Updated By SaraNextGen

On March 11, 2024, 11:35 AM

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, 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 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,

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 (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.

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 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 ?

Answer:

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?

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 .

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/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.

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 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?

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 

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).

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.

Question 5.13:

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 

Now, 1 molecule of  contains 14 electrons.

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

Answer:

Avogadro number = 6.02 × 10²³

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 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 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).

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 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.

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 pV²T 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 CO₂.

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.

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, 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 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,

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 (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.

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 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 ?

Answer:

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?

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 .

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/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.

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 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?

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 

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).

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.

Question 5.13:

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 

Now, 1 molecule of  contains 14 electrons.

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

Answer:

Avogadro number = 6.02 × 10²³

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 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 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).

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 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.

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 pV²T 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 CO₂.

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.