. Behaviour Of Real Gases: Deviation From Ideal Gas Behaviour
  9. Liquefaction Of Gases
  10. Liquid State
    • Vapour Pressure
    • Surface Tension and Viscosity.

Access the answers of NCERT Solutions for Class 11 Chemistry Chapter 5

Q1. What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 at 30°C?

Answer:

Initial pressure, P 1 = 1 bar

Initial volume, V 1 = 500 d m 3 dm^{ 3 }

Final volume, V 2 = 200 d m 3 dm^{ 3 }

As the temperature remains the 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 = P 1 V 1 V 2 \frac{P_{1}V_{1}}{V_{2}}

= 1 × 500 200 \frac{1 \; \times \; 500}{200}

= 2.5 bar

∴ theminimum pressure required to compress is 2.5 bar.

Q2. 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?

Answer:

Initial pressure, P 1 = 1.2 bar

Initial volume, V 1 = 120 mL

Final volume, V 2 = 180 mL

As the temperature remains the 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 = P 1 V 1 V 2 \frac{P_{1}V_{1}}{V_{2}}

= 1.2 × 120 180 \frac{1.2 \; \times \; 120}{180}

= 0.8 bar

Therefore, the min pressure required is 0.8 bar.

Q3. Using the equation of state pV=nRT; show that at a given temperature density of a gas is proportional to gas pressure p.

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

n V \frac{n}{V} = p R T \frac{p}{RT}

Replace n with m M \frac{m}{M} , therefore,

m M V \frac{m}{MV} = p R T \frac{p}{RT} ……..(2)

Where, m = mass

M = molar mass

But, m V \frac{m}{V} = d

Where, d = density

Therefore, from equation (2), we get

d M \frac{d}{M} = p R T \frac{p}{RT}

d = ( M R T \frac{M}{RT} ) p

d ∝ \propto p

Therefore, at a given temp, the density of the gas (d) is proportional to its pressure (p).

Q4. At 0°C, the density of a certain oxide of a gas at 2 bar is the same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?

Answer:

Density (d) of the substance at temp (T) can be given by,

d = M p R T \frac{Mp}{RT}

Now, density of oxide (d1) is as given,

d 1 d_{1} = M 1 p 1 R T \frac{M_{1}p_{1}}{RT}

Where, M1 = mass of the oxide

p1 = pressure of the oxide

Density of dinitrogen gas (d2) is as given,

d 2 d_{2} = M 1 p 2 R T \frac{M_{1}p_{2}}{RT}

Where, M2 = mass of the oxide

p2 = pressure of the oxide

Acc to the question,

d1 = d2

Therefore, M 1 p 1 = M 2 p 2 M_{1}p_{1} = M_{2}p_{2}

Given:

p 1 p_{1} = 2 bar p 2 p_{2} = 5 bar

Molecular mass of nitrogen, M 2 M_{2} = 28 g/mol

Now, M 1 M_{1}

= M 2 p 2 p 1 \frac{M_{ 2 }p_{2}}{p_{ 1 }}

= 28 × 5 2 \frac{ 28 × 5 }{ 2 }

= 70 g/mol

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

Q5. The pressure of 1 g of an ideal gas A at 27 °C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at the same temperature the pressure becomes 3 bar. Find a relationship between their molecular masses.

Answer:

For ideal gas A, the ideal gas equation is given by,

p X V = n X R T p_{X}V = n_{X}RT ……(1)

Where p X p_{X} and n X n_{X} represents 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 R T p_{Y}V = n_{Y}RT ……(2)

Where, p Y p_{Y} and n Y 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 = m X M X p_{ X }V = \frac{m_{ X }}{M_{ X }} RT p X M X m X \frac{p_{ X }M_{ X }}{m_{ X }} = R T V \frac{ R T}{ V } ……(3)

From equation (2),

p Y V = m Y M Y p_{ Y }V =\frac{m_{ Y }}{M_{ Y }} RT p Y M Y m Y \frac{p_{ Y }M_{ Y }}{m_{ Y }} = R T V \frac{ R T}{V} …… (4)

Where, M X M_{ X } and M Y M_{ Y } are the molecular masses of gases X and Y respectively.

Now, from equation (3) and (4),

p X M X m X \frac{p_{ X }M_{ X }}{m_{ X }} = p Y M Y m Y \frac{p_{ Y }M_{ Y }}{m_{ Y }} ….. (5)

Given,

m X m_{ X } = 1 g p X p_{ X } = 2 bar m Y m_{ Y } = 2 g p Y p_{ Y } = (3 – 2) = 1 bar (Since total pressure is 3 bar)

Substituting these values in equation (5),

2 × M X 1 \frac{2 \; \times \; M_{X} }{1} = 1 × M Y 2 \frac{1 \; \times \; M_{Y} }{2}

4 M X M_{ X } = M Y M_{ Y }

Therefore, the relationship between the molecular masses of X and Y is,

4 M X M_{ X } = M Y M_{ Y }

Q6. 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?

Answer:

The reaction of aluminum with caustic soda is as given below:

2Al + 2NaOH + 2H2O → \rightarrow 2NaAlO2 + 3H2

At Standard Temperature Pressure (273.15 K and 1 atm), 54 g ( 2 × 27 g) of Al gives 3 ×22400 mL of H2.

Therefore, 0.15 g Al gives:

= 3 × 22400 × 0.15 54 \frac{3 \; \times \; 22400 \; \times \; 0.15}{54} mL of H2

= 186.67 mL of H2

At Standard Temperature Pressure,

p 1 p_{ 1 } = 1 atm V 1 V_{ 1 } = 186.67 mL T 1 T_{ 1 } = 273.15 K

Let the volume of dihydrogen be V 2 V_{ 2 } at p 2 p_{ 2 } = 0.987 atm (since 1 bar = 0.987 atm) and T 2 T_{ 2 } = 2 0 ∘ 20^{\circ} C = (273.15 + 20) K = 293.15 K.

Now,

p 1 V 1 T 1 \frac{p_{ 1 }V_{ 1 }}{T_{ 1 }} = p 2 V 2 T 2 \frac{p_{ 2 }V_{ 2 }}{T_{ 2 }} V 2 = p 1 V 1 T 2 p 2 T 1 V_{ 2 } = \frac{p_{ 1 }V_{ 1 }T_{ 2 }}{p_{ 2 }T_{ 1 }}

= 1 × 186.67 × 293.15 0.987 × 273.15 \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. 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 dm3 flask at 27 °C?

Answer:

It is known that,

p = m M \frac{m}{M} R T V \frac{RT}{V}

For methane (CH4),

p C H 4 p_{CH_{ 4 }}

= 3.2 16 \frac{ 3.2 }{ 16 } × 8.314 × 300 9 × 1 0 − 3 \frac{8.314 \; \times \;300 }{9 \; \times \; 10^{-3 }} [Since 9 dm3 = 9 × 1 0 − 3 9 \; \times \; 10^{-3} m3 2 7 ∘ 27^{\circ} C = 300 K]

= 5.543 × 1 0 4 10^{ 4 } Pa

For carbon dioxide (CO2),

p C O 2 p_{CO_{ 2 }}

= 4.4 44 \frac{ 4.4 }{ 44 } × \frac{8.314 \; \times \;300 }{9 \; \times \; 10^{-3}

= 2.771 × 1 0 4 10^{4} Pa

Total pressure exerted by the mixture can be calculated as:

p = p C H 4 p_{CH_{ 4 }} + p C O 2 p_{CO_{ 2 }}

= (5.543 × 1 0 4 10^{ 4 } + 2.771 × 1 0 4 10^{4} ) Pa

= 8.314 × 1 0 4 10^{ 4 } Pa

Q8. 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?

Answer:

Let the partial pressure of H 2 {H_{ 2 }} in the container be p H 2 p_{H_{ 2 }} .

Now,

p 1 {p_{ 1 }} = 0.8 bar p 2 {p_{ 2 }} = p H 2 p_{H_{ 2 }} V 1 {V_{ 1 }} = 0.5 L V 2 {V_{ 2 }} = 1 L

It is known that,

p 1 {p_{ 1 }} V 1 {V_{ 1 }} = p 2 {p_{ 2 }} V 2 {V_{ 2 }} p 2 {p_{ 2 }} = p 1 × V 1 V 2 \frac{p_{ 1 } \; \times \; V_{ 1 }}{V_{ 2 }} p H 2 p_{H_{ 2 }} = 0.8 × 0.5 1 \frac{ 0.8  \; \times \; 0.5 }{ 1 }

= 0.4 bar

Now, let the partial pressure of O2 in the container be p O 2 p_{O_{ 2 }} .

Now,

p 1 {p_{ 1 }} = 0.7 bar p 2 {p_{ 2 }} = p O 2 p_{O_{ 2 }} V 1 {V_{ 1 }} = 2.0 L V 2 {V_{ 2 }} = 1 L p 1 {p_{ 1 }} V 1 {V_{ 1 }} = p 2 {p_{ 2 }} V 2 {V_{ 2 }} p 2 {p_{ 2 }} = p 1 × V 1 V 2 \frac{p_{ 1 } \; \times \; V_{ 1 }}{V_{ 2 }} p O 2 p_{O_{ 2 }} = 0.7 × 20 1 \frac{ 0.7  \; \times \; 20 }{ 1 }

= 1.4 bar

Total pressure of the gas mixture in the container can be obtained as:

p t o t a l p_{total} = p H 2 p_{H_{ 2 }} + p O 2 p_{O_{ 2 }}

= 0.4 + 1.4

= 1.8 bar

Q9. The density of a gas is found to be 5.46 g/dm3 at 27 °C at 2 bar pressure. What will be its density at STP?

Answer:

Given,

d1 = 5.46 g/dm3

p1 = 2 bar

T1 = 2 7 ∘ 27^{\circ} C = (27 + 273) K = 300 K

p2 = 1 bar

T2 = 273 K

d2 = ?

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

d = M p R T \frac{Mp}{RT} d 1 d 2 \frac{d_{1}}{d_{2}} = M p 1 R T 1 M p 2 R T 2 \frac{\frac{M \;p_{ 1 }}{R \; T_{ 1 }}}{\frac{M \; p_{ 2 }}{R \; T_{ 2 }}} d 1 d 2 \frac{d_{1}}{d_{2}} = p 1 T 2 p 2 T 1 \frac{p_{1} \; T_{2}}{p_{2} \; T_{1}}

d2 = p 2 T 1 d 1 p 1 T 2 \frac{p_{2} \; T_{1} \; d_{1}}{p_{1} \; T_{2}}

= 1 × 300 × 5.46 2 × 273 \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 weighs 0.0625 g at 546 °C and 0.1 bar pressure. What is the molar mass of phosphorus?

Answer:

Given,

p = 0.1 bar

V = 34.05 mL = 34.05 × 10-3 d m 3 dm^{ 3 }

R = 0.083 bar d m 3 dm^{ 3 } at K-1 mol-1

T = 54 6 ∘ C 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 = p V R T \frac{ pV }{ RT }

= 0.1 × 34.05 × 1 0 − 3 0.083 × 819 \frac{  0.1 \times 34.05 × 10^{-3}  }{  0.083 \times 819 }

= 5.01 × 10-5 mol

Therefore, molar mass of phosphorus = 0.0625 5.01 × 1 0 − 5 \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 2 7 ∘ 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 47 7 ∘ 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 2 7 ∘ 27^{\circ} C is V.

Now,

V1 = V

T1 = 2 7 ∘ 27^{\circ} C = 300 K V2 = ?

T2 = 47 7 ∘ 477^{\circ} C = 750 K

Acc to Charles's law,

V 1 T 1 \frac{V_{1}}{T_{1}} = V 2 T 2 \frac{V_{2}}{T_{2}} V 1 V_{ 1 } = V 1 T 2 T 1 \frac{V_{ 1 }T_{ 2 }}{T_{ 1 }}

= 750 V 300 \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

= 1.5 V 2.5 V \frac{1.5V }{ 2.5V }

= 3 5 \frac{3 }{ 5 }

Q12. Calculate the temperature of 4.0 mol of a gas occupying 5 dm3 at 3.32 bar. (R = 0.083 bar dm3 K–1 mol–1).

Given,

N= 4.0 mol

V = 5 d m 3 dm^{ 3 }

p = 3.32 bar

R = 0.083 bar d m 3 dm^{ 3 } at K-1 mol-1

The temp (T) can be calculated using the ideal gas equation as:

pV = nRT

T = p V n R \frac{ pV }{ nR }

= 3.32 × 5 4 × 0.083 \frac{3.32 \; \times \; 5 }{ 4 \;\times \; 0.083 }

= 50 K

Therefore, the required temp is 50 K.

Q13. Calculate the total number of electrons present in 1.4 g of dinitrogen gas.

Answer:

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

Thus, 1.4 g of N2

= 1.4 28 \frac{ 1.4 }{ 28 }

= 0.05 mol

= 0.05 × 6.02 × 1023 no. of molecules

= 3.01 × 1023 no. of molecules

Now, 1 molecule of N2 has 14 electrons.

Therefore, 3.01 × 1023 molecules of N2 contains,

= 14 × 3.01 × 1023

= 4.214 × 1023 electrons

Q14. How much time would it take to distribute one Avogadro number of wheat grains, if 1010 grains are distributed each second?

Answer:

Avogadro no. = 6.02 × 1023

Therefore, time taken

= 6.02 × 1 0 23 1 0 10 s \frac{6.02 \; \times \; 10^{23}}{10^{10}} s

= 6.02 × 1013 s

= 6.02 × 1 0 23 60 × 60 × 24 × 365 y e a r s \frac{6.02 \; \times \; 10^{23}}{60 \; \times \; 60 \; \times \; 24 \; \times \; 365 } years

= 1.909 × 106 years

Therefore, the time taken would be 1.909 × 106 years.

Q15. 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

Answer:

Given:

Mass of O2 = 8 g

No. of moles

= 8 32 \frac{ 8 }{ 32 }

= 0.25 mole

Mass of H2 = 4 g

No. of moles

= 4 2 \frac{ 4 }{ 2 }

= 2 mole

Hence, total no. of moles in the mixture

= 0.25 + 2

= 2.25 mole

Given:

V = 1 d m 3 dm^{ 3 }

n = 2.25 mol

R = 0.083 bar d m 3 dm^{ 3 } at K-1 mol-1

T = 2 7 ∘ 27^{\circ} C = 300 K

Total pressure :
pV = nRT

p = n R T V \frac{ nRT }{ V }

= 225 × 0.083 × 300 1 \frac{225 \; \times \; 0.083 \; \times \; 300 }{ 1 }

= 56.025 bar

Therefore, the total pressure of the mixture is 56.025 bar.

Q16. Payload is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the payload 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–1 mol–1)

Answer:

Given:

r = 10 m

Therefore, volume of the balloon

= 4 3 \frac{4}{3} πr3

= 4 3 × 22 7 × 1 0 3 \frac{ 4 }{ 3 }\; \times \; \frac{ 22 }{ 7 } \; \times \; 10^{3}

= 4190.5 m3 (approx.)

Therefore, the volume of the displaced air

= 4190.5 × 1.2 kg

= 5028.6 kg

Mass of helium,

= M p V R T \frac{ MpV }{ RT }

Where, M = 4 × 10-3 kg mol-1

p = 1.66 bar

V = volume of the balloon

= 4190.5 m3

R = 0.083 0.083 bar d m 3 dm^{ 3 } at K-1 mol-1

T = 27 °C = 300 K

Then,

m = 4 × 1 0 − 3 × 1.66 × 4190.5 × 1 0 3 0.083 × 300 \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. Calculate the volume occupied by 8.8 g of CO2 at 31.1°C and 1 bar pressure. R = 0.083 bar dm3 K–1 mol–1.

Answer:

pVM = mRT

V = m R T M p \frac{mRT}{Mp}

Given:

m = 8.8 g

R = 0.083 bar d m 3 dm^{ 3 } at K-1 mol-1.

T = 31.1 °C = 304.1 K

M = 44 g

p = 1 bar

Thus, Volume (V),

= 8.8 × 0.083 × 304.1 44 × 1 \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 °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?.

Answer:

Volume,

V = m R T M p \frac{mRT}{Mp}

= 0.184 × R × 290 2 × p \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,

= m R T M p \frac{mRT}{Mp}

= 2.9 × R × 368 M × p \frac{2.9  \; \times \; R \; \times \; 368}{ M \; \times \; p}

According to the ques,

0.184 × R × 290 2 × p \frac{0.184  \; \times \; R \; \times \; 290}{ 2 \; \times \; p} = 2.9 × R × 368 M × p \frac{2.9  \; \times \; R \; \times \; 368}{ M \; \times \; p} 0.184 × 290 2 \frac{0.184  \; \times \; 290}{ 2 } = 2.9 × 368 M \frac{2.9  \;  \times \; 368}{ M }

M = 2.9 × 368 × 2 0.184 × 290 \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 dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate 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 (nH2),

= 20 2 \frac{20}{2}

= 10 moles

No. of moles of dioxygen (nO2),

= 80 32 \frac{80}{32}

= 2.5 moles

Given:

ptotal = 1 bar

Therefore, partial pressure of dihydrogen (pH2),

= n H 2 n H 2 + n O 2 \frac{ n_{H_{2}} }{ n_{H_{2}} \; + \; n_{O_{2}} } × ptotal

= 10 10 + 2.5 × 1 \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 p V 2 T 2 n \frac{ pV^{ 2 }T^{ 2 } }{ n } ?

Answer:

SI unit of pressure, p = N m − 2 Nm^{ -2 }

SI unit of volume, V = m 3 m^{ 3 }

SI unit of temp, T = K

SI unit of number of moles, n = mol

Hence, SI unit of p V 2 T 2 n \frac{ pV^{ 2 }T^{ 2 } }{ n } is,

= ( N m − 2 ) ( m 3 ) 2 ( K ) 2 m o l \frac{(Nm^{ -2 }) \; (m^{3})^{2} \; (K)^{2}}{mol}

= N m 4 K 2 m o l − 1 Nm^{ 4 }K^{ 2 }mol^{ -1 }

Q21. In terms of Charles' law explain why − 27 3 ∘ -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.

Charles' law

It was found that for all gases (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 temp-axis at − 27 3 ∘ -273^{\circ} C. That is the volume of any gas at − 27 3 ∘ -273^{\circ} C is 0. This happens because all gasses get transferred into liquid form before reaching − 27 3 ∘ -273^{\circ} C.

Therefore, it can be said that − 27 3 ∘ -273^{\circ} C is the lowest possible temp.

Q22. 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?

Answer:

If the critical temperature 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 CO2 intermolecular forces of attraction are stronger.

Q23. Explain 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'.

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We at BYJU'S strive to create high-quality content and to make it available to all knowledge seekers. The NCERT Solutions for Class 11 Chemistry (Chapter 5) of the second term provided here have been carefully constructed while keeping the needs of CBSE Class 11 students in mind. These NCERT Solutions are crafted to be concept focused and can, therefore, be used by students to revise the key topics in the chapter. The Chemistry subject experts who created these solutions have provided simple yet detailed solutions to every NCERT intext and exercise question. Students can also reach out to our support team to clear any concept related doubts they may face.

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What is the meaning of thermal energy according to NCERT Solutions for Class 11 Chemistry Chapter 5?

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