States of Matter (Old NCERT) NCERT Solutions | Chemistry Class 11 - NEET PDF Download

Question 5.1: What will be the minimum pressure required to compress 500 dm³ of air at 1 bar to 200 dm³ 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,

Therefore, the minimum pressure required is 2.5 bar.

Question 5.2: 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,

Therefore, the pressure would be 0.8 bar.

Question 5.3: 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,

Replacing n with , we have

Where,
m → Mass of gas
M → Molar mass of gas
But, (d = density of gas)
Thus, from equation (ii), we have

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

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

Question 5.4: 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 by the expression,

d =

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

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

Density of dinitrogen gas (d₂) is given by,

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

According to the given question,

Molecular mass of nitrogen, M₂ = 28 g/mol

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

Question 5.5: 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.
Ans:- For ideal gas A, the ideal gas equation is given by,

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,

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,

(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

.

Question 5.6: 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 can be represented as:

At STP (273.15 K and 1 atm), 54 g (2 × 27 g) of Al gives 3 × 22400 mL of H_2..

0.15 g Al gives i.e., 186.67 mL of H_2.

At STP,

Let the volume of dihydrogen be at p₂ = 0.987 atm (since 1 bar = 0.987 atm) and T₂ = 20°C = (273.15 20) K = 293.15 K._.

Therefore, 203 mL of dihydrogen will be released.

Question 5.7: 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 ?
Ans:- It is known that,

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.

Question 5.8: 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
Now,
p₁ = 0.8 bar, p₂ = p_H2 = ?
V₁ = 0.5 L, V₂ = 1L
It is known that,

Now, let the partial pressure of O₂ in the vessel be Po₂.

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.

Question 5.9 : Density of a gas is found to be 5.46g/dm³ at 27°C at 2 bar pressure.What will be its density at STP?
Ans:- 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.

Question 5.10: 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:- Given,
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
The number of moles (n) can be calculated using the ideal gas equation as:

Therefore, molar mass of phosphorus = 1247.5 g mol^–1

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

Question 5.11: 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:- 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

Question 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).
Ans:- 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.

Question 5.13: Calculate the total number of electrons present in 1.4 g of dinitrogen gas.
Ans:- Molar mass of dinitrogen (N₂) = 28 g mol^–1

Thus, 1.4 g of

Now, 1 molecule of contains 14 electrons.

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

= 4.214 × 10²³ electrons

Question 5.14: How much time would it take to distribute one Avogadro number of wheat grains, if 10¹⁰ grains are distributed each second?
Ans:- Avogadro number = 6.02 × 10²³

Thus, time required

= 

Hence, the time taken would be 1.9090 X 10⁶ years..

Question 5.15: 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:- Given,
Mass of dioxygen (O₂) = 8 g
Thus, number of moles of
Mass of dihydrogen (H₂) = 4 g
Thus, number of moles of
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.

Question 5.16: 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). 
Ans:- Given,
Radius of the balloon, r = 10 m
Volume of the balloon

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,

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.

Question 5.17: 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:- 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

Hence, the volume occupied is 5.05 L.

Question 5.18: 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 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.

Question 5.19: A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.
Ans:- Let the weight of dihydrogen be 20 g and the weight of dioxygen be 80 g.

Then, the number of moles of dihydrogen,      and the number of moles of dioxygen, .
Given,
Total pressure of the mixture, p_total = 1 bar
Then, partial pressure of dihydrogen,

Hence, the partial pressure of dihydrogen is.

Question 5.20: What would be the SI unit for the quantity pV²T ²/n?
Ans:- The SI unit for pressure, p is Nm^–2.
The SI unit for volume, V is m^3.
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        is given by,

Question 5.21: 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.

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.

Question 5.22: 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:- 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.

Question 5.23: Explain the physical significance of Van der Waals parameters.
Ans:- 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.