CLASS 11 PHYSICS CHAPTER 12 – KINETIC THEORY
NCERT CLASS 11 PHYSICS
CHAPTER 12 – KINETIC THEORY
IN-TEXT / CONCEPT-BASED QUESTIONS AND ANSWERS
1. What is kinetic theory of gases?
Answer:
Kinetic theory of gases explains the behaviour of gases on the basis of the motion of their molecules. According to this theory, a gas consists of a large number of tiny molecules that are in continuous random motion.
The pressure and temperature of a gas are explained in terms of the motion and kinetic energy of its molecules.
2. What is meant by the molecular nature of matter?
Answer:
The molecular nature of matter means that all substances are made up of extremely small particles called atoms and molecules. These particles are always in motion and have spaces between them.
3. State Avogadro's law.
Answer:
Avogadro's law states that equal volumes of all gases, at the same temperature and pressure, contain equal numbers of molecules.
Mathematically,
V ∝ n
where V is volume and n is the number of moles.
4. What is an ideal gas?
Answer:
An ideal gas is a theoretical gas that obeys the ideal gas equation at all temperatures and pressures.
The ideal gas equation is:
PV = nRT
No real gas behaves as a perfectly ideal gas under all conditions.
5. Under what conditions do real gases behave approximately like ideal gases?
Answer:
Real gases behave approximately like ideal gases at:
Low pressure.
High temperature.
Under these conditions, the molecules are far apart and intermolecular forces become negligible.
6. State Boyle's law.
Answer:
Boyle's law states that for a fixed mass of gas at constant temperature, the pressure of the gas is inversely proportional to its volume.
P ∝ 1/V
Therefore,
PV = constant
7. State Charles' law.
Answer:
Charles' law states that at constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature.
V ∝ T
Therefore,
V/T = constant
8. What is Dalton's law of partial pressures?
Answer:
Dalton's law states that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures exerted by the individual gases.
P = P₁ + P₂ + P₃ + ...
9. What is Boltzmann's constant? Write its value.
Answer:
Boltzmann's constant is the universal constant that relates the average kinetic energy of a molecule to the absolute temperature of a gas.
Its value is:
kB = 1.38 × 10⁻²³ J K⁻¹
10. What is meant by the root mean square speed of gas molecules?
Answer:
The root mean square speed is the square root of the mean of the squares of the speeds of all the molecules of a gas.
It is represented by vrms.
vrms = √(3kBT/m)
where m is the mass of one molecule.
11. What is the average translational kinetic energy of a gas molecule?
Answer:
The average translational kinetic energy of a gas molecule is:
Average kinetic energy = (3/2)kBT
where kB is Boltzmann's constant and T is the absolute temperature.
12. What is the law of equipartition of energy?
Answer:
The law of equipartition of energy states that at thermal equilibrium, energy is equally distributed among all independent degrees of freedom of a molecule.
Each degree of freedom contributes an average energy of:
(1/2)kBT per molecule.
13. What are degrees of freedom?
Answer:
Degrees of freedom are the independent ways in which a molecule can possess energy.
For example, a monatomic gas molecule has three translational degrees of freedom.
14. What is specific heat capacity of a gas?
Answer:
Specific heat capacity of a gas is the amount of heat required to raise the temperature of unit mass of the gas by one kelvin.
For gases, specific heat capacity may be measured at constant volume (Cv) or constant pressure (Cp).
15. What is the relation between Cp and Cv for an ideal gas?
Answer:
For one mole of an ideal gas, Mayer's relation is:
Cp − Cv = R
where Cp is molar specific heat capacity at constant pressure, Cv is molar specific heat capacity at constant volume and R is the universal gas constant.
16. What is mean free path?
Answer:
Mean free path is the average distance travelled by a gas molecule between two successive collisions.
It is represented by λ.
λ = 1/(√2 πd²n)
where d is the molecular diameter and n is the number of molecules per unit volume.
17. Why does the pressure of a gas increase when its temperature increases at constant volume?
Answer:
When the temperature of a gas increases, the average kinetic energy and speed of its molecules increase. As a result, molecules collide more frequently and forcefully with the walls of the container.
Therefore, the pressure of the gas increases.
18. Why do gas molecules exert pressure on the walls of a container?
Answer:
Gas molecules are in continuous random motion. When they collide with the walls of the container, their momentum changes. This produces a force on the walls.
Force per unit area is called pressure.
19. Why is the average kinetic energy of gas molecules independent of the nature of the gas?
Answer:
The average translational kinetic energy of a gas molecule is:
E = (3/2)kBT
This expression depends only on the absolute temperature and Boltzmann's constant. It does not depend on the mass or nature of the gas molecule.
Therefore, at the same temperature, all ideal gases have the same average translational kinetic energy per molecule.
20. Why is the root mean square speed of hydrogen greater than that of oxygen at the same temperature?
Answer:
The root mean square speed is given by:
vrms = √(3RT/M)
At the same temperature, the root mean square speed is inversely proportional to the square root of molar mass.
Hydrogen has a much smaller molar mass than oxygen. Therefore, hydrogen molecules have a greater root mean square speed.
EXERCISE QUESTIONS AND ANSWERS
Q1. Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3 Å.
Answer:
Given:
Diameter of oxygen molecule = 3 Å
Radius, r = 1.5 Å = 1.5 × 10⁻¹⁰ m
Avogadro's number, NA = 6.02 × 10²³
Molar volume at STP = 22.4 L = 22.4 × 10⁻³ m³
Step 1: Calculate the volume of one oxygen molecule.
Volume = (4/3)πr³
= (4/3) × 3.14 × (1.5 × 10⁻¹⁰)³
= 1.41 × 10⁻²⁹ m³
Step 2: Calculate the total molecular volume of one mole.
Total molecular volume = NA × volume of one molecule
= 6.02 × 10²³ × 1.41 × 10⁻²⁹
= 8.49 × 10⁻⁶ m³
Step 3: Calculate the required fraction.
Fraction = Total molecular volume / Actual volume
= (8.49 × 10⁻⁶) / (22.4 × 10⁻³)
= 3.79 × 10⁻⁴
Final Answer:
Fraction of molecular volume to actual volume = 3.8 × 10⁻⁴ approximately.
Q2. Molar volume is the volume occupied by 1 mol of any ideal gas at standard temperature and pressure (STP: 1 atmospheric pressure, 0 °C). Show that it is 22.4 litres.
Answer:
Given:
Number of moles, n = 1
Temperature, T = 273 K
Pressure, P = 1.013 × 10⁵ Pa
Gas constant, R = 8.31 J mol⁻¹ K⁻¹
Formula:
PV = nRT
Therefore,
V = nRT/P
Substituting the values:
V = (1 × 8.31 × 273)/(1.013 × 10⁵)
V = 2268.63/(1.013 × 10⁵)
V = 0.0224 m³
Since 1 m³ = 1000 litres,
V = 0.0224 × 1000
V = 22.4 litres
Final Answer:
The molar volume of an ideal gas at STP is 22.4 litres.
Q3. Figure 12.8 shows a plot of PV/T versus P for 1.00 × 10⁻³ kg of oxygen gas at two different temperatures. Answer the following questions: (a) What does the dotted plot signify? (b) Which is true: T₁ greater than T₂ or T₁ less than T₂? (c) What is the value of PV/T where the curves meet on the y-axis? (d) If similar plots are obtained for 1.00 × 10⁻³ kg of hydrogen, will the curves meet at the same value of PV/T on the y-axis? If not, what mass of hydrogen will give the same value?
Answer:
(a) What does the dotted plot signify?
For an ideal gas:
PV/T = nR
This value remains constant at all pressures for a fixed amount of gas.
Therefore, the dotted horizontal line represents the behaviour of an ideal gas.
(b) Which is true: T₁ greater than T₂ or T₁ less than T₂?
A real gas behaves more like an ideal gas at higher temperatures.
The curve corresponding to T₁ is closer to the dotted ideal-gas line.
Therefore,
T₁ > T₂
(c) Calculate the value of PV/T where the curves meet the y-axis.
At very low pressure, the gas behaves approximately as an ideal gas.
PV/T = nR
Number of moles:
n = mass/molar mass
n = (1.00 × 10⁻³)/0.032
n = 0.03125 mol
Therefore,
PV/T = 0.03125 × 8.31
= 0.2597 J K⁻¹
(d) Will hydrogen give the same value? If not, calculate the required mass.
No. The same mass of hydrogen contains more moles than oxygen because hydrogen has a smaller molar mass.
For the same value of PV/T, the number of moles must be the same.
Required mass of hydrogen:
m = n × M
= 0.03125 × 2.02 × 10⁻³ kg
= 6.31 × 10⁻⁵ kg
Final Answers:
(a) The dotted line represents ideal gas behaviour.
(b) T₁ > T₂.
(c) PV/T = 0.26 J K⁻¹.
(d) No. Required mass of hydrogen = 6.3 × 10⁻⁵ kg.
Q4. An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder. Take R = 8.31 J mol⁻¹ K⁻¹ and molecular mass of oxygen = 32 u.
Answer:
Given:
Volume = 30 L = 0.030 m³
Initial gauge pressure = 15 atm
Final gauge pressure = 11 atm
Initial temperature = 27 °C = 300 K
Final temperature = 17 °C = 290 K
Molar mass of oxygen = 32 g/mol = 0.032 kg/mol
R = 8.31 J mol⁻¹ K⁻¹
Atmospheric pressure = 1.013 × 10⁵ Pa
Step 1: Convert gauge pressure into absolute pressure.
Initial absolute pressure:
P₁ = 15 + 1 = 16 atm
Final absolute pressure:
P₂ = 11 + 1 = 12 atm
Step 2: Calculate the initial number of moles.
Using:
PV = nRT
n₁ = P₁V/RT
n₁ = (16 × 1.013 × 10⁵ × 0.030)/(8.31 × 300)
n₁ ≈ 19.5 mol
Step 3: Calculate the final number of moles.
n₂ = P₂V/RT
n₂ = (12 × 1.013 × 10⁵ × 0.030)/(8.31 × 290)
n₂ ≈ 15.1 mol
Step 4: Calculate the moles of oxygen removed.
Moles removed = n₁ − n₂
= 19.5 − 15.1
= 4.4 mol
Step 5: Calculate the mass of oxygen removed.
Mass = Number of moles × Molar mass
= 4.4 × 0.032
= 0.1408 kg
Final Answer:
Mass of oxygen taken out = approximately 0.14 kg or 140 g.
Q5. An air bubble of volume 1.0 cm³ rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C?
Answer:
Given:
Initial volume, V₁ = 1.0 cm³
Depth, h = 40 m
Initial temperature, T₁ = 12 + 273 = 285 K
Final temperature, T₂ = 35 + 273 = 308 K
Density of water, ρ = 1000 kg/m³
Atmospheric pressure, P₀ = 1.013 × 10⁵ Pa
g = 9.8 m/s²
Step 1: Calculate the pressure at the bottom of the lake.
P₁ = P₀ + ρgh
P₁ = 1.013 × 10⁵ + (1000 × 9.8 × 40)
P₁ = 1.013 × 10⁵ + 3.92 × 10⁵
P₁ = 4.933 × 10⁵ Pa
Step 2: Calculate the pressure at the surface.
P₂ = P₀
P₂ = 1.013 × 10⁵ Pa
Step 3: Apply the combined gas law.
P₁V₁/T₁ = P₂V₂/T₂
Therefore,
V₂ = (P₁V₁T₂)/(P₂T₁)
Substituting:
V₂ = (4.933 × 10⁵ × 1.0 × 308)/(1.013 × 10⁵ × 285)
V₂ ≈ 5.26 cm³
Final Answer:
The volume of the air bubble at the surface is approximately 5.3 cm³.
Q6. Estimate the total number of air molecules, including oxygen, nitrogen, water vapour and other constituents, in a room of capacity 25.0 m³ at a temperature of 27 °C and pressure of 1 atm.
Answer:
Given:
Volume, V = 25.0 m³
Temperature, T = 27 + 273 = 300 K
Pressure, P = 1.013 × 10⁵ Pa
Boltzmann's constant, kB = 1.38 × 10⁻²³ J K⁻¹
Formula:
PV = NkBT
Therefore,
N = PV/kBT
Substituting:
N = (1.013 × 10⁵ × 25.0)/(1.38 × 10⁻²³ × 300)
N = (2.5325 × 10⁶)/(4.14 × 10⁻²¹)
N ≈ 6.12 × 10²⁶
Final Answer:
Total number of air molecules = 6.1 × 10²⁶ molecules approximately.
Q7. Estimate the average thermal energy of a helium atom at (i) room temperature of 27 °C, (ii) the temperature on the surface of the Sun, 6000 K, and (iii) the temperature of 10 million kelvin, which is the typical core temperature of a star.
Answer:
Helium is a monatomic gas.
The average translational kinetic energy of one atom is:
E = (3/2)kBT
where kB = 1.38 × 10⁻²³ J K⁻¹.
(i) At room temperature, 27 °C
T = 27 + 273 = 300 K
E = (3/2) × 1.38 × 10⁻²³ × 300
E = 6.21 × 10⁻²¹ J
(ii) At the surface temperature of the Sun, 6000 K
T = 6000 K
E = (3/2) × 1.38 × 10⁻²³ × 6000
E = 1.242 × 10⁻¹⁹ J
(iii) At 10 million kelvin
T = 10⁷ K
E = (3/2) × 1.38 × 10⁻²³ × 10⁷
E = 2.07 × 10⁻¹⁶ J
Final Answers:
(i) At 300 K = 6.2 × 10⁻²¹ J
(ii) At 6000 K = 1.24 × 10⁻¹⁹ J
(iii) At 10⁷ K = 2.07 × 10⁻¹⁶ J
Q8. Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal numbers of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is vrms the largest?
Answer:
Part 1: Do the vessels contain equal numbers of molecules?
According to the ideal gas equation:
PV = NkBT
Therefore,
N = PV/kBT
All three vessels have the same pressure, volume and temperature.
Hence, the number of molecules in all three vessels is equal.
Part 2: Is the root mean square speed the same?
The root mean square speed is:
vrms = √(3RT/M)
where M is the molar mass.
The molar masses are approximately:
Neon = 20.2 g/mol
Chlorine (Cl₂) = 71 g/mol
Uranium hexafluoride (UF₆) = 352 g/mol
At the same temperature, the root mean square speed is inversely proportional to the square root of molar mass.
Therefore, the gas with the smallest molar mass has the highest rms speed.
Final Answer:
All three vessels contain equal numbers of molecules.
Their root mean square speeds are not equal.
Neon has the highest root mean square speed because it has the smallest molar mass.
Q9. At what temperature is the root mean square speed of an argon atom in a cylinder equal to the rms speed of a helium atom at −20 °C? Take the atomic mass of argon as 39.9 u and that of helium as 4.0 u.
Answer:
Given:
Temperature of helium:
THe = −20 + 273 = 253 K
Molar mass of helium = 4.0 g/mol
Molar mass of argon = 39.9 g/mol
The root mean square speed is:
vrms = √(3RT/M)
Since both gases have equal rms speeds:
√(3RTAr/MAr) = √(3RTHe/MHe)
Squaring both sides:
TAr/MAr = THe/MHe
Therefore,
TAr = THe × MAr/MHe
Substituting:
TAr = 253 × 39.9/4.0
TAr = 2523.7 K
Final Answer:
The required temperature of argon is approximately 2.52 × 10³ K.
Q10. Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm pressure and temperature of 17 °C. Take the radius of a nitrogen molecule to be approximately 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions. Take the molecular mass of nitrogen as 28.0 u.
Answer:
Given:
Pressure, P = 2.0 atm
P = 2 × 1.013 × 10⁵ Pa
P = 2.026 × 10⁵ Pa
Temperature, T = 17 + 273 = 290 K
Radius of nitrogen molecule = 1.0 Å
r = 1.0 × 10⁻¹⁰ m
Diameter, d = 2r
d = 2 × 10⁻¹⁰ m
Molar mass of nitrogen = 28 g/mol
M = 0.028 kg/mol
R = 8.31 J mol⁻¹ K⁻¹
kB = 1.38 × 10⁻²³ J K⁻¹
Step 1: Calculate the number density of molecules.
Using:
P = nkBT
Therefore,
n = P/kBT
n = (2.026 × 10⁵)/(1.38 × 10⁻²³ × 290)
n ≈ 5.06 × 10²⁵ m⁻³
Step 2: Calculate the mean free path.
Formula:
λ = 1/(√2 πd²n)
Substituting:
λ = 1/[1.414 × 3.14 × (2 × 10⁻¹⁰)² × 5.06 × 10²⁵]
λ ≈ 1.11 × 10⁻⁷ m
Step 3: Calculate the rms speed of nitrogen molecules.
vrms = √(3RT/M)
vrms = √[(3 × 8.31 × 290)/0.028]
vrms ≈ 508 m/s
Step 4: Calculate the collision frequency.
Collision frequency = vrms/λ
= 508/(1.11 × 10⁻⁷)
≈ 4.58 × 10⁹ s⁻¹
Step 5: Calculate the time between successive collisions.
Time between collisions = λ/vrms
= (1.11 × 10⁻⁷)/508
≈ 2.18 × 10⁻¹⁰ s
Step 6: Estimate the duration of a collision.
Approximate collision time:
t = d/vrms
= (2 × 10⁻¹⁰)/508
≈ 3.94 × 10⁻¹³ s
Comparison:
Time between collisions ≈ 2.18 × 10⁻¹⁰ s
Collision duration ≈ 3.94 × 10⁻¹³ s
The time between collisions is much greater than the collision duration.
Final Answer:
Mean free path = 1.1 × 10⁻⁷ m approximately.
Collision frequency = 4.6 × 10⁹ s⁻¹ approximately.
Time between collisions = 2.2 × 10⁻¹⁰ s approximately.
Collision duration = 3.9 × 10⁻¹³ s approximately.
A nitrogen molecule spends most of its time moving freely between collisions.
IMPORTANT QUESTIONS AND ANSWERS
1. State the main assumptions of the kinetic theory of an ideal gas.
Answer:
The main assumptions are:
A gas consists of a very large number of tiny molecules.
The molecules are in continuous random motion.
The size of the molecules is very small compared with the distance between them.
The molecules exert negligible intermolecular forces except during collisions.
Collisions between molecules and with the walls of the container are perfectly elastic.
The time taken during a collision is very small compared with the time between successive collisions.
Molecules obey Newton's laws of motion.
The motion of molecules is random, so there is no preferred direction.
2. Derive the expression for pressure exerted by an ideal gas.
Answer:
Consider a cubical container of side length L containing N molecules of a gas.
Let:
m = Mass of each molecule
vx = Velocity of a molecule along the x-direction
A molecule strikes the wall perpendicular to the x-axis.
Step 1: Change in momentum.
Initial momentum = mvx
Final momentum = −mvx
Change in momentum = −mvx − mvx
= −2mvx
Magnitude of change in momentum = 2mvx
Step 2: Time between two successive collisions with the same wall.
Distance travelled = 2L
Time = 2L/vx
Step 3: Force exerted by one molecule.
Force = Change in momentum/Time
F = 2mvx/(2L/vx)
F = mvx²/L
Step 4: Total force due to all molecules.
F = (m/L)Σvx²
Step 5: Calculate pressure.
Pressure = Force/Area
Area = L²
P = (m/L³)Σvx²
Since volume V = L³,
P = (m/V)Σvx²
For random molecular motion:
Σvx² = (1/3)Σv²
Therefore,
P = (m/3V)Σv²
For N molecules:
P = (Nm/3V)vrms²
Since mass density ρ = Nm/V,
P = (1/3)ρvrms²
This is the required expression for the pressure of an ideal gas.
3. Derive the relation between pressure and kinetic energy of gas molecules.
Answer:
From the kinetic theory of gases:
P = (1/3)ρvrms²
We know:
ρ = Nm/V
Therefore,
P = (1/3)(Nm/V)vrms²
PV = (1/3)Nmvrms²
Multiplying both sides by 3/2:
(3/2)PV = (1/2)Nmvrms²
But,
(1/2)mvrms² = Average translational kinetic energy per molecule.
Therefore,
(3/2)PV = N × Average kinetic energy
or,
PV = (2/3) × Total translational kinetic energy
This shows that the pressure of an ideal gas is related to the total translational kinetic energy of its molecules.
4. Derive the expression for the average kinetic energy of a gas molecule in terms of temperature.
Answer:
According to kinetic theory:
PV = (1/3)Nmvrms²
From the ideal gas equation:
PV = NkBT
Equating both equations:
(1/3)Nmvrms² = NkBT
Cancelling N:
(1/3)mvrms² = kBT
Multiplying by 3/2:
(1/2)mvrms² = (3/2)kBT
But,
(1/2)mvrms² = Average translational kinetic energy
Therefore,
Average kinetic energy = (3/2)kBT
This is the required expression.
5. Derive the expression for the root mean square speed of gas molecules.
Answer:
We know that the average translational kinetic energy is:
(1/2)mvrms² = (3/2)kBT
Cancelling 1/2:
mvrms² = 3kBT
Therefore,
vrms² = 3kBT/m
Taking the square root:
vrms = √(3kBT/m)
For one mole of gas, using m = M/NA and R = NAkB:
vrms = √(3RT/M)
This is the required expression for root mean square speed.
6. Explain the law of equipartition of energy.
Answer:
The law of equipartition of energy states that at thermal equilibrium, energy is equally distributed among all independent degrees of freedom of a molecule.
Each degree of freedom contributes an average energy of:
(1/2)kBT per molecule.
For one mole, the energy per degree of freedom is:
(1/2)RT
If a molecule has f degrees of freedom, its average energy is:
E = (f/2)kBT
For one mole:
U = (f/2)RT
7. Explain the degrees of freedom of a molecule.
Answer:
Degrees of freedom are the number of independent coordinates or modes required to describe the motion or energy of a molecule.
The main types are:
Translational degrees of freedom: Motion along the x, y and z axes.
Rotational degrees of freedom: Rotation about suitable axes.
Vibrational degrees of freedom: Vibrations of atoms within a molecule.
Examples:
Monatomic gas: 3 translational degrees of freedom.
Diatomic gas at ordinary temperatures: approximately 5 active degrees of freedom.
A non-linear polyatomic gas at ordinary temperatures: approximately 6 active degrees of freedom.
The number of active degrees of freedom can depend on temperature.
8. Derive the relation between Cp and Cv for an ideal gas.
Answer:
For one mole of an ideal gas:
PV = RT
The first law of thermodynamics gives:
dQ = dU + dW
At constant volume:
dQv = Cv dT
Since volume is constant:
dW = 0
Therefore,
dU = Cv dT
At constant pressure:
dQp = Cp dT
Also,
dQp = dU + P dV
Substituting dU:
Cp dT = Cv dT + P dV
Dividing by dT:
Cp = Cv + P(dV/dT)
From the ideal gas equation:
PV = RT
At constant pressure:
P(dV/dT) = R
Therefore,
Cp = Cv + R
Hence,
Cp − Cv = R
This is known as Mayer's relation.
9. What is the relation between the ratio of specific heats and degrees of freedom?
Answer:
For an ideal gas having f degrees of freedom:
Cv = (f/2)R
Using Mayer's relation:
Cp = Cv + R
Cp = (f/2)R + R
Cp = [(f + 2)/2]R
The ratio of specific heats is:
γ = Cp/Cv
γ = [(f + 2)/2]R ÷ [(f/2)R]
Therefore,
γ = (f + 2)/f
This is the required relation.
10. Explain mean free path and write its expression.
Answer:
Mean free path is the average distance travelled by a gas molecule between two successive collisions.
It is represented by λ.
The expression for mean free path is:
λ = 1/(√2 πd²n)
where:
d = Diameter of a gas molecule
n = Number of molecules per unit volume
The SI unit of mean free path is metre (m).
IMPORTANT FORMULAE – CHAPTER 12
Ideal gas equation:
PV = nRT
Molecular form of ideal gas equation:
PV = NkBT
Number of moles:
n = m/M
Number of molecules:
N = nNA
Boltzmann's constant:
kB = R/NA
Pressure according to kinetic theory:
P = (1/3)ρvrms²
Root mean square speed:
vrms = √(3kBT/m)
Molar form of rms speed:
vrms = √(3RT/M)
Average translational kinetic energy per molecule:
E = (3/2)kBT
Total translational kinetic energy:
E = (3/2)NkBT
Energy per degree of freedom:
E = (1/2)kBT
Internal energy for f degrees of freedom:
U = (f/2)nRT
Mayer's relation:
Cp − Cv = R
Ratio of specific heats:
γ = Cp/Cv
Ratio of specific heats in terms of degrees of freedom:
γ = (f + 2)/f
Mean free path:
λ = 1/(√2 πd²n)
Number density:
n = N/V
Collision frequency (approximate):
Collision frequency = vrms/λ
Time between successive collisions (approximate):
t = λ/vrms
Density of an ideal gas:
ρ = PM/RT
IMPORTANT CONSTANTS AND SI UNITS
| Quantity | Value / SI Unit |
|---|---|
| Boltzmann's constant (kB) | 1.38 × 10⁻²³ J K⁻¹ |
| Universal gas constant (R) | 8.31 J mol⁻¹ K⁻¹ |
| Avogadro's number (NA) | 6.02 × 10²³ mol⁻¹ |
| Atmospheric pressure | 1.013 × 10⁵ Pa |
| Temperature | Kelvin (K) |
| Pressure | Pascal (Pa) |
| Volume | Cubic metre (m³) |
| Mean free path | Metre (m) |
| Molecular speed | Metre per second (m/s) |
| Kinetic energy | Joule (J) |
| Specific heat capacity | J kg⁻¹ K⁻¹ |
IMPORTANT POINTS TO REMEMBER
Kinetic theory explains the behaviour of gases using molecular motion.
Gas molecules are in continuous random motion.
Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.
Real gases behave approximately like ideal gases at low pressure and high temperature.
The pressure of a gas is caused by molecular collisions with the walls of its container.
Average translational kinetic energy is directly proportional to absolute temperature.
At the same temperature, all ideal gases have the same average translational kinetic energy per molecule.
Lighter gas molecules have higher rms speeds at the same temperature.
The mean free path increases when the pressure decreases, if temperature and molecular size remain constant.
Cp is greater than Cv for an ideal gas.
For one mole of an ideal gas, Cp − Cv = R.
Temperature must be converted into kelvin when using gas equations.
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