## CBSE NCERT Class XI (11th) | Chemistry

Chapter 5 - States of Matter

Q1 :

Answer :

Given,

Initial pressure,

*p*1 = 1 bar
Initial volume,

*V*1 = 500 dm3
Final volume,

*V*2 = 200 dm3
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.

Q2 :

Answer :

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.

Q3 :

Answer :

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*

*Ã¢â€ ’*Pressure of gas

*V*Ã¢â€ ’ Volume of gas

*n*Ã¢â€ ’ Number of moles of gas

R Ã¢â€ ’ Gas constant

*T*Ã¢â€ ’ Temperature of gas

From equation (i) we have,

Replacing

*n*with , we have
Where,

*m*Ã¢â€ ’ Mass of gas

*M*Ã¢â€ ’ 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)*
Q4 :

Answer :

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.

Q5 :

Answer :

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, MA and MB 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

.

Q6 :

Answer :

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..

0.15 g Al gives i.e., 186.67 mL of H2.

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.

Q7 :

Answer :

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,

For methane (CH 4),

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.

Q8 :

Answer :
p1V1=p2

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.

Now,

= ?

It is known that,

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

Q9 :

Answer :

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 (

*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 dmâ€“3.

Q10 :

Answer :

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 × 10â€“3 L = 34.05 × 10â€“3 dm3

R = 0.083 bar dm3 Kâ€“1 molâ€“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 molâ€“1

Hence, the molar mass of phosphorus is 1247.5 g molâ€“1.

Q11 :

Answer :

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*â€“*V*= 1.5*V*
Hence, fraction of air expelled out

Q12 :

Answer :

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 Kâ€“1 molâ€“1

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

Hence, the required temperature is 50 K.

Q13 :

Answer :

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

Answer :

Molar mass of dinitrogen (N2) = 28 g molâ€“1

Thus, 1.4 g of

Now, 1 molecule of contains 14 electrons.

Therefore, 3.01 × 1023 molecules of N2 contains = 1.4 × 3.01 × 1023

= 4.214 × 1023 electrons

Q14 :

Answer :

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

Answer :

Avogadro number = 6.02 × 1023

Thus, time required

=

Hence, the time taken would be.

Q15 :

Answer :

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

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 Kâ€“1 molâ€“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.

Q16 :

Answer :

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
Volume of the balloon

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

Given,

Density of air = 1.2 kg mâ€“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 â€“ 1217.5) kg

= 3811.1 kg

Hence, the pay load of the balloon is 3811.1 kg.

Q17 :

Answer :

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.

Answer :

It is known that,

Here,

*m*= 8.8 g

R = 0.083 bar LKâ€“1 molâ€“1

*T*= 31.1°C = 304.1 K

*M*= 44 g

*p*= 1 bar

Hence, the volume occupied is 5.05 L.

Q18 :

Answer :

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 molâ€“1.

Q19 :

Answer :

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.

Q20 :

Answer :

What would be the SI unit for the quantity

*pV*2*T*2/*n?*Answer :

The SI unit for pressure,

*p*is Nmâ€“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 is given by,

Q21 :

Answer :

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.

Q22 :

Answer :

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.

Q23 :

Answer :

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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