Molecular Theory of gases - Physics II- For HSC Part 2 / XII / Class 12 (Science Group) - Section D: Numerical & Worked Exmples

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Unit 15: Molecular Theory of gases
Physics II
For HSC Part 2 / XII / Class 12 (Science Group)

SECTION D
Numerical

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

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

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)
DATA:
Given:
Number of moles = n = 1 gram mole
Temperature = 0 °C = 0 + 273 = 273 K
Pressure = P = 1 atm = 1.01 x 10⁵ N/m²
Universal gas Constant = R = 8.314 J mol^-1 K^-1

Required:
Volume = V = ?

Formula:
PV = nRT

Solution:

PV = nRT
1.01 x 10⁵ x V = 1 x 8.314 x 273

V = 0.2247 m³
V = 0.2247 x 1000 = 22.47 L /mole

Result: The volume occupied is 22.47 liters / mole Answer

15.4. An air storage tank whose volume is 112 liters contain 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)
DATA:
Given:
Volume = V₁ = V₂ = 112 liters
Mass of air = m₁ = 3 kg
Pressure = P₁ = 18 atm
Pressure = P₂ = 21 atm
Temperature = T (There is no change in temperature)

Required:
Mass of air = m₂ = ?

Formula:

• Δm = m₁ - m₂
• PV = nRT

Solution:

Δm = m₁ - m₂
Δm = 3.5 - 3 = 0.5 kg

Result: 0.5 kg air would have to be forced. Answer

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 x 10⁵ Pa)

15.6. The molar mass of nitrogen gas N₂ is 28 gmol^-1. For 100 g of nitrogen, calculate.
(a) the number of moles. (Ans: (a) 3.57 mole)
(b) the volume occupied at room temperature (20 °C) and pressure of 1.01 x 10⁵ Pa. (Ans: (b) 0.086 m³ or 86 dm³)
DATA:
Given:
Molar mass of Nitrogen = M = 28 gmol^-1
Mass of Nitrogen= m = 100 g
Temperature = T = 20 °C = 20 + 273 = 293 K
Pressure = P = 1.01 x 10⁵ Pa
Universal gas Constant = R = 8.314 J mol^-1 K^-1

Required:
(a) Number of moles = n = ?
(b) Volume = V = ?

Formula:

1. Number of moles = mass / Molar mass
2. PV = nRT

Solution:

Result: (a) Number of moles are 3.57 moles
(b) The volume occupied is 0.086 m³ or 86 dm³

15.7. A sample of a gas contains 3.0 x 10²⁴ atoms. Calculate the volume of the gas at a temperature of 300 K and a pressure of 120 KPa. (Ans: 0.104 m³)
DATA:
Given:
Number of Atoms = N = 3.0 x 10²⁴ atoms
Avogadro's Number = N_A = 6.022 x 10²³
Temperature = T = 00 K
Pressure = P = 120 KPa = 120 x 10³ Pa

Required:
Volume of the gas = V = ?

Formula:

• Number of moles = Number of atoms / Avogadro's number
• PV = nRT

Solution:

Result: The volume of gas is 0.104 dm³ Answer

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 x10^-2 kg/m³. (Ans: 1835.86 ms^-1)
DATA:
Given:
Temperature = T = 0°C = 0 + 273 = 273 K
Pressure = 1 atm = 1.01 x 10⁵ Pa
Density = = 8.99 x10^-2 kg/m³

Required:
Root mean square speed = V_rms = ?

Formula:
P = ¹/₃ 𝝆 V²

Solution:

Result: The root mean square speed is 1835.8 m/s Answer

15.9. Calculate the root mean square speed of hydrogen molecule at 500 K (mass of proton = 1.67 x 10^-27 kg and K = 1.38 x 10^-23 J/molecule.-K) (Ans: 2489.49ms^-I)
DATA:
Given:
Temperature = T = 500 K
Mass of proton = 1.67 x 10^-27 kg
Mass of hydrogen molecule = H₂ = m = 2 x 1.67 x 10^-27 = 3.34 x 10^-27 kg
Boltzmann constant = K = 1.38 x 10^-23 J/molecule.k
Pressure = 1 atm = 1.01 x 10⁵ Pa

Required:
Root mean square speed = V_rms = ?

Formula:
V² = 3KT / m

Solution:

Result: Root mean square speed is 2489.50 m/s Answer

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: 5.86x10^-21 J, 6.48x10^-21 J)
(b) What is the Kinetic energy per mole of an ideal gas at these temperatures? (Ans: 3526.57 J, 3901 J)
DATA:
Given:
Temperature = T = 10 °C = 10 + 273 = 283 K
Temperature = T = 40 °C = 40 + 273 = 313 K
Boltzmann constant = K = 1.38 x 10^-23 J/molecule.k
Number of moles = n = 1 mole
Universal gas Constant = R = 8.314 J mol^-1 K^-1

Required:
(a) Average value of Kinetic Energy = K.E_avg = ?
(b) Kinetic energy per mole of ideal gas = K.E = ?
Formula:

• (a) K.E_avg = ³/₂ KT
• K.E (per mole) = ³/₂ nRT

Solution:

Result:
(a-i) 5.858 x 10^-21
(a-ii) 6.479 x 10^-21

(b-i) 3529.3 J
(b-ii) 3903.4 J Answer

Worked Examples

Numerical From Self Assessment Questions

1. Convert each of the following temperature from the centigrade scale to Kelvin scale and Fahrenheit 0°C, 20°C, 120°C, 500°C, —23°C, 200°C.
Ans: Centigrade (Celsius) To Kelvin Scale:
FORMULA:

T_K = T_°C + 273

i) 0 °C
Solution:
T_K = T_°C + 273
⇒ T_K = 0 + 273 = 273 K Ans.

ii) 20 °C
Solution:
T_K = T_°C + 273
⇒ T_K = 20 + 273 = 293 K Ans.

iii) 120 °C
Solution:
T_K = T_°C + 273
⇒ T_K = 120 + 273 = 393 K Ans.

iv) 500 °C
Solution:
T_K = T_°C + 273
⇒ T_K = 500 + 273 = 773 K Ans.

v) -23 °C
Solution:
T_K = T_°C + 273
⇒ T_K = -23 + 273 = 250 K Ans.

vi) 200 °C
Solution:
T_K = T_°C + 273
⇒ T_K = 200 + 273 = 493 K Ans.

Centigrade (Celsius) To Fahrenheit Scale:

2. Convert each of the following temperature from the Kelvin scale to the Celsius scale and Fahrenheit O K, 20 K, 100 K, 300 K, 373 K, and 500 K.
Ans: Kelvin Scale To Centigrade (Celsius):
FORMULA:

T_°C = T_K - 273

i) 0 K
Solution:
T_°C = T_K - 273
⇒ T_°C = 0 - 273 = -273 °C Ans.

ii) 20 K
Solution:
T_°C = T_K - 273
⇒ T_°C = 20 - 273 = -253 °C Ans.

iii) 100 K
Solution:
T_°C = T_K - 273
⇒ T_°C = 100 - 273 = -173 °C Ans.

iv) 300 K
Solution:
T_°C = T_K - 273
⇒ T_°C = 300 - 273 = 27 °C Ans.

v) 373 K
Solution:
T_°C = T_K - 273
⇒ T_°C = 373 - 273 = 100 °C Ans.

vi) 500 K
Solution:
T_°C = T_K - 273
⇒ T_°C = 500 - 273 = 227 °C Ans.

Kelvin Scale To Fahrenheit:

i) 0 K
Solution:
T_°F = 1.8 (T_K - 273) + 32
⇒ T_°F = 1.8 (0 - 273) + 32
⇒ T_°F = 1.8 (-273) + 32 ⇒ T_°F = - 491.4 + 32 = -459.4 °F Ans.

ii) 20 K
Solution:
T_°F = 1.8 (T_K - 273) + 32
⇒ T_°F = 1.8 (20 - 273) + 32
⇒ T_°F = 1.8 (-253) + 32 ⇒ T_°F = - 455.4 + 32 = -423.4 °F Ans.

iii) 100 K
Solution:
T_°F = 1.8 (T_K - 273) + 32
⇒ T_°F = 1.8 (100 - 273) + 32
⇒ T_°F = 1.8 (-173) + 32 ⇒ T_°F = - 311.4 + 32 = -279.4 °F Ans.

iv) 300 K
Solution:
T_°F = 1.8 (T_K - 273) + 32
⇒ T_°F = 1.8 (300 - 273) + 32
⇒ T_°F = 1.8 (27) + 32 ⇒ T_°F = - 48.6 + 32 = 80.6 °F Ans.

v) 373 K
Solution:
T_°F = 1.8 (T_K - 273) + 32
⇒ T_°F = 1.8 (373 - 273) + 32
⇒ T_°F = 1.8 (100) + 32 ⇒ T_°F = 180 + 32 = 212 °F Ans.

vi) 500 K
Solution:
T_°F = 1.8 (T_K - 273) + 32
⇒ T_°F = 1.8 (500 - 273) + 32
⇒ T_°F = 1.8 (227) + 32 ⇒ T_°F = 408.6 + 32 = 440.6 °F Ans.