Maharashtra State Board · Class 12 · Physics · Chapter 3

Kinetic Theory of Gases and Radiation — Formula Sheet

Board Formulas

16 formulas

1.Ideal Gas Equation
$P V = n R T = N k_{B} T$
$P$ : 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)
n = number of moles, N = number of molecules. k_B = R/N_A = 1.38 × 10⁻²³ J K⁻¹. A real gas behaves like an ideal gas at low pressure and high temperature.
2.Mean Free Path
$λ = \frac{1}{2 π d ^{2} ( N / V )}$
$λ$ : Mean free path (m) · $d$ : Diameter of a molecule (m) · $N / V$ : Number of molecules per unit volume (m⁻³)
Average distance travelled by a molecule between two successive collisions. It is smaller for a denser gas and for bigger molecules.
3.Pressure Exerted by an Ideal Gas★
$P = \frac{1}{3} \frac{N}{V} m v_{r m s}^{2} = \frac{1}{3} ρ v_{r m s}^{2}$
$m$ : Mass of one molecule (kg) · $v_{r m s}$ : Root mean square speed (m s⁻¹) · $ρ$ : Density of the gas (kg m⁻³)
Pressure comes from the change in momentum of molecules colliding with the walls. Here ρ = Nm/V is the density of the gas.
4.Root Mean Square Speed★
$v_{r m s} = \frac{3 R T}{M _{0}} = \frac{3 k _{B} T}{m} = \frac{3 P}{ρ}$
$M_{0}$ : Molar mass of the gas (kg mol⁻¹) · $m$ : Mass of one molecule (kg) · $T$ : Absolute temperature (K)
v_rms ∝ √T and ∝ 1/√M₀. Lighter gases move faster at the same temperature. M₀ must be in kg mol⁻¹.
5.Kinetic Interpretation of Temperature★
$\frac{1}{2} m v_{r m s}^{2} = \frac{3}{2} k_{B} T$
$k_{B}$ : Boltzmann constant (J K⁻¹) · $T$ : Absolute temperature (K)
The average translational KE of a molecule depends only on the absolute temperature, not on the nature of the gas. For one mole: E = (3/2)RT. At T = 0 K the translational KE would be zero.
6.Law of Equipartition of Energy
$E_{per degree of freedom} = \frac{1}{2} k_{B} T$
$k_{B}$ : Boltzmann constant (J K⁻¹) · $T$ : Absolute temperature (K)
In thermal equilibrium, each degree of freedom of a molecule has average energy ½k_BT (per mole: ½RT). A vibrational mode counts twice (kinetic + potential), contributing k_BT.
7.Mayer's Relation★
$C_{P} - C_{V} = R$
$C_{P}$ : 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)
C_P and C_V are molar specific heats. C_P > C_V because, at constant pressure, part of the heat supplied goes into work done in expansion. If heat is measured in calories, C_P − C_V = R/J.
8.Ratio of Specific Heats
$γ = \frac{C _{P}}{C _{V}}$
$γ$ : Adiabatic ratio (no unit)
Always greater than 1. Its value depends on the atomicity of the gas through the number of degrees of freedom.
9.Specific Heats of a Monatomic Gas
$C_{V} = \frac{3}{2} R, C_{P} = \frac{5}{2} R, γ = \frac{5}{3} \approx 1.67$
Monatomic gases (He, Ne, Ar) have 3 translational degrees of freedom, so U = (3/2)RT per mole.
10.Specific Heats of a Diatomic Gas
$C_{V} = \frac{5}{2} R, C_{P} = \frac{7}{2} R, γ = \frac{7}{5} = 1.4$
Rigid diatomic molecule (O₂, N₂ at ordinary temperatures): 3 translational + 2 rotational degrees of freedom. If vibration is also excited: C_V = (7/2)R, C_P = (9/2)R, γ = 9/7 ≈ 1.29.
11.Absorption, Reflection and Transmission Coefficients
$a + r + t_{r} = 1$
$a$ : Coefficient of absorption (absorptive power) · $r$ : Coefficient of reflection (reflectance) · $t_{r}$ : Coefficient of transmission (transmittance)
a, r, t_r are the fractions of the incident radiant energy that are absorbed, reflected and transmitted. Perfect blackbody: a = 1, r = t_r = 0. Athermanous (opaque) body: t_r = 0, so a + r = 1.
12.Emissive Power and Coefficient of Emission
$R = \frac{Q}{A t}, e = \frac{R}{R _{B}}$
$R$ : 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)
R is the radiant energy emitted per unit time per unit area (W m⁻²). Do not confuse it with the gas constant. e is the emissivity: e = 1 for a perfect blackbody and 0 < e < 1 for ordinary bodies. Here t is time (the transmission coefficient is written t_r).
13.Kirchhoff's Law of Heat Radiation
$\frac{R}{a} = R_{B} \Leftrightarrow a = e$
$a$ : Coefficient of absorption of the body · $e$ : Coefficient of emission of the body
At a given temperature, the ratio of emissive power to absorptive power is the same for all bodies and equals the emissive power of a perfect blackbody. Good absorbers are good emitters.
14.Wien's Displacement Law★
$λ_{ma x} T = b$
$λ_{ma x}$ : 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
As temperature increases, the wavelength of maximum emission shifts to shorter wavelengths. b = 2.897 × 10⁻³ m K. Used to estimate the surface temperature of stars.
15.Stefan–Boltzmann Law★
$R_{B} = σ T^{4}, R = e σ T^{4}$
$σ$ : Stefan–Boltzmann constant = 5.67 × 10⁻⁸ W m⁻² K⁻⁴ · $T$ : Absolute temperature of the body (K) · $e$ : Emissivity of the body
The energy radiated per unit time per unit area is proportional to T⁴. Doubling the absolute temperature multiplies the emissive power by 16. Total power from area A: P = eσAT⁴.
16.Net Rate of Loss of Heat by Radiation
$\frac{Q}{t} = e σ A (T^{4} - T_{0}^{4})$
$Q / t$ : 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)
A body at temperature T in surroundings at T₀ emits eσAT⁴ and absorbs eσAT₀⁴. If T < T₀ the result is negative, meaning a net gain of heat.