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States Of Matter

Class 11th Chemistry Part I CBSE Solution
Exercise
  1. What will be the minimum pressure required to compress 500 dm^3 of air at 1 bar to 200…
  2. A vessel of 120 mL capacity contains a certain amount of gas at 35°C and 1.2 bar pressure.…
  3. Using the equation of state pV = nRT; show that at a given temperature density of a gas is…
  4. At 0°C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at…
  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…
  6. The drain cleaner, Drainex contains small bits of aluminium which react with caustic soda…
  7. What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon…
  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…
  9. Density of a gas is found to be 5.46 g/dm^3 at 27°C at 2 bar pressures. What will be its…
  10. 34.05 mL of phosphorus vapour weighs 0.0625 g at 546°C and 0.1 bar pressure. What is the…
  11. A student forgot to add the reaction mixture to the round bottomed flask at 27°C but…
  12. Calculate the temperature of 4.0 mol of a gas occupying 5 dm^3 at 3.32 bar. (R = 0.083 bar…
  13. Calculate the total number of electrons present in 1.4 g of dinitrogen gas.…
  14. How much time would it take to distribute one Avogadro number of wheat grains, if 10^10…
  15. Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen…
  16. Pay load is defined as the difference between the mass of displaced air and the mass of…
  17. Calculate the volume occupied by 8.8 g of CO2 at 31.1°C and 1 bar pressure. R = 0.083 bar…
  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…
  19. A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of…
  20. What would be the SI unit for the quantity pV2T2/n?
  21. In terms of Charles’ law explain why -273°C is the lowest possible temperature.…
  22. Critical temperature for carbon dioxide and methane are 31.1°C and -81.9°C respectively.…
  23. Explain the physical significance of van der Waals parameters.

Exercise
Question 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 that P1 = 1 bar.

V1 = 500 dm3


V2 = 200dm3


As temperature is constant, the final pressure can be calculated by Boyle's law,


P1V1 = P2V2



⇒ P2 =


⇒ P2 = 2.5 bar


Therefore, minimum pressure required is 2.5 bar.



Question 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:

According to Boyle’s Law P1V1= P2V2

Here the temperature is constant. Therefore

1.2 X 120 = P2X 180

OR

P2= 1.2 X 120 /180 = 0.8 bar


Question 3.

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 ……….. (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 = m/M we have



because the number of moles is given as:


Where,


M - Mass of gas


M - Molar mass of gas


But,


(where, d = density of gas)


Thus, from equation (ii), we have




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


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



Question 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,


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



Where, M1 and P1 are mass and pressure of the oxide respectively.


Density of dinitrogen gas (d2) is given by,


⇒ d2 =


Where,M2 and P2 are the mass and pressure of the oxide respectively.


According to the given question,


⇒ d1 = d2


Molecular mass of nitrogen, M2 = 28 g/mol


⇒ M1P1 = M2P2


Therefore, we can write


Hence molar mass



⇒ M = 70 g/mol


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



Question 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,

⇒ PaV = naRT


Where, Pa and na represent the pressure and number of moles of gas A.


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


⇒ PbV = nbRT


Where,Pa and nb represent the pressure and number of moles of gas B.


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


Therefore, we have,



Where, Ma and Mb are the molecular masses of gases A and B respectively.


Now, we have


Substituting the values


ma = 1 g


Pa = 2 bar


mb = 2 g


Pb = 1 bar.


Given,


(Since total pressure is 3 bar)


Substituting these values in equation we have



Thus, a relationship between the molecular masses of A and B is given by


⇒ 4Ma = Mb



Question 6.

The drain cleaner, Drainex contains small bits of aluminium 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 aluminium 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 Hydrogen


⇒ 0.15 g Al gives i.e. 3×22400×0.15/54 = 186.67 ml of Hydrogen.


At STP,


Let the volume of dihydrogen V2 be at p2 = 0.987 atm (since 1 bar = 0.987 atm) and T2 = 293.15K.


⇒ 20°C = (273.15 + 20) K = 293.15 K.


= constant


So,



⇒ V2 = 203ml


Therefore, 203 ml of dihydrogen will be released.



Question 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: Given:
The volume of the container: 9 dm3
The temperature of the container: 300 K
The atomic mass of the CH4 =12+4×1=16
the mole of CH4 =
The atomic mass of CO2 =44g
The moles of CO2=
The total number of mole = 0.2+ 0.1 mole = 0.3 mole
The ideal gas equation:
PV = n R T
Where,
P is the pressure of the gas
V is the volume of the gas
n is the no. of mole of gas
R is the gas constant
T is the temperature in K
Putting the values in the above equation, we get
P × 9 dm3 = 0.3 ×8.314×104× 300 K



Question 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 hydrogen in the vessel be Pa.

P1 = 0.8 bar,


V1 = 0.5L


V2 = 1L


P2 = Pa


Now, at constant temperature


⇒ P1V1 = P2V2



Pa = 0.4 bar


Similarly, for oxygen we get the partial pressures of oxygen as



⇒ Pb = 1.4 bar


It is known that,


The total pressure exerted is the sum of partial pressures so total pressure is (1.4+0.4) = 1.8 bar.



Question 9.

Density of a gas is found to be 5.46 g/dm3 at 27°C at 2 bar pressures. What will be its density at STP?


Answer:

For an ideal gas

Density ρ = p x M/ RT

F rom the given data,

5.46 = 2 x M/ 300 x R ………(1)

ALSO ρSTP = 1 x M/ 273 x R ………..(2)

FROM (1) & (2) ,WE GET

ρSTP = (1 x M/ 273 x R) x (300 x R/ 2 x M)

= 300 / 273 X 2 = 3.00 g/dm3


Question 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 × 0.001 L = 34.05 × 0.001dm3


R = 0.083 bar dm3/K mol


T = 546°C = (546 + 273) K = 819 K


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


PV = nRT


Where, P - Pressure of gas


V - Volume of gas


n - Number of moles of gas


R- Gas constant


T -Temperature of gas



n = 5.01×10-5 mol


Therefore, molar mass of phosphorus = mass/moles



⇒ M = 1247.5 g/mol


Hence, the molar mass of phosphorus is 1247.5 g/mol.



Question 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:

Given:

Initial temperature, T1 =27°C = (27 + 273) K = 300K


Final temperature, T2 = 477°C (477 + 273) K = 750K


Let the volume of the round bottomed flask be V.


Then, the volume of air inside the flask at 300K is, V1 = V.


Let the volume of air inside the flask at 750K is = V2


According to Charlee’s law we know that: -





(∵ V1 = V)



∴ Volume of air expelled out = 2.5 V – V = 1.5 V


Hence, the fraction of the air expelled out is given by: -



∴ The fraction of the air expelled out is: 3/5.



Question 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:

Number of moles, n = 4


Volume, V = 5 dm3


Pressure, P = 3.32 bar


Universal gas constant, R = 0.083 bar dm3 K–1 mol–1


Let the temperature be T.


Then, from ideal gas equation we know that: -


PV = nRT




= 50K


Hence, the temperature of 4.0 mol of a gas occupying 5 dm3 at 3.32 bar is 50K.



Question 13.

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


Answer:

Given:

Mass of dinitrogen = 1.4 g


We know that molar mass of dinitrogen = 28 g mol-1


∴ The number of moles of dinitrogen in 1.4g = 1.4/28 = 0.05mol


Now, the number of molecules in 1.4g of dinitrogen: -


= 0.05 x 6.022 × 1023


= 3.01 x 1023


We know that 1 molecule of dinitrogen contains 14 electrons.


∴ 3.01 × 1023 molecules of dinitrogen contain:


= 14 × 3.01 × 1023


= 4.214 × 1023 electrons.


Hence, the total number of electrons present in 1.4 g of dinitrogen gas is 4.214 × 1023.



Question 14.

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


Answer:

We know that Avogadro number = 6.02 × 1023.

Given that rate of distribution = 1010 grains per second.


Thus, time required to distribute one Avogadro number of wheat grains: -


= Avogadro number / rate of distribution



= 6.023 x 1013 s



=1.909 x 106 years


Hence, the time taken to distribute one Avogadro number of wheat grains is 1.909 x 106 years.



Question 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 K1 mol1.


Answer:

Given:

Mass of dioxygen= 8 g


Mass of dihydrogen= 4 g


Now, molar mass of dioxygen = 32g


Thus, number of moles of dioxygen = 8/32 = 0.25 mole


Also, molar mass of dihydrogen = 2g


Thus, number of moles of dihydrogen = 4/2 = 2 mole


Therefore, total number of moles in the mixture:


= 0.25 + 2 = 2.25 mole


Given: -


V = volume = 1 dm3,


n = number of moles = 2.25 mol


R = universal gas constant = 0.083 bar dm3 K-1 mol-1


T = temperature = 27°C = 300 K


Let the total pressure be P.


From ideal gas equation we know that -


PV = nRT




= 56.025 bar.


Hence, total pressure of the mixture is 56.025 bar.



Question 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 m3 and R = 0.083 bar dm3 K1 mol1).


Answer:

Given

Radius of the balloon, r = 10 m


∴ Volume of the balloon = 4/3 x πr3


= 4/3 x 22/7 x 103


= 4190.5 m3


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


Also, given that


Density of air = 1.2 kg m-3


Then, mass of displaced air = 4190.5 × 1.2 kg


= 5028.6 kg


From ideal gas equation we know that: -


PV = nRT


⇒ PV = m/M x RT


Where, P = pressure = 1.66bar


V= volume = 4190.5 m3


R = universal gas constant = 0.083 bar dm3 K–1 mol–1


T= temperature = 27 + 273 = 300K


m = mass of helium inside the balloon


M = molar mass of helium = g




=1117.5kg


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


∴The pay load of the balloon is 3811.1 kg.



Question 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.


Answer:

From ideal gas equation we know that: -

PV = nRT


⇒ PV = m/M x RT


Where, P = pressure = 1 bar


R = universal gas constant = 0.083 bar dm3 K–1 mol–1


T= temperature = 31.1 +273 = 304.1 K


m = mass of CO2 = 8.8 g


M = molar mass CO2 = 44g


Let the volume be V




= 5.05L


∴ The volume occupied is 5.05 L.



Question 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:

From ideal gas equation we know that: -

PV = nRT


⇒ PV = m/M x RT



Where, P = pressure of dihydrogen


R = Universal gas constant


T= temperature of dihydrogen


m = mass of dihydrogen


M = molar mass of dihydrogen


V = volume of dihydrogen



Let M be the molar mass of the unknown gas, pressure remains constant. Volume (V) occupied by the unknown gas can be calculated in the same way as above:




According to the question,





= 40gmol-1


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



Question 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(H2) = 20/2 = 10 moles


The number of moles of dioxygen:


n(O2) = 80/32 = 2.5 moles


Given that,


Total pressure of the mixture, p(total) = 1 bar


Then, partial pressure of dihydrogen:




= 0.8 bar


Hence, the partial pressure of dihydrogen is 0.8 bar.



Question 20.

What would be the SI unit for the quantity pV2T2/n?


Answer:

We know that

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 pV2T2/n is: -



= Nm4K2mol-1.



Question 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.



From the above graph between V and T we can find that at constant pressure if temperature is lowered, there is also a decrease in the volume at constant pressure. The volume should therefore become zero at –273°C. Any further lowering is impossible as it will imply negative volume. Hence, –273°C is the lowest possible temperature.



Question 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:

The Critical temperature of any gas gives us an idea regarding the intermolecular forces of attraction between the molecules of that gas. Higher the intermolecular forces of attraction, higher will be the critical temperature. Hence, intermolecular forces of attraction are stronger in the case of CO2 as critical temperature for carbon dioxide is comparatively higher.



Question 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. This implies that if a molecule has a large value of a then the intermolecular forces of attraction between its molecules will be large and hence it could be easily liquefied.


Physical significance of ‘b’:


‘b’ is a measure of the volume of a gas molecule. It gives us the volume occupied by all the gaseous molecules contained in a given vessel which was earlier neglected according to kinetic theory and this correction term becomes more significant at high pressure as the molecules are themselves incompressible and occupy an appreciable fraction of the total volume.