Is it correct to say: Using the temperature in °C in v_rms, (3/2)k_BT or Stefan's law?

Every kinetic-theory and radiation formula needs the absolute temperature in kelvin: T(K) = t(°C) + 273.

Is it correct to say: Using molar mass in g/mol in v_rms = √(3RT/M₀)?

Convert M₀ to kg/mol (divide by 1000). Otherwise v_rms comes out √1000 ≈ 31.6 times too small.

Is it correct to say: Writing the net radiation loss as eσA(T − T₀)⁴?

The body emits eσAT⁴ and absorbs eσAT₀⁴, so the net loss is eσA(T⁴ − T₀⁴).

Is it correct to say: Thinking λ_max increases as a body gets hotter?

λ_max = b/T. A hotter body peaks at a shorter wavelength, which is why a heated iron rod glows red, then orange, then white.

Is it correct to say: Assuming a good reflector is also a good emitter?

By Kirchhoff's law a = e. A good reflector is a poor absorber, so it is also a poor emitter. A dull black surface is both a good absorber and a good emitter.

Is it correct to say: Taking C_P − C_V = R for specific heats per unit mass?

C_P − C_V = R is for molar specific heats. Per unit mass (principal specific heats), the difference is R/M₀.

📄 Formula sheet (PDF)

Kinetic Theory of Gases and Radiation

Ideal gas, mean free path, pressure and rms speed of gas molecules, equipartition of energy, specific heats, blackbody radiation, Kirchhoff's, Wien's and Stefan–Boltzmann laws — Maharashtra HSC Physics Ch 3

📐 16 formulas✏️ 4 examples🎯 6 practice🏫 Maharashtra State Board📚 Class 12✓ 2025–26 syllabus

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Board Exam Tips

→The derivation of the pressure of an ideal gas, P = ⅓(N/V)m v²_rms, is a standard long answer. Follow a clear sequence: momentum change per collision, time between collisions, force from one molecule, then sum over N molecules and use v²_rms = 3⟨v_x²⟩.
→Always convert temperature to kelvin and molar mass to kg/mol (32 g/mol = 0.032 kg/mol) before using v_rms = √(3RT/M₀).
→Use k_B for one molecule and R for one mole. Average KE per molecule is (3/2)k_BT and per mole is (3/2)RT; mixing them up gives answers that are wrong by a factor of 6 × 10²³.
→In the radiation part, the Maharashtra textbook uses a, r and t_r for the coefficients of absorption, reflection and transmission, and e for the coefficient of emission (emissivity). Use the same symbols.
→In Stefan's law, use T⁴ − T₀⁴ for the net loss, never (T − T₀)⁴, and keep both temperatures in kelvin.

📐 Formulas(16)

Ideal Gas Equation

P V = n R T = N k_{B} T

Symbol	Meaning
$P$	Pressure (Pa)
$V$	Volume (m³)
$n$	Number of moles
$R$	Universal gas constant = 8.314 J mol⁻¹ K⁻¹
$N$	Number of molecules
$k_{B}$	Boltzmann constant = 1.38 × 10⁻²³ J K⁻¹
$T$	Absolute temperature (K)

Mean Free Path

λ = \frac{1}{2 π d ^{2} ( N / V )}

Symbol	Meaning
$λ$	Mean free path (m)
$d$	Diameter of a molecule (m)
$N / V$	Number of molecules per unit volume (m⁻³)

Pressure Exerted by an Ideal Gas★ Board fav

P = \frac{1}{3} \frac{N}{V} m v_{r m s}^{2} = \frac{1}{3} ρ v_{r m s}^{2}

Symbol	Meaning
$m$	Mass of one molecule (kg)
$v_{r m s}$	Root mean square speed (m s⁻¹)
$ρ$	Density of the gas (kg m⁻³)

Root Mean Square Speed★ Board fav

v_{r m s} = \frac{3 R T}{M _{0}} = \frac{3 k _{B} T}{m} = \frac{3 P}{ρ}

Symbol	Meaning
$M_{0}$	Molar mass of the gas (kg mol⁻¹)
$m$	Mass of one molecule (kg)
$T$	Absolute temperature (K)

Kinetic Interpretation of Temperature★ Board fav

\frac{1}{2} m v_{r m s}^{2} = \frac{3}{2} k_{B} T

Symbol	Meaning
$k_{B}$	Boltzmann constant (J K⁻¹)
$T$	Absolute temperature (K)

Law of Equipartition of Energy

E_{per degree of freedom} = \frac{1}{2} k_{B} T

Symbol	Meaning
$k_{B}$	Boltzmann constant (J K⁻¹)
$T$	Absolute temperature (K)

Mayer's Relation★ Board fav

C_{P} - C_{V} = R

Symbol	Meaning
$C_{P}$	Molar specific heat at constant pressure (J mol⁻¹ K⁻¹)
$C_{V}$	Molar specific heat at constant volume (J mol⁻¹ K⁻¹)
$R$	Universal gas constant (J mol⁻¹ K⁻¹)

Ratio of Specific Heats

γ = \frac{C _{P}}{C _{V}}

Symbol	Meaning
$γ$	Adiabatic ratio (no unit)

Specific Heats of a Monatomic Gas

C_{V} = \frac{3}{2} R, C_{P} = \frac{5}{2} R, γ = \frac{5}{3} \approx 1.67

Specific Heats of a Diatomic Gas

C_{V} = \frac{5}{2} R, C_{P} = \frac{7}{2} R, γ = \frac{7}{5} = 1.4

Absorption, Reflection and Transmission Coefficients

a + r + t_{r} = 1

Symbol	Meaning
$a$	Coefficient of absorption (absorptive power)
$r$	Coefficient of reflection (reflectance)
$t_{r}$	Coefficient of transmission (transmittance)

Emissive Power and Coefficient of Emission

R = \frac{Q}{A t}, e = \frac{R}{R _{B}}

Symbol	Meaning
$R$	Emissive power of the body (W m⁻²)
$Q$	Radiant energy emitted (J)
$A$	Surface area (m²)
$t$	Time (s)
$R_{B}$	Emissive power of a perfect blackbody at the same temperature (W m⁻²)
$e$	Coefficient of emission or emissivity (no unit)

Kirchhoff's Law of Heat Radiation

\frac{R}{a} = R_{B} \Leftrightarrow a = e

Symbol	Meaning
$a$	Coefficient of absorption of the body
$e$	Coefficient of emission of the body

Wien's Displacement Law★ Board fav

λ_{ma x} T = b

Symbol	Meaning
$λ_{ma x}$	Wavelength at which the blackbody emits the most energy (m)
$T$	Absolute temperature of the blackbody (K)
$b$	Wien's constant = 2.897 × 10⁻³ m K

Stefan–Boltzmann Law★ Board fav

R_{B} = σ T^{4}, R = e σ T^{4}

Symbol	Meaning
$σ$	Stefan–Boltzmann constant = 5.67 × 10⁻⁸ W m⁻² K⁻⁴
$T$	Absolute temperature of the body (K)
$e$	Emissivity of the body

Net Rate of Loss of Heat by Radiation

\frac{Q}{t} = e σ A (T^{4} - T_{0}^{4})

Symbol	Meaning
$Q / t$	Net rate of loss of heat (W)
$T$	Temperature of the body (K)
$T_{0}$	Temperature of the surroundings (K)
$A$	Surface area of the body (m²)

✏️ Solved Examples

1Solved Exampleeasy3 steps

Find the rms speed of oxygen molecules at 27 °C. (Molar mass of oxygen = 32 g/mol, R = 8.314 J mol⁻¹ K⁻¹)

Convert the units: T = 27 + 273 = 300 K, M₀ = 32 g/mol = 0.032 kg/mol.

T = 300 K, M_{0} = 0.032 kg mol^{- 1}

2Solved Exampleboard3 steps

The spectrum of a star peaks at a wavelength of 500 nm. Treating the star as a blackbody, estimate its surface temperature. (Wien's constant b = 2.897 × 10⁻³ m K)

Wien's displacement law: λ_max T = b.

T = \frac{b}{λ _{ma x}}

3Solved Exampleboard3 steps

A body of surface area 0.5 m² and emissivity 0.6 is kept at 727 °C in surroundings at 27 °C. Find the net rate at which it loses heat by radiation. (σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴)

Convert to kelvin: T = 727 + 273 = 1000 K, T₀ = 27 + 273 = 300 K.

T^{4} - T_{0}^{4} = (1000)^{4} - (300)^{4} = 1 0^{12} - 8.1 \times 1 0^{9} = 9.919 \times 1 0^{11} K^{4}

4Solved ExampleHOTS3 steps

At what temperature will the rms speed of hydrogen molecules equal the rms speed of oxygen molecules at 47 °C? (Molar masses: H₂ = 2 g/mol, O₂ = 32 g/mol)

v_rms = √(3RT/M₀). Equal speeds require equal T/M₀.

\frac{T _{H}}{M _{H}} = \frac{T _{O}}{M _{O}}

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Using the temperature in °C in v_rms, (3/2)k_BT or Stefan's law
✓Every kinetic-theory and radiation formula needs the absolute temperature in kelvin: T(K) = t(°C) + 273.
2
Using molar mass in g/mol in v_rms = √(3RT/M₀)
✓Convert M₀ to kg/mol (divide by 1000). Otherwise v_rms comes out √1000 ≈ 31.6 times too small.
3
Writing the net radiation loss as eσA(T − T₀)⁴
✓The body emits eσAT⁴ and absorbs eσAT₀⁴, so the net loss is eσA(T⁴ − T₀⁴).
4
Thinking λ_max increases as a body gets hotter
✓λ_max = b/T. A hotter body peaks at a shorter wavelength, which is why a heated iron rod glows red, then orange, then white.
5
Assuming a good reflector is also a good emitter
✓By Kirchhoff's law a = e. A good reflector is a poor absorber, so it is also a poor emitter. A dull black surface is both a good absorber and a good emitter.
6
Taking C_P − C_V = R for specific heats per unit mass
✓C_P − C_V = R is for molar specific heats. Per unit mass (principal specific heats), the difference is R/M₀.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
Find the average translational kinetic energy of a gas molecule at 27 °C. (k_B = 1.38 × 10⁻²³ J K⁻¹)
Q2
A body absorbs 70% and reflects 20% of the radiant energy falling on it. Find its coefficient of transmission.
Q3
Two perfect blackbodies are at 300 K and 600 K. Find the ratio of their emissive powers (hotter : cooler).
Q4
For a gas, C_V = 20.8 J mol⁻¹ K⁻¹. Find C_P and γ, and identify whether the gas is monatomic or diatomic. (R = 8.314 J mol⁻¹ K⁻¹)
Q5
The rms speed of molecules of a gas of density 1.2 kg/m³ is 500 m/s. Find the pressure of the gas.
Q6
The temperature of a gas is 27 °C. To what temperature must it be heated so that the rms speed of its molecules doubles?

📝 Notes

Kinetic Theory of Gases and Radiation — Maharashtra HSC Overview

Chapter 3 of the Maharashtra Board (Balbharati) Std XII Physics textbook has two parts. The first explains gas behaviour from molecular motion. The second covers heat radiation and the blackbody.

From molecules to pressure and temperature

The whole kinetic theory section follows from one derivation. Molecules bouncing off the walls give $P = \tfrac{1}{3}(N/V)m\,v_{rms}^2$ . Comparing this with $PV = Nk_BT$ gives $\tfrac{1}{2}m\,v_{rms}^2 = \tfrac{3}{2}k_BT$ , so temperature is a measure of average molecular kinetic energy. Rearranging gives $v_{rms} = \sqrt{3RT/M_0}$ . Most numericals use one of these three lines.

Equipartition and specific heats

Each degree of freedom carries $\tfrac{1}{2}k_BT$ . Counting the degrees of freedom gives $C_V$ , Mayer's relation $C_P - C_V = R$ gives $C_P$ , and their ratio is $\gamma$ :

Monatomic: $\gamma = 5/3$ .
Rigid diatomic: $\gamma = 7/5$ .
Diatomic with vibration: $\gamma = 9/7$ .

Heat radiation

For radiation falling on a body, $a + r + t_r = 1$ . A perfect blackbody has $a = 1$ and is also the best emitter (Kirchhoff's law, $a = e$ ). The blackbody spectrum is described by two laws:

Wien's law: the peak shifts to shorter wavelengths as $T$ rises ( $\lambda_{max}T = b$ ).
Stefan–Boltzmann law: the total emitted power rises as $T^4$ ( $R = e\sigma T^4$ ).

Approach for numericals

Write all temperatures in kelvin and molar masses in kg/mol. For ratio questions (rms speeds, emissive powers), set up the proportionality ( $v_{rms} \propto \sqrt{T/M_0}$ , $R \propto T^4$ , $\lambda_{max} \propto 1/T$ ) instead of computing each value separately. This is faster and avoids rounding errors.