Question

NCERT Solutions Class 11 Chemistry Chapter 5 States of Matter: In this chapter, we have discussed the NCERT Solutions of the chapter States of Matter. There are 4 states of matter namely solid, liquid, gas, and plasma.

In the NCERT Solutions Class 11, we have provided solutions with in-depth analysis of every particular topic. The topics mentioned in the chapter are ­-

• States of Matter – different forms in which matter can exist in nature is called the state of matter. There is a total of five states in which matter exists and those are solid-liquid, gas, and plasma.
• Intermolecular Forces - the forces of attraction between atoms, molecules, and ions when they are placed close to each other are called intermolecular forces.
• Thermal Energy – the intrinsic energy possessed by a system in the state of thermodynamic equilibrium because of its temperature is called thermal energy.
• The Gaseous State – In the gaseous state of matter the particles are far away from each other, have higher kinetic energy, and the particles are fast-moving. In the gaseous state, the particles are not arranged in any particular way.
• Gas Law - the gas law can be defined as, at the constant volume, the pressure of the gas is directly proportional to the temperature of the gas. 
• Ideal Gas Equation – According to ideal gas law the product of the pressure and the volume of one gram molecule of an ideal gas is equal to the product of the absolute temperature of the gas and the universal gas constant.
• Kinetic Energy and Molecular Speeds - in any fluid we have observed that if the kinetic energy of a molecule is high it has more molecular speed.
• Kinetic Molecular Theory of Gases – there are three main assumptions of the kinetic theory of gases.
  • When molecules collide there is no gain or loss of any energy.
  • A particular molecule of any gas-only occupies a very small amount of space with respect to the container.
  • Due to the energy possessed by the gaseous molecule, these molecules are always in linear motion.
• Liquid State – the state of matter in which its molecules flow but do not disperse and possess a high degree of incompressibility.

Our NCERT Solutions Class 11 Chemistry is the study material one should refer to when preparing for the important exams like CBSE board or any competitive exams like JEE Mains, JEE Advance, Olympiad, and other similar exams.

Answer 1

NCERT Solutions Class 11 Chemistry Chapter 5 States of Matter

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

Answer:

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,

Therefore, the minimum pressure required is 2.5 bar.

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?

Answer:

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,

Therefore, the pressure would be 0.8 bar.

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

Answer:

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,

\(\frac{n}{V} = \frac p {RT}\)

Replacing n with \(\frac mM\) , we have

\(\frac mM=\frac p {RT}\)  ............. (ii)

Where, m → Mass of gas 

M → Molar mass of gas

But, \(\frac m V =d\)(d = density of gas) 

Thus, from equation (ii), we have

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

⇒ d = \((\frac M {RT})p\)

Molar mass (M) of a gas is always constant and therefore, at constant temperature

(T),\(\frac M {RT}\) = constant.

d = (constant) p

⇒ d \(\propto\) p

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

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?

Answer:

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

d = \(\frac{Mp}{RT}\)

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

d₁ = \(\frac{M_1p_1}{RT}\)

Where, M₁ and p₁ are the mass and pressure of the oxide respectively. 

Density of dinitrogen gas (d₂) is given by,

d₂ = \(\frac{M_2 p_2}{RT}\)

Where, M₂ and p₂ are the mass and pressure of the oxide respectively. 

According to the given question,

d₁ = d₂

\(\therefore\) M₁p₁ = M₂p₂

GIven,

p₁ = 2 bar

p₂ = 5 bar

Molecular mass of nitrogen, M₂ = 28 g/mol

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

5. 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,

p_AV = n_ART   ..........(i)

Where, p_A and n_A represent the pressure and number of moles of gas A. 

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

p_BV = n_BRT   ..........(ii)

Where, p_B and n_B represent the pressure and number of moles of gas B. 

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

From equation (i), we have

From equation (ii), we have

Where, M_A and M_B are the molecular masses of gases A and B respectively. 

Now, from equations (iii) and (iv), we have

Given,

m_A = 1 g

p_A = 2 bar

m_B = 2 g

p_B = (3 - 2) = 1 bar

(Since total pressure is 3 bar) 

Substituting these values in equation (v), we have

Thus, a relationship between the molecular masses of A and B is given by 4M_A = M_B .

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?

Answer:

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

At STP (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 \(\frac{3\times22400\times0.15}{54}\) mL of H₂  

i.e., 186.67 mL of H₂.

At STP,

p₁ = 1 atm

V₁ = 186.67 mL

T₁ = 273.15 K

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.

Now,

Therefore, 203 mL of dihydrogen will be released.

Answer 2

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 dm³ flask at 27 °C ?

Answer:

It is known that,

p = \(\frac m M \frac{RT} V\)

For methane (CH₄),

For carbon dioxide (CO₂),

Total pressure exerted by the mixture can be obtained as:

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

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?

Answer:

Let the partial pressure of H₂ in the vessel be P_^H2. 

Now,

p₁ = 0.8 bar

p₂ = \(P_{H_2}\)

V₁ = 0.5 L

V₂ = 1L

It is known that,

Now, let the partial pressure of O₂ in the vessel be \(P_{O_2}\).

Now,

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

Hence, the total pressure of the gaseous mixture in the vessel is 1.8 bar.

9. 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,

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

Hence, the density of the gas at STP will be 3 g dm^–3.

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?

Answer:

Given, 

p = 0.1 bar 

V = 34.05mL 

= 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 

The number of moles (n) can be calculated using the ideal gas equation as:

Therefore, molar mass of phosphorus = \(\frac{0.0625}{5.01\times10^{-5}}\) = 1247.5 g mol^–1

Hence, the molar mass of phosphorus is 1247.5 g mol^–1.

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?

Answer:

Let the volume of the round bottomed flask be V. 

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

Now,

V₁ = V 

T₁ = 27°C = 300 K 

V₂  =? 

T₂ = 477° C = 750 K

According to Charles’s law,

Therefore, volume of air expelled out = 2.5 V – V = 1.5 V

Hence, fraction of air expelled out = \(\frac{1.5V}{2.5V}=\frac{3}{5}\)

12. 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).

Answer:

Given, 

n = 4.0 mol 

V = 5 dm³ 

p = 3.32 bar 

R = 0.083 bar dm³ K^–1 mol^–1

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

Hence, the required temperature is 50 K.

13. 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 = \(\frac{1.4}{28}\) = 0.05 mol

= 0.05 x 6.02 x 10²³ numbers of molecules

= 3.01 x 10²³ numbers of molecules

Now, 1 molecule of N₂ contains 14 electrons.

Therefore, 3.01 × 10²³ molecules of N₂ contains = 14 × 3.01 × 1023 

= 4.214 × 10²³ electrons

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

Answer:

Avogadro number = 6.02 × 10²³ 

Thus, time required

Hence, the time taken would be 1.909 x 10⁶ years.

Answer 3

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.

Answer:

Given, 

Mass of dioxygen (O₂) = 8 g

Thus, number of moles of O₂ = \(\frac{8}{32}\) = 0.25 mole

Mass of dihydrogen (H₂) = 4 g

Thus, number of moles of H₂ = \(\frac{4}{2}\) = 2 mole

Therefore, total number of moles in the mixture = 0.25 + 2 = 2.25 mole 

Given, 

V = 1 dm³ 

n = 2.25 mol 

R = 0.083 bar dm³ K^–1 mol^–1 

T = 27°C = 300 K

Total pressure (p) can be calculated as: 

pV = nRT

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

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^–1 mol^–1).

Answer:

Given, Radius of the balloon, r = 10 m

\(\therefore\) Volume of the balloon = \(\frac{4}{3}\pi r^3\)

= \(\frac{4}{3} \times\frac{22}{7}\times10^3\)

= 4190.5 m³

Thus, the volume of the displaced air is 4190.5 m³. 

Given, 

Density of air = 1.2 kg m^–3 

Then, mass of displaced air = 4190.5 × 1.2 kg 

= 5028.6 kg 

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

m = \(\frac{MpV}{RT}\)

Here,

M = 4 x 10^-3 kg mol^-1

p = 1.66 bar

V = Volume of the balloon

= 4190.5 m³

R= 0.083 bar dm³ K^-1 mol^-1

T = 27° C = 300K

Then, 

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.

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

R = 0.083 bar LK^–1 mol^–1.

Answer:

It is known that,

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

Thus volume (V) = \(\cfrac{8.8\times0.083\times304.1}{44\times1}\)

= 5.04806 L

= 5.05 L

Hence, the volume occupied is 5.05 L.

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?

Answer:

Volume (V) occupied by dihydrogen is given by,

Let M be the molar mass of the unknown gas. Volume (V) occupied by the unknown gas can be calculated as:

According to the question,

Hence, the molar mass of the gas is 40 g mol^–1.

19. 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 and the weight of dioxygen be 80 g.

Then, the number of moles of dihydrogen, \(n_{H_2} = \frac{20}2=10 \, moles\) and the number of moles of dioxygen,

\(n_{O_2}=\frac{80}{32}=2.5 \,moles\)

Given, 

Total pressure of the mixture, p_total = 1 bar 

Then, partial pressure of dihydrogen,

Hence, the partial pressure of dihydrogen is 0.8 bar.

20. What would be the SI unit for the quantity pV² T² /n?

Answer:

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.

Therefore, the SI unit for quantity \(\frac{pV^2T^2}{n}\) is given by,

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

Answer:

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

It was found that for all gases (at any given pressure), the plots of volume vs. temperature (in °C) is a straight line. If this line is extended to zero volume, then it intersects the temperature-axis at – 273°C. In other words, the volume of any gas at – 273°C is zero. 

This is because all gases get liquefied before reaching a temperature of – 273°C. Hence, it can be concluded that – 273°C is the lowest possible temperature.

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?

Answer:

Higher is the critical temperature of a gas, easier is its liquefaction. This means that the intermolecular forces of attraction between the molecules of a gas are directly proportional to its critical temperature. Hence, intermolecular forces of attraction are stronger in the case of CO₂.

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

Answer:

Physical significance of ‘a’: 

‘a’ is a measure of the magnitude of intermolecular attractive forces within a gas. 

Physical significance of ‘b’: 

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