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

NCERT Solutions for Class 11 Chemistry: Chapter 5 (States of Matter) are provided on this page for the perusal of class 11 chemistry students. Detailed, step-by-step solutions to every question listed in chapter 5 of the NCERT class 11 chemistry textbook can be found here. Also, these NCERT solutions can be downloaded in a PDF format (for free) by clicking the download button provided above.

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NCERT Solutions for Chemistry – Class 11, 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.

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, P1 = 1 bar

Initial volume, V1 = 500 dm3dm^{ 3 }

Final volume, V2 = 200 dm3dm^{ 3 }

As the temperature remains same,the final pressure (P2) can be calculated with the help of Boyle’s law.

Acc. Boyle’s law,

P1V1 = P2V2

P2 = P1V1V2\frac{P_{1}V_{1}}{V_{2}}

= 1  ×  500200\frac{1 \; \times \; 500}{200}

= 2.5 bar

∴ the minimum 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, P1 = 1.2 bar

Initial volume, V1 = 120 mL 

Final volume, V2 = 180 mL

As the temperature remains same, final pressure (P2) can be calculated with the help of Boyle’s law.

According to the Boyle’s law,

P1V1 = P2V2

P2 = P1V1V2\frac{P_{1}V_{1}}{V_{2}}

= 1.2  ×  120180\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

nV\frac{n}{V} = pRT\frac{p}{RT}

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

mMV\frac{m}{MV} = pRT\frac{p}{RT}……..(2)

Where, m = mass

M = molar mass

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

Where, d = density

Therefore, from equation (2), we get

dM\frac{d}{M} = pRT\frac{p}{RT}

d = (MRT\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°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?

Answer:

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

d = MpRT\frac{Mp}{RT}

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

d1d_{1} = M1p1RT\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,

d2d_{2} = M1p2RT\frac{M_{1}p_{2}}{RT}

Where, M2 = mass of the oxide

p2 = pressure of the oxide

Acc to the question,

d1 = d2

Therefore, M1p1=M2p2M_{1}p_{1} = M_{2}p_{2}

Given:

p1p_{1} = 2 bar

p2p_{2} = 5 bar

Molecular mass of nitrogen, M2M_{2} = 28 g/mol

Now, M1M_{1}

= M2p2p1\frac{M_{ 2 }p_{2}}{p_{ 1 }}

= 28×52\frac{ 28 × 5 }{ 2 }

= 70 g/mol

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

 

Q5. 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 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,

pXV=nXRTp_{X}V = n_{X}RT……(1)

Where pXp_{X} and nXn_{X} represent the pressure and number of moles of gas X.

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

pYV=nYRTp_{Y}V = n_{Y}RT……(2)

Where, pYp_{Y} and nYn_{Y} represent the pressure and number of moles of gas Y.

[V and T are constants for gases X and Y]

From equation (1),

pXV=mXMXp_{ X }V = \frac{m_{ X }}{M_{ X }} RT

pXMXmX\frac{p_{ X }M_{ X }}{m_{ X }} = RTV\frac{ R T}{ V } ……(3)

From equation (2),

pYV=mYMYp_{ Y }V =\frac{m_{ Y }}{M_{ Y }} RT

pYMYmY\frac{p_{ Y }M_{ Y }}{m_{ Y }} = RTV\frac{ R T}{V} …… (4)

Where, MXM_{ X } and MYM_{ Y } are the molecular masses of gases X and Y respectively.

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

pXMXmX\frac{p_{ X }M_{ X }}{m_{ X }} = pYMYmY\frac{p_{ Y }M_{ Y }}{m_{ Y }} ….. (5)

Given,

mXm_{ X } = 1 g

pXp_{ X } = 2 bar

mYm_{ Y } = 2 g

pYp_{ Y } = (3 – 2) = 1 bar (Since total pressure is 3 bar)

Substituting these values in equation (5),

2  ×  MX1\frac{2 \; \times \; M_{X} }{1} = 1  ×  MY2\frac{1 \; \times \; M_{Y} }{2}

4 MXM_{ X } = MYM_{ Y }

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

4 MXM_{ X } = MYM_{ 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  ×  0.1554\frac{3 \; \times \; 22400 \; \times \; 0.15}{54} mL of H2

= 186.67 mL of H2

At Standard Temperature Pressure,

p1p_{ 1 } = 1 atm

V1V_{ 1 } = 186.67 mL

T1T_{ 1 } = 273.15 K

Let the volume of dihydrogen be V2V_{ 2 } at p2p_{ 2 } = 0.987 atm (since 1 bar = 0.987 atm) and T2T_{ 2 } = 2020^{\circ} C = (273.15 + 20) K = 293.15 K.

Now,

p1V1T1\frac{p_{ 1 }V_{ 1 }}{T_{ 1 }} = p2V2T2\frac{p_{ 2 }V_{ 2 }}{T_{ 2 }} V2=p1V1T2p2T1V_{ 2 } = \frac{p_{ 1 }V_{ 1 }T_{ 2 }}{p_{ 2 }T_{ 1 }}

= 1  ×  186.67  ×  293.150.987  ×  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 = mM\frac{m}{M} RTV\frac{RT}{V}

For methane (CH4),

pCH4p_{CH_{ 4 }}

= 3.216\frac{ 3.2 }{ 16 } × 8.314  ×  3009  ×  10–           3\frac{8.314 \; \times \;300 }{9 \; \times \; 10^{ –            3 }} [Since 9 dm3 = 9  ×  1039 \; \times \; 10^{ -3 } m3         2727^{\circ}C = 300 K]

= 5.543 × 10410^{ 4 } Pa

For carbon dioxide (CO2),

pCO2p_{CO_{ 2 }}

= 4.444\frac{ 4.4 }{ 44 } × 8.314  ×  3009  ×  10–           3\frac{8.314 \; \times \;300 }{9 \; \times \; 10^{ –            3 }}

= 2.771 × 10410^{4} Pa

Total pressure exerted by the mixture can be calculated as:

p = pCH4p_{CH_{ 4 }} + pCO2p_{CO_{ 2 }}

= (5.543 × 10410^{ 4 } + 2.771 × 10410^{4}) Pa

= 8.314 × 10410^{ 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 H2{H_{ 2 }} in the container be pH2p_{H_{ 2 }}.

Now,

p1{p_{ 1 }} = 0.8 bar

p2{p_{ 2 }} = pH2p_{H_{ 2 }} V1{V_{ 1 }} = 0.5 L

V2{V_{ 2 }} = 1 L

It is known that,

p1{p_{ 1 }} V1{V_{ 1 }} = p2{p_{ 2 }} V2{V_{ 2 }} p2{p_{ 2 }} = p1  ×  V1V2\frac{p_{ 1 } \; \times \; V_{ 1 }}{V_{ 2 }} pH2p_{H_{ 2 }} = 0.8   ×  0.51\frac{ 0.8  \; \times \; 0.5 }{ 1 }

= 0.4 bar

Now, let the partial pressure of O2 in the container be pO2p_{O_{ 2 }}.

Now,

p1{p_{ 1 }} = 0.7 bar

p2{p_{ 2 }} = pO2p_{O_{ 2 }} V1{V_{ 1 }} = 2.0 L

V2{V_{ 2 }} = 1 L

p1{p_{ 1 }} V1{V_{ 1 }} = p2{p_{ 2 }} V2{V_{ 2 }} p2{p_{ 2 }} = p1  ×  V1V2\frac{p_{ 1 } \; \times \; V_{ 1 }}{V_{ 2 }} pO2p_{O_{ 2 }} = 0.7   ×  201\frac{ 0.7  \; \times \; 20 }{ 1 }

= 1.4 bar

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

ptotalp_{total} = pH2p_{H_{ 2 }} + pO2p_{O_{ 2 }}

= 0.4 + 1.4

= 1.8 bar

 

Q9. 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 = 2727^{\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 = MpRT\frac{Mp}{RT} d1d2\frac{d_{1}}{d_{2}} = M  p1R  T1M  p2R  T2\frac{\frac{M \;p_{ 1 }}{R \; T_{ 1 }}}{\frac{M \; p_{ 2 }}{R \; T_{ 2 }}} d1d2\frac{d_{1}}{d_{2}} = p1  T2p2  T1\frac{p_{1} \; T_{2}}{p_{2} \; T_{1}}

d2 = p2  T1  d1p1  T2\frac{p_{2} \; T_{1} \; d_{1}}{p_{1} \; T_{2}}

= 1  ×300  ×5.462  ×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 dm3dm^{ 3 }

R = 0.083 bar dm3dm^{ 3 } at  K-1 mol-1

T = 546C546^{\circ} C = (546 + 273) K = 819 K
The no of moles (n) can be calculated using the ideal gas equation as:

pV = nRT

n = pVRT\frac{ pV }{ RT }

=  0.1×34.05×103  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 = 0.06255.01 ×105\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 2727^{\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 477477^{\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 2727^{\circ} C is V.

Now,

V1 = V

T1 = 2727^{\circ} C = 300 K V2 = ?

T2 = 477477^{\circ} C = 750 K

Acc to Charles’s law,

V1T1\frac{V_{1}}{T_{1}} = V2T2\frac{V_{2}}{T_{2}} V1V_{ 1 } = V1T2T1\frac{V_{ 1 }T_{ 2 }}{T_{ 1 }}

= 750V300\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

= 1.5V2.5V\frac{1.5V }{ 2.5V }

= 35\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 dm3dm^{ 3 }

p = 3.32 bar

R = 0.083 bar dm3dm^{ 3 } at  K-1 mol-1

 

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

pV = nRT

T = pVnR\frac{ pV }{ nR }

= 3.32  ×  54  ×  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

= 1.428\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  ×  10231010s\frac{6.02 \; \times \; 10^{23}}{10^{10}} s

= 6.02 × 1013 s

= 6.02  ×  102360  ×  60  ×  24  ×  365years\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

= 832\frac{ 8 }{ 32 }

= 0.25 mole

 

Mass of H2 = 4 g

No. of moles

= 42\frac{ 4 }{ 2 }

= 2 mole

 

Hence, total no of moles in the mixture

= 0.25 + 2

= 2.25 mole

 

Given:

V = 1 dm3dm^{ 3 }

n = 2.25 mol

R = 0.083 bar dm3dm^{ 3 } at K-1 mol-1

T = 2727^{\circ} C = 300 K

 

Total pressure :
pV = nRT

p = nRTV\frac{ nRT }{ V }

= 225  ×  0.083  ×  3001\frac{225 \; \times \; 0.083 \; \times \; 300 }{ 1 }

= 56.025 bar

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

 

Q16. 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–1 mol–1)

Answer:

Given:

r = 10 m

Therefore, volume of the balloon

= 43\frac{4}{3} πr3

= 43  ×  227  ×  103\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,

= MpVRT\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 dm3dm^{ 3 } at K-1 mol-1

T = 27 °C = 300 K

 

Then,

m = 4  ×  103  ×  1.66  ×  4190.5  ×  1030.083  ×  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 = mRTMp\frac{mRT}{Mp}

Given:

m = 8.8 g

R = 0.083 bar dm3dm^{ 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  ×  304.144  ×  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 = mRTMp\frac{mRT}{Mp}

= 0.184   ×  R  ×  2902  ×  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,

= mRTMp\frac{mRT}{Mp}

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

According to the ques,

0.184   ×  R  ×  2902  ×  p\frac{0.184  \; \times \; R \; \times \; 290}{ 2 \; \times \; p} = 2.9   ×  R  ×  368M  ×  p\frac{2.9  \; \times \; R \; \times \; 368}{ M \; \times \; p} 0.184   ×  2902\frac{0.184  \; \times \; 290}{ 2 } = 2.9    ×  368M\frac{2.9  \;  \times \; 368}{ M }

M = 2.9  ×  368  ×  20.184  ×  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),

= 202\frac{20}{2}

= 10 moles

No. of moles of dioxygen (nO2),

= 8032\frac{80}{32}

= 2.5 moles

Given:

ptotal = 1 bar

Therefore, partial pressure of dihydrogen (pH2),

= nH2nH2  +  nO2\frac{ n_{H_{2}} }{ n_{H_{2}} \; + \; n_{O_{2}} } × ptotal

= 1010  +  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 pV2T2n\frac{ pV^{ 2 }T^{ 2 } }{ n }?

Answer:

SI unit of pressure, p = Nm2Nm^{ -2 }

SI unit of volume, V = m3m^{ 3 }

SI unit of temp, T = K

SI unit of number of moles, n = mol

Hence, SI unit of pV2T2n\frac{ pV^{ 2 }T^{ 2 } }{ n } is,

= (Nm2)  (m3)2  (K)2mol\frac{(Nm^{ -2 }) \; (m^{3})^{2} \; (K)^{2}}{mol}

= Nm4K2mol1Nm^{ 4 }K^{ 2 }mol^{ -1 }

 

Q21. In terms of Charles’ law explain why 273-273^{\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 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 temp-axis at 273-273^{\circ} C. That is the volume of any gas at273-273^{\circ} C is 0. This happens because all gasses get transferred into liquid form before reaching 273-273^{\circ} C.

Therefore, it can be said that273-273^{\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 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 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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