Class 12 - Unit # 15: Molecular Theory of Gases - Solved Numericals

Problem 15.1

The freezing point of mercury is -39°C. Convert it into °F and the comfort level temperature of 20°C into Kelvin. (Ans: -38.2°F, 293K)

Given Data:

a) Freezing Point of Mercury:

• Temperature in Celsius (T_C) = -39°C
• Temperature in Fahrenheit (T_F) = ?

b) Comfort Level Temperature:

• Temperature in Celsius (T_C) = 20°C
• Temperature in Kelvin (T_K) = ?

Solution:

Part (a): Convert Celsius to Fahrenheit

Using the conversion formula:

T_F =

9

5

T_C + 32

Substituting T_C = -39°C:

T_F = (

9

5

× -39 ) + 32

T_F = -70.2 + 32

T_F = -38.2°F

Part (b): Convert Celsius to Kelvin

Using the conversion formula:

T_K = T_C + 273

Substituting T_C = 20°C:

T_K = 20 + 273

T_K = 293K

Result:

The freezing point of mercury is -38.2°F and the comfort level temperature is 293K.

Problem 15.2

The boiling point of liquid nitrogen is -321°F. Change it into equivalent Kelvin temperature. (Ans: 77K)

Given Data:

• Temperature in Fahrenheit (T_F) = -321°F
• Temperature in Kelvin (T_K) = ?

Solution:

First, convert the Fahrenheit temperature into Celsius using the formula:

T_C =

5

9

(T_F - 32)

Substituting T_F = -321°F:

T_C =

5

9

(-321 - 32)

T_C =

5

9

(-353)

T_C = -196.1°C

Now, convert Celsius to Kelvin:

T_K = T_C + 273

T_K = -196.1 + 273

T_K ≈ 77K (76.9K)

Result:

The boiling point of liquid nitrogen is 77K.

Problem 15.3

Calculate the volume occupied by a gram-mole of a gas at 0°C and a pressure of 1.0 atmosphere. (Ans: 22.4 liters/mole)

Given Data:

• Number of moles (n) = 1 mol
• Temperature (T) = 0°C = 273 K
• Pressure (P) = 1 atm = 1.01 × 10⁵ Pa
• Universal Gas Constant (R) = 8.314 J/(mol·K)
• Volume of Gas (V) = ?

Solution:

According to the General Gas Equation (Ideal Gas Law):

PV = nRT

Rearranging the equation to solve for Volume (V):

V =

nRT

P

Substituting the values:

V =

1 × 8.314 × 273

1.01 × 10⁵

V = 0.0224 m³

To convert the volume from cubic meters (m³) to liters (or dm³):

V = 0.0224 × 1000 Liters

V = 22.4 Liters (or dm³)

Result:

The volume occupied by the gas is 22.4 liters/mole.

Problem 15.4

An air storage tank whose volume is 112 liters contains 3 kg of air at a pressure of 18 atmospheres. How much air would have to be forced into the tank to increase the pressure to 21 atmospheres, assuming no change in temperature? (Ans: 0.5 kg)

Given Data:

• Volume of Gas (V) = 112 Liters
• Initial Mass of Gas (m₁) = 3 kg
• Initial Pressure (P₁) = 18 atm
• Final Pressure (P₂) = 21 atm
• Change in Mass of Gas (Δm) = ?

Solution:

Since Volume (V) and Temperature (T) are constant, pressure is directly proportional to the mass of the gas (P ∝ m):

P₁

m₁

=

P₂

m₂

Rearranging the equation to solve for final mass (m₂):

m₂ =

P₂ × m₁

P₁

Substituting the values:

m₂ =

21 × 3

18

m₂ = 3.5 kg

Now, calculate the additional mass (Δm) that must be forced into the tank:

Δm = m₂ - m₁

Δm = 3.5 kg - 3 kg

Δm = 0.5 kg

Result:

The additional mass of air to be forced into the tank is 0.5 kg.

Problem 15.5

A balloon contains 0.04 m³ of air at a pressure of 120 kPa. Calculate the pressure required to reduce its volume to 0.025 m³ at constant temperature. (Ans: 1.9 × 10⁵ Pa)

Given Data:

• Initial Volume (V₁) = 0.04 m³
• Initial Pressure (P₁) = 120 kPa = 1.2 × 10⁵ Pa
• Final Volume (V₂) = 0.025 m³
• Final Pressure (P₂) = ?

Solution:

According to Boyle's Law (at constant temperature):

P₁V₁ = P₂V₂

Rearranging the formula to find P₂:

P₂ =

P₁V₁

V₂

Substituting the values:

P₂ =

120 × 10³ × 0.04

0.025

P₂ = 192,000 Pa

P₂ = 1.92 × 10⁵ Pa

Result:

The pressure required to reduce the volume of the balloon is 1.9 × 10⁵ Pa.

Problem 15.6

The molar mass of nitrogen gas (N₂) is 28 g/mol. For 100 g of nitrogen, calculate:
(a) The number of moles. (Ans: 3.57 mole)
(b) The volume occupied at room temperature (20°C) and pressure of 1.01 × 10⁵ Pa. (Ans: 0.086 m³ or 86 dm³)

Given Data:

Part (a):

• Mass of Nitrogen (m) = 100 g
• Molar Mass of Nitrogen (M) = 28 g/mol
• Number of Moles (n) = ?

Part (b):

• Temperature (T) = 20°C + 273 = 293 K
• Pressure (P) = 1.01 × 10⁵ Pa
• Universal Gas Constant (R) = 8.314 J/(mol·K)
• Volume (V) = ?

Solution:

Part (a): Find the Number of Moles

n =

m

M

Substituting values:

n =

100

28

n = 3.57 moles

Part (b): Find the Volume Occupied

Using the General Gas Equation (PV = nRT):

V =

nRT

P

Substituting values (using n = 3.57):

V =

3.57 × 8.314 × 293

1.01 × 10⁵

V = 0.086 m³

Converting to dm³ (Liters) by multiplying by 1000:

V = 86 dm³ (or 86 Liters)

Result:

The number of moles present is 3.57 moles and the volume occupied is 0.086 m³ (86 dm³).

Problem 15.7

A sample of a gas contains 3.0 × 10²⁴ atoms. Calculate the volume of the gas at a temperature of 300K and a pressure of 120 kPa. (Ans: 0.104 m³)

Given Data:

• Number of Atoms (N) = 3.0 × 10²⁴ atoms
• Temperature (T) = 300 K
• Pressure (P) = 120 kPa = 1.2 × 10⁵ Pa
• Avogadro's Number (N_A) = 6.02 × 10²³ atoms/mol
• Volume of Gas (V) = ?

Solution:

First, calculate the number of moles (n) of gas:

n =

N

N_A

n =

3.0 × 10²⁴

6.02 × 10²³

n = 4.98 moles

Now, use the General Gas Equation (PV = nRT) to find volume (V):

V =

nRT

P

V =

4.98 × 8.314 × 300

1.2 × 10⁵

V = 0.104 m³

Result:

The volume of the gas is 0.104 m³.

Problem 15.8

Calculate the root mean square speed of hydrogen molecules at 0°C and 1.0 atm pressure, assuming hydrogen to be an ideal gas. The density of hydrogen is 8.99 × 10^-2 kg/m³. (Ans: 1835.86 m/s)

Given Data:

• Temperature (T) = 0°C = 273 K
• Pressure (P) = 1 atm = 1.01 × 10⁵ Pa
• Density of Hydrogen (ρ) = 8.99 × 10^-2 kg/m³
• Root Mean Square Speed (v_rms) = ?

Solution:

The formula for root mean square speed in terms of pressure and density is:

v_rms = √

3P

ρ

Substituting the given values:

v_rms = √

3 × (1.01 × 10⁵)

8.99 × 10^-2

v_rms = √3,370,411.57

v_rms = 1835.87 m/s

Result:

The root mean square speed of hydrogen molecules is 1835.87 m/s.

Problem 15.9

Calculate the root mean square speed of a hydrogen molecule at 500K (mass of proton = 1.67 × 10^-27 kg and Boltzmann constant, k = 1.38 × 10^-23 J/K). (Ans: 2489.49 m/s)

Given Data:

• Temperature (T) = 500 K
• Mass of a Hydrogen Molecule (m) = 2 × (Mass of proton) = 2 × 1.67 × 10^-27 kg = 3.34 × 10^-27 kg
• Boltzmann Constant (k) = 1.38 × 10^-23 J/K
• Root Mean Square Speed (v_rms) = ?

Solution:

The formula for root mean square speed using molecular parameters is:

v_rms = √

3kT

m

Substituting the given values:

v_rms = √

3 × (1.38 × 10^-23) × 500

3.34 × 10^-27

v_rms = √

2.07 × 10^-20

3.34 × 10^-27

v_rms = √6,197,604.79

v_rms = 2489.49 m/s

Result:

The root mean square speed of the hydrogen molecule is 2489.49 m/s.

Problem 15.10

(a) Determine the average value of the kinetic energy of the particles of an ideal gas at 10°C and at 40°C. (Ans: 5.86 × 10^-21 J, 6.48 × 10^-21 J)
(b) What is the kinetic energy per mole of an ideal gas at these temperatures? (Ans: 3526.57 J, 3901 J)

Given Data:

• First Temperature (T₁) = 10°C + 273 = 283 K
• Second Temperature (T₂) = 40°C + 273 = 313 K
• Boltzmann Constant (k) = 1.38 × 10^-23 J/K
• Universal Gas Constant (R) = 8.314 J/(mol·K)

Solution:

Part (a): Average Kinetic Energy per Molecule

The formula for the average kinetic energy of gas molecules is:

K.E._avg =

3

2

kT

At T₁ = 283 K:

K.E.₁ =

3

2

× (1.38 × 10^-23) × 283

K.E.₁ = 5.86 × 10^-21 J

At T₂ = 313 K:

K.E.₂ =

3

2

× (1.38 × 10^-23) × 313

K.E.₂ = 6.48 × 10^-21 J

Part (b): Kinetic Energy per Mole

The formula for kinetic energy per mole of an ideal gas is:

K.E._mole =

3

2

RT

At T₁ = 283 K:

K.E._mole1 =

3

2

× 8.314 × 283

K.E._mole1 = 3529.23 J (approx 3526.57 J using N_A calculation)

At T₂ = 313 K:

K.E._mole2 =

3

2

× 8.314 × 313

K.E._mole2 = 3903.42 J (approx 3901 J)

Result:

(a) The average kinetic energies are 5.86 × 10^-21 J at 10°C and 6.48 × 10^-21 J at 40°C.
(b) The kinetic energy per mole is 3526.57 J at 10°C and 3901 J at 40°C.