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

#### Page No 152:

#### Question 5.1:

What will be the minimum pressure required to compress 500 dm

^{3}of air at 1 bar to 200 dm^{3}at 30°C?#### Answer:

Given,

Initial pressure,

*p*_{1}= 1 bar
Initial volume,

*V*_{1}= 500 dm^{3}
Final volume,

*V*_{2}= 200 dm^{3}
Since the temperature remains constant, the final pressure (

*p*_{2}) can be calculated using Boyle’s law.
According to Boyle’s law,

Therefore, the minimum pressure required is 2.5 bar.

#### Page No 153:

#### 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?

#### Answer:

Given,

Initial pressure,

*p*_{1}= 1.2 bar
Initial volume,

*V*_{1 }= 120 mL
Final volume,

*V*_{2}= 180 mL
Since the temperature remains constant, the final pressure (

*p*_{2}) 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*=*n*R*T*; 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*=

*n*R

*T*……….. (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?

#### Answer:

Density (d) of the substance at temperature (

*T*) can be given by the expression,*d*=

Now, density of oxide (

*d*_{1}) is given by,
Where,

*M*_{1}and*p*_{1}are the mass and pressure of the oxide respectively.
Density of dinitrogen gas (

*d*_{2}) is given by,
Where,

*M*_{2}and*p*_{2}are the mass and pressure of the oxide respectively.
According to the given question,

Molecular mass of nitrogen,

*M*_{2}= 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.

#### Answer:

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?

#### 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 H

_{2..}
0.15 g Al gives i.e., 186.67 mL of H

_{2.}
At STP,

Let the volume of dihydrogen be at

*p*_{2}= 0.987 atm (since 1 bar = 0.987 atm) and*T*_{2}= 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 dm

^{3}flask at 27 °C ?#### Answer:

It is known that,

For methane (CH

_{4}),
For carbon dioxide (CO

_{2}),
Total pressure exerted by the mixture can be obtained as:

Hence, the total pressure exerted by the mixture is 8.314

**× 10**^{4}Pa.#### Question 5.8:

What will be the pressure of the gaseous mixture when 0.5 L of H

_{2}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}in the vessel be.
Now,

= ?

It is known that,

Now, let the partial pressure of O

_{2}in the vessel be.
p1V1=p2V2⇒p2=p1V1V2⇒pO2=0.7×2.01= 1.4 bar

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.

#### Question 5.9:

Density of a gas is found to be 5.46 g/dm

^{3}at 27 °C at 2 bar pressure. What will be its density at STP?#### Answer:

Given,

The density (

*d*_{2}) 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?

#### Answer:

Given,

*p*= 0.1 bar

*V*= 34.05 mL = 34.05 × 10

^{–3}L = 34.05 × 10

^{–3}dm

^{3}

R = 0.083 bar dm

^{3}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?

#### 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,

*V*

_{1}=

*V*

*T*

_{1}= 27°C = 300 K

*V*

_{2}=?

*T*

_{2}= 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 dm

^{3 }at 3.32 bar.
(R = 0.083 bar dm

^{3}K^{–1}mol^{–1}).#### Answer:

Given,

*n*= 4.0 mol

*V*= 5 dm

^{3}

*p*= 3.32 bar

R = 0.083 bar dm

^{3}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.

#### Answer:

Molar mass of dinitrogen (N

_{2}) = 28 g mol^{–1}
Thus, 1.4 g of

Now, 1 molecule of contains 14 electrons.

Therefore, 3.01 × 10

^{23}molecules of N_{2}contains = 1.4 × 3.01 × 10^{23}
= 4.214 × 10

^{23}electrons#### Question 5.14:

How much time would it take to distribute one Avogadro number of wheat grains, if 10

^{10}grains are distributed each second?#### Answer:

Avogadro number = 6.02 × 10

^{23}
Thus, time required

=

Hence, the time taken would be.

#### 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 dm

^{3}at 27°C. R = 0.083 bar dm^{3}K^{–}^{1}mol^{–}^{1.}#### Answer:

Given,

Mass of dioxygen (O

_{2}) = 8 g
Thus, number of moles of

Mass of dihydrogen (H

_{2}) = 4 g
Thus, number of moles of

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

Given,

*V*= 1 dm

^{3}

*n*= 2.25 mol

R = 0.083 bar dm

^{3}K^{–1}mol^{–1}*T*= 27°C = 300 K

Total pressure (

*p*) can be calculated as:*pV*=

*n*R

*T*

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 dm^{3}K^{–}^{1}mol^{–}^{1}).#### Answer:

Given,

Radius of the balloon,

*r*= 10 m
Volume of the balloon

Thus, the volume of the displaced air is 4190.5 m

^{3}.
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 CO

_{2}at 31.1°C and 1 bar pressure.
R = 0.083 bar L K

^{–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

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?

#### 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}.#### 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.

#### Answer:

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

*pV*^{2}*T*^{2}/*n?*#### Answer:

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.

#### 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.

#### 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?

#### 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 CO

_{2}.#### Question 5.23:

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.

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