Skip to content
CalibratePhysics

Chapter 3

Kinetic Theory of Gases and Radiation

Maharashtra HSC · 14 subtopics · textbook pages 56–74

Formula SheetEvery formula this chapter can be examined on, in one table.

Gas Laws and Equation of State

FormulaSymbols, units and conditions
PV=nRT=NkBTPV = nRT = Nk_{B}T
  • PP pressure (Pa), VV volume (m3^3), TT absolute temperature (K), nn number of moles, NN number of molecules. Valid for an ideal gas at all PP and TT; for a real gas only at low pressure and high temperature.
P1V1T1=P2V2T2\dfrac{P_{1}V_{1}}{T_{1}} = \dfrac{P_{2}V_{2}}{T_{2}}
  • Combined gas law for a fixed mass of gas. TT must be in kelvin.
n=NNA=MM0n = \dfrac{N}{N_{A}} = \dfrac{M}{M_{0}}
  • MM = mass of the gas taken, M0M_{0} = molar mass (mass of 1 mole), NAN_{A} = Avogadro number.
R=NAkBR = N_{A}k_{B}
  • Universal gas constant R=8.314R = 8.314 J mol−1^{-1} K−1^{-1} =8314= 8314 J kmol−1^{-1} K−1^{-1} (numericals often quote 83208320 J kmol−1^{-1} K−1^{-1} or R=8300R = 8300 J kg−1^{-1} K−1^{-1} in the older units). Boltzmann constant kB=1.38×10−23k_{B} = 1.38 \times 10^{-23} J K−1^{-1}. NA=6.022×1023N_{A} = 6.022 \times 10^{23} mol−1^{-1} =6.03×1026= 6.03 \times 10^{26} kmol−1^{-1}.
  • Individual gas laws: Boyle V∝1/PV \propto 1/P at constant TT; Charles V∝TV \propto T at constant PP; Gay-Lussac P∝TP \propto T at constant VV. All for a fixed mass.
  • S.T.P. / N.T.P. values: T=273T = 273 K, P=1.013×105P = 1.013 \times 10^{5} N m−2^{-2} (=1= 1 atm), molar volume =22.4= 22.4 litre mol−1^{-1}.

Mean Free Path

FormulaSymbols, units and conditions
λ=12 πd2(NV)\lambda = \dfrac{1}{\sqrt{2}\,\pi d^{2}\left(\dfrac{N}{V}\right)}
  • λ\lambda = mean free path (m), dd = diameter of a molecule (m), N/VN/V = number of molecules per unit volume (number density, m−3^{-3}).
λ=kBT2 πd2P\lambda = \dfrac{k_{B}T}{\sqrt{2}\,\pi d^{2}P}
  • Same relation with N/V=P/kBTN/V = P/k_{B}T substituted from PV=NkBTPV = Nk_{B}T. Use this form whenever PP and TT are given instead of the density.
  • Dependences to quote: λ∝1N/V\lambda \propto \dfrac{1}{N/V}, λ∝1d2\lambda \propto \dfrac{1}{d^{2}}, λ∝T\lambda \propto T at constant PP, λ∝1P\lambda \propto \dfrac{1}{P} at constant TT.

Pressure of Ideal Gas

FormulaSymbols, units and conditions
P=13NVmv2‾P = \dfrac{1}{3}\dfrac{N}{V}m\overline{v^{2}}
  • mm = mass of one molecule (kg), v2‾\overline{v^{2}} = mean square speed (m2^2 s−2^{-2}). Derived for a cubical box of side LL, but valid for a container of any shape.
P=13ρ v2‾P = \dfrac{1}{3}\rho\,\overline{v^{2}}
  • ρ=NmV\rho = \dfrac{Nm}{V} is the density of the gas (kg m−3^{-3}).
P=23NV(12mv2‾)P = \dfrac{2}{3}\dfrac{N}{V}\left(\dfrac{1}{2}m\overline{v^{2}}\right)
  • Pressure written in terms of the average translational kinetic energy of one molecule. Use this form to deduce Boyle's law.
v2‾=vx2‾+vy2‾+vz2‾,vx2‾=vy2‾=vz2‾=13v2‾\overline{v^{2}} = \overline{v_{x}^{2}} + \overline{v_{y}^{2}} + \overline{v_{z}^{2}}, \qquad \overline{v_{x}^{2}} = \overline{v_{y}^{2}} = \overline{v_{z}^{2}} = \dfrac{1}{3}\overline{v^{2}}
  • Symmetry result used in the derivation: the molecules have no preferred direction.

Root Mean Square (rms) Speed

FormulaSymbols, units and conditions
v2‾=3PVNm\overline{v^{2}} = \dfrac{3PV}{Nm}
  • Rearranged from the pressure expression.
vrms=v2‾=3RTM0=3kBTm=3Pρv_{rms} = \sqrt{\overline{v^{2}}} = \sqrt{\dfrac{3RT}{M_{0}}} = \sqrt{\dfrac{3k_{B}T}{m}} = \sqrt{\dfrac{3P}{\rho}}
  • M0M_{0} = molar mass in kg mol−1^{-1} (or kg kmol−1^{-1} if RR is in J kmol−1^{-1} K−1^{-1}), mm = mass of one molecule. Use whichever form matches the data given.
vrms∝T,vrms∝1M0,v1v2=T1M1×M2T2v_{rms} \propto \sqrt{T}, \qquad v_{rms} \propto \dfrac{1}{\sqrt{M_{0}}}, \qquad \dfrac{v_{1}}{v_{2}} = \sqrt{\dfrac{T_{1}}{M_{1}} \times \dfrac{M_{2}}{T_{2}}}
  • The comparison form for "compare the rms speeds of two gases" numericals.
vˉ=v1+v2+⋯+vNN,vrms=v12+v22+⋯+vN2N\bar{v} = \dfrac{v_{1}+v_{2}+\cdots+v_{N}}{N}, \qquad v_{rms} = \sqrt{\dfrac{v_{1}^{2}+v_{2}^{2}+\cdots+v_{N}^{2}}{N}}
  • For a small listed set of molecular speeds. The two are not equal; vrms>vˉv_{rms} > \bar{v} unless all speeds are equal.
vs=γRTM0v_{s} = \sqrt{\dfrac{\gamma RT}{M_{0}}}
  • Speed of sound in a gas; γ=CP/CV\gamma = C_{P}/C_{V}. Since γ<3\gamma < 3, sound is always slower than vrmsv_{rms}, though of the same order.
  • Standard values quoted in the text: vrms=517v_{rms} = 517 m s−1^{-1} for N2_2 and 483483 m s−1^{-1} for O2_2, both at 300300 K.

Interpretation of Temperature in Kinetic Theory

FormulaSymbols, units and conditions
PV=23N(12mv2‾)=23EPV = \dfrac{2}{3}N\left(\dfrac{1}{2}m\overline{v^{2}}\right) = \dfrac{2}{3}E
  • EE = total internal (purely kinetic) energy of the gas. For an ideal gas there is no potential energy term.
E=32NkBTandEN=32kBTE = \dfrac{3}{2}Nk_{B}T \qquad \text{and} \qquad \dfrac{E}{N} = \dfrac{3}{2}k_{B}T
  • Average kinetic energy per molecule is 32kBT\dfrac{3}{2}k_{B}T, and depends only on the absolute temperature — not on the nature of the gas.
12mv2‾=32kBT\dfrac{1}{2}m\overline{v^{2}} = \dfrac{3}{2}k_{B}T
  • Average translational kinetic energy of one molecule.
K.E. per mole=32RT,K.E. per unit mass=32RTM0,K.E. per unit volume=32P\text{K.E. per mole} = \dfrac{3}{2}RT, \qquad \text{K.E. per unit mass} = \dfrac{3}{2}\dfrac{RT}{M_{0}}, \qquad \text{K.E. per unit volume} = \dfrac{3}{2}P
  • The three forms the board asks for. For a mass MM of gas, K.E. =32MM0RT= \dfrac{3}{2}\dfrac{M}{M_{0}}RT.
P∝1V at constant TP \propto \dfrac{1}{V} \ \text{at constant } T
  • Boyle's law, deduced from kinetic theory.

Law of Equipartition of Energy

FormulaSymbols, units and conditions
Average energy per quadratic term=12kBT\text{Average energy per quadratic term} = \dfrac{1}{2}k_{B}T
  • Statement of the law, valid for a gas in thermal equilibrium at temperature TT; fails at very low temperatures where quantum effects matter.
12mvx2‾=12mvy2‾=12mvz2‾=12kBT\dfrac{1}{2}m\overline{v_{x}^{2}} = \dfrac{1}{2}m\overline{v_{y}^{2}} = \dfrac{1}{2}m\overline{v_{z}^{2}} = \dfrac{1}{2}k_{B}T
  • Each translational component carries 12kBT\dfrac{1}{2}k_{B}T; the three together give 32kBT\dfrac{3}{2}k_{B}T.
Erot=12Izωz2+12Iyωy2E_{\text{rot}} = \dfrac{1}{2}I_{z}\omega_{z}^{2} + \dfrac{1}{2}I_{y}\omega_{y}^{2}
  • Rotational kinetic energy of a diatomic molecule lying along the xx-axis, about the two perpendicular axes.
Evib=12mu2+12kr2E_{\text{vib}} = \dfrac{1}{2}mu^{2} + \dfrac{1}{2}kr^{2}
  • uu = vibrational speed, rr = separation of the atoms, kk = force constant. Two quadratic terms, so each vibrational mode contributes 2×12kBT=kBT2 \times \dfrac{1}{2}k_{B}T = k_{B}T.
E per molecule=f2kBTE \ \text{per molecule} = \dfrac{f}{2}k_{B}T
  • ff = total number of quadratic terms (degrees of freedom, with each vibrational mode counted twice).
  • MoleculeTranslational dofRotational dofVibrational contributionEnergy per molecule
  • Monatomic30032kBT\dfrac{3}{2}k_{B}T
  • Diatomic (rigid)32052kBT\dfrac{5}{2}k_{B}T
  • Diatomic (non-rigid)32kBTk_{B}T72kBT\dfrac{7}{2}k_{B}T
  • Polyatomic (non-linear)33f kBTf\,k_{B}T(3+f)kBT(3+f)k_{B}T

Specific Heat Capacity

FormulaSymbols, units and conditions
dQ1=dE=CV dT,dQ2=dE+dW=CP dT,dW=P dVdQ_{1} = dE = C_{V}\,dT, \qquad dQ_{2} = dE + dW = C_{P}\,dT, \qquad dW = P\,dV
  • Heat supplied at constant volume and at constant pressure to one mole.
CP−CV=RC_{P} - C_{V} = R
  • Mayer's relation, for one mole, with heat and work in the same (mechanical) units.
CP−CV=RJC_{P} - C_{V} = \dfrac{R}{J}
  • Form used when heat is measured in calories and work in joules; JJ = mechanical equivalent of heat (≈4.15\approx 4.15 J cal−1^{-1} from the textbook worked example).
SP−SV=RM0J,CP=M0SP,CV=M0SVS_{P} - S_{V} = \dfrac{R}{M_{0}J}, \qquad C_{P} = M_{0}S_{P}, \qquad C_{V} = M_{0}S_{V}
  • SPS_{P}, SVS_{V} = principal specific heats (per unit mass); CPC_{P}, CVC_{V} = molar specific heats.
γ=CPCV\gamma = \dfrac{C_{P}}{C_{V}}
  • Adiabatic ratio; always greater than 1 since CP>CVC_{P} > C_{V}.
  • GasCVC_{V}CPC_{P}γ\gamma
  • Monatomic32R\dfrac{3}{2}R52R\dfrac{5}{2}R53\dfrac{5}{3}
  • Diatomic (rigid rotator)52R\dfrac{5}{2}R72R\dfrac{7}{2}R75\dfrac{7}{5}
  • Diatomic (non-rigid, vibrating)72R\dfrac{7}{2}R92R\dfrac{9}{2}R97\dfrac{9}{7}
  • Polyatomic (ff vibrational modes)(3+f)R(3+f)R(4+f)R(4+f)R4+f3+f\dfrac{4+f}{3+f}

Absorption, Reflection and Transmission of Heat Radiation

FormulaSymbols, units and conditions
Q=Qa+Qr+QtQ = Q_{a} + Q_{r} + Q_{t}—
a+r+tr=1a + r + t_{r} = 1—
a=QaQ,r=QrQ,tr=QtQa = \dfrac{Q_{a}}{Q}, \qquad r = \dfrac{Q_{r}}{Q}, \qquad t_{r} = \dfrac{Q_{t}}{Q}
  • aa = coefficient of absorption, rr = coefficient of reflection, trt_{r} = coefficient of transmission. All three are ratios of energies, hence dimensionless and unitless.
  • Special cases: a=1a = 1 perfect blackbody; r=1r = 1 perfect reflector; tr=1t_{r} = 1 perfect transmitter (completely transparent); tr=0t_{r} = 0 opaque body, for which a+r=1a + r = 1.

Emission of Heat Radiation

FormulaSymbols, units and conditions
R=QAtR = \dfrac{Q}{At}
  • Emissive power (radiant power) = heat radiated per unit area per unit time. SI unit J m−2^{-2} s−1^{-1} or W m−2^{-2}; dimensions [L0M1T−3][L^{0}M^{1}T^{-3}].
e=RRBe = \dfrac{R}{R_{B}}
  • Coefficient of emission (emissivity): ratio of the emissive power of the surface to that of a perfect blackbody at the same temperature. e=1e = 1 for a perfect blackbody, e=0e = 0 for a perfect reflector, 0<e<10 < e < 1 for an ordinary body (e.g. copper e=0.3e = 0.3).

Kirchhoff's Law of Heat Radiation

FormulaSymbols, units and conditions
Ra=RBat a given temperature, for all wavelengths\dfrac{R}{a} = R_{B} \qquad \text{at a given temperature, for all wavelengths}—
a=eora(λ)=e(λ)a = e \qquad \text{or} \qquad a(\lambda) = e(\lambda)
  • RR = emissive power of the body, aa = its coefficient of absorption, RBR_{B} = emissive power of a perfect blackbody at the same temperature. Good absorbers are good emitters.

Spectral Distribution and Wien's Displacement Law

FormulaSymbols, units and conditions
λmax∝1T⇒λmaxT=b\lambda_{max} \propto \dfrac{1}{T} \qquad \Rightarrow \qquad \lambda_{max}T = b
  • λmax\lambda_{max} = wavelength at which the emissive power of a blackbody is maximum (m), TT = absolute temperature (K), Wien's constant b=2.897×10−3b = 2.897 \times 10^{-3} m K.
  • λmax\lambda_{max} is the wavelength of maximum intensity, not the maximum wavelength emitted. Area under the RλR_{\lambda} versus λ\lambda curve = total energy radiated per unit area per unit time at that temperature.

Stefan-Boltzmann Law of Radiation

FormulaSymbols, units and conditions
R=σT4R = \sigma T^{4}
  • Perfect blackbody: power radiated per unit area is proportional to the fourth power of the absolute temperature. Stefan's constant σ=5.67×10−8\sigma = 5.67 \times 10^{-8} J m−2^{-2} s−1^{-1} K−4^{-4} (W m−2^{-2} K−4^{-4}); dimensions [L0M1T−3K−4][L^{0}M^{1}T^{-3}K^{-4}]. Some numericals use σ=5.7×10−8\sigma = 5.7 \times 10^{-8}.
R=eσT4R = e\sigma T^{4}
  • Ordinary body of emissivity ee.
Q=σAtT4andQ=eσAtT4Q = \sigma A t T^{4} \qquad \text{and} \qquad Q = e\sigma A t T^{4}
  • Energy radiated in time tt by area AA; blackbody and ordinary body respectively.
dQdt=σA(T4−T04),dQdt=eσA(T4−T04)\dfrac{dQ}{dt} = \sigma A\left(T^{4} - T_{0}^{4}\right), \qquad \dfrac{dQ}{dt} = e\sigma A\left(T^{4} - T_{0}^{4}\right)
  • Net rate of loss of heat by a body at temperature TT in surroundings at T0T_{0}. If T<T0T < T_{0} the same expression with (T04−T4)\left(T_{0}^{4} - T^{4}\right) gives the net rate of gain.
R1R2=T14−T04T24−T04\dfrac{R_{1}}{R_{2}} = \dfrac{T_{1}^{4} - T_{0}^{4}}{T_{2}^{4} - T_{0}^{4}}
  • Ratio form for "compare the rates of loss of heat" numericals; ee, σ\sigma and AA cancel.
P=σAT4=σ(4πr2)T4P = \sigma A T^{4} = \sigma\left(4\pi r^{2}\right)T^{4}
  • For a spherical radiator (star, sphere) of radius rr. Comparing two spheres: P1P2=r12T14r22T24\dfrac{P_{1}}{P_{2}} = \dfrac{r_{1}^{2}T_{1}^{4}}{r_{2}^{2}T_{2}^{4}}.
Short NotesThe chapter subtopic by subtopic, in the textbook's own order.14 subtopics
  1. 1Behaviour of a Gas
  2. 2Ideal Gas and Real Gas
  3. 3Mean Free Path
  4. 4Pressure of Ideal Gas1 board question
  5. 5Root Mean Square (rms) Speed4 board questions
  6. 6Interpretation of Temperature in Kinetic Theory6 board questions
  7. 7Law of Equipartition of Energy
  8. 8Specific Heat Capacity2 board questions
  9. 9Absorption, Reflection and Transmission of Heat Radiation1 board question
  10. 10Perfect Blackbody6 board questions
  11. 11Emission of Heat Radiation1 board question
  12. 12Kirchhoff's Law of Heat Radiation and its Theoretical Proof2 board questions
  13. 13Spectral Distribution of Blackbody Radiation5 board questions
  14. 14Stefan-Boltzmann Law of Radiation4 board questions
The whole chapter in brief

3.1 Introduction — Equation of State

  • The state of a gas is specified by PP, VV, TT, internal energy EE. The relation between them is the equation of state.
  • Boyle: V∝1/PV \propto 1/P at constant TT. Charles: V∝TV \propto T at constant PP. Gay-Lussac: P∝TP \propto T at constant VV. All for a fixed mass of enclosed gas.
  • Combining: PV∝TPV \propto T, i.e. P1V1T1=P2V2T2\dfrac{P_{1}V_{1}}{T_{1}} = \dfrac{P_{2}V_{2}}{T_{2}}, and PV=nRT=NkBTPV = nRT = Nk_{B}T with R=NAkBR = N_{A}k_{B}.
  • The three gas laws hold for real gases only if the pressure is not too high and the temperature is not close to the liquefaction temperature.
  • A gas obeying PV=nRTPV = nRT at all pressures and temperatures is an ideal gas.

3.2 Behaviour of a Gas

  • PP, VV, TT are a macroscopic description. There are about 102310^{23} particles per m3^3, so describing the gas through the motion of a single molecule (as one would a thrown stone with Newton's laws) is futile.
  • Approach of kinetic theory: make assumptions about the molecules, take averages of physical quantities over the enormous number of particles, and relate those averages to the macroscopic parameters.

3.3 Ideal Gas and Real Gas

Ideal gasReal gas
Obeys PV=nRTPV = nRT at all PP and TTObeys it only at low pressure and high temperature
No intermolecular interactionAtoms/molecules do interact
Molecules are point particles (size neglected)Molecules have finite size
Internal energy purely kineticHas potential energy of interaction as well
Does not exist in realityAll actual gases
  • A real gas is said to be in the ideal state when its molecules are so far apart that intermolecular interaction is practically absent — that is, at sufficiently low density, meaning low pressure or high temperature, with TT well above the liquefaction temperature.
  • Deviation is largest at high pressure and low temperature because the molecules are then close together, so both their finite size and their mutual attraction matter.
  • Molecules are uniformly dispersed in the container and execute random motion. On approach they repel, behaving as small hard spheres, so collisions are elastic and both speed and direction change abruptly. Between two successive collisions a molecule travels in a straight line with constant velocity; force acts only during a collision.

3.4 Mean Free Path

  • Definition: the average distance traversed by a molecule with constant velocity between two successive collisions.

λ=12 πd2(NV)=kBT2 πd2P\lambda = \dfrac{1}{\sqrt{2}\,\pi d^{2}\left(\dfrac{N}{V}\right)} = \dfrac{k_{B}T}{\sqrt{2}\,\pi d^{2}P}

  • λ\lambda varies inversely with the number density N/VN/V: higher density, more collisions, smaller λ\lambda.
  • λ\lambda varies inversely with d2d^{2}, not dd, because collision depends on the cross section of a molecule.
  • Application: in metal-film coating, the enclosure is evacuated so that λ\lambda of the air molecules exceeds the size of the enclosure; the metal vapour atoms then reach the target without colliding with air molecules and get deposited.
  • Textbook result: for nitrogen at 0 ∘0\,^{\circ}C and 1 atm with d=324d = 324 pm, λ=8.0×10−8\lambda = 8.0 \times 10^{-8} m, about 247 molecular diameters.

3.5 Pressure of Ideal Gas

P=13NVmv2‾=13ρv2‾P = \dfrac{1}{3}\dfrac{N}{V}m\overline{v^{2}} = \dfrac{1}{3}\rho\overline{v^{2}}

  • Set-up: nn moles in a cubical box of side LL, volume V=L3V = L^{3}, walls at constant temperature TT, walls parallel to the coordinate planes. Intermolecular collisions neglected; only elastic collisions with the walls counted.
  • Key steps to state: Δpx=−2mvx\Delta p_{x} = -2mv_{x}, momentum given to the wall =+2mvx= +2mv_{x}; distance between two collisions with the same wall =2L= 2L, so Δt=2L/vx\Delta t = 2L/v_{x}; average force from one molecule =mvx12/L= mv_{x1}^{2}/L; divide total force by area L2L^{2}; then use vx2‾=13v2‾\overline{v_{x}^{2}} = \dfrac{1}{3}\overline{v^{2}}.
  • The result is valid for a container of any shape, and intermolecular collisions do not change it, because in a gas in equilibrium every molecule that loses a particular velocity is replaced by another acquiring it.

3.6 Root Mean Square (rms) Speed

vrms=3RTM0=3kBTm=3Pρ⇒vrms∝Tv_{rms} = \sqrt{\dfrac{3RT}{M_{0}}} = \sqrt{\dfrac{3k_{B}T}{m}} = \sqrt{\dfrac{3P}{\rho}} \qquad \Rightarrow \qquad v_{rms} \propto \sqrt{T}

  • Mean square speed v2‾∝T\overline{v^{2}} \propto T; rms speed ∝T\propto \sqrt{T} and ∝1/M0\propto 1/\sqrt{M_{0}}. At the same temperature, lighter molecules move faster.
  • Maxwell's distribution of molecular speeds: a plot of nvn_{v} against vv; the shaded strip nv dvn_{v}\,dv is the number of molecules with speeds between vv and v+dvv + dv. Speeds range from zero to infinity; averages such as v2‾\overline{v^{2}} follow from this distribution.
  • Speed of sound vs=γRT/M0v_{s} = \sqrt{\gamma RT/M_{0}} is of the same order as, but always less than, vrmsv_{rms} (since γ<3\gamma < 3). The molecules are the medium that transports sound energy.

3.7 Interpretation of Temperature in Kinetic Theory

PV=23E,E=32NkBT,EN=32kBTPV = \dfrac{2}{3}E, \qquad E = \dfrac{3}{2}Nk_{B}T, \qquad \dfrac{E}{N} = \dfrac{3}{2}k_{B}T

  • The internal energy of an ideal gas is purely kinetic — there is no potential energy term because the molecules are non-interacting.
  • Average energy per molecule is proportional to the absolute temperature and is the same for all gases at a given TT. This is the equation that links the macroscopic parameter TT to molecular motion.
  • Boyle's law from kinetic theory: P=23NV(12mv2‾)P = \dfrac{2}{3}\dfrac{N}{V}\left(\dfrac{1}{2}m\overline{v^{2}}\right). At a fixed temperature 12mv2‾\dfrac{1}{2}m\overline{v^{2}} is constant and NN is fixed, so P=constant/VP = \text{constant}/V, i.e. P∝1/VP \propto 1/V.
  • Heating a gas: the supplied heat increases the average kinetic energy of the molecules; since 12mv2‾=32kBT\dfrac{1}{2}m\overline{v^{2}} = \dfrac{3}{2}k_{B}T, the temperature rises.

3.8 Law of Equipartition of Energy

  • Statement: for a gas in thermal equilibrium at temperature TT, the average energy per molecule associated with each quadratic term (in velocity or in position) is 12kBT\dfrac{1}{2}k_{B}T.
  • Degrees of freedom: the total number of coordinates or independent quantities required to describe the position and configuration of the system completely.
  • Degrees of freedom depend on the atomicity of the molecule (number of atoms), its geometric structure or shape, and the temperature.
MoleculeTranslationalRotationalVibrational modes
Monatomic (He)300
Diatomic (O2_2, N2_2, CO), rigid320
Diatomic, non-rigid321 (contributes kBTk_{B}T)
Linear polyatomic32ff
Non-linear polyatomic (NH3_3)33ff
  • A diatomic molecule lying along the xx-axis rotates about the yy- and zz-axes only; rotation about the xx-axis is not a rotation, as it does not change the positions of the atoms.
  • Each vibrational mode contributes 2×12kBT=kBT2 \times \dfrac{1}{2}k_{B}T = k_{B}T, because vibrational energy has both a kinetic term 12mu2\dfrac{1}{2}mu^{2} and a potential term 12kr2\dfrac{1}{2}kr^{2}, each quadratic.
  • The law holds at high temperatures and fails at extremely low temperatures where quantum effects become important.

3.9 Specific Heat Capacity

  • Two specific heats are defined for a gas because a small rise in temperature causes a considerable change in both volume and pressure; heating at constant volume and at constant pressure are physically different.
  • Mayer's relation: CP−CV=RC_{P} - C_{V} = R for one mole, or CP−CV=R/JC_{P} - C_{V} = R/J if heat is in calories and work in joules. In terms of principal specific heats, SP−SV=RM0JS_{P} - S_{V} = \dfrac{R}{M_{0}J}.
  • CP>CVC_{P} > C_{V} always: at constant pressure part of the heat is spent as work P dVP\,dV in moving the piston, so more heat is needed for the same rise in temperature.
Gasdof usedCVC_{V}CPC_{P}γ\gamma
Monatomic332R\dfrac{3}{2}R52R\dfrac{5}{2}R1.671.67
Diatomic (rigid)552R\dfrac{5}{2}R72R\dfrac{7}{2}R1.401.40
Diatomic (vibrating)772R\dfrac{7}{2}R92R\dfrac{9}{2}R1.291.29
Polyatomic6+2f6 + 2f(3+f)R(3+f)R(4+f)R(4+f)R4+f3+f\dfrac{4+f}{3+f}

3.10 Absorption, Reflection and Transmission of Heat Radiation

  • Radiation is the fastest mode of heat transfer and the only one that needs no material medium; physical contact is not required. Thermal radiation is electromagnetic radiation emitted by bodies at a temperature higher than the surroundings.
  • Q=Qa+Qr+QtQ = Q_{a} + Q_{r} + Q_{t}, hence a+r+tr=1a + r + t_{r} = 1. All three coefficients are dimensionless.
  • Coefficient of absorption aa: ratio of the amount of heat absorbed to the total heat incident.
  • Coefficient of reflection rr: ratio of the radiant energy reflected to the total energy incident.
  • Coefficient of transmission trt_{r}: ratio of the radiant energy transmitted to the total energy incident.
Diathermanous substanceAthermanous substance
Heat radiation can pass through itLargely opaque to heat radiation
tr≠0t_{r} \neq 0tr=0t_{r} = 0, so a+r=1a + r = 1
Neither a good absorber nor a good reflectorAbsorbs or reflects the incident radiation
Glass, quartz, sodium chloride, hydrogen, oxygen, dry airWater, wood, iron, copper, moist air, benzene
  • aa, rr and trt_{r} depend on the wavelength of the incident radiation as well as on the material: the same body can be diathermanous at one wavelength and a good absorber at another.

3.11 Perfect Blackbody

  • Definition: a body which absorbs the entire radiant energy incident on it, so a=1a = 1.
  • Lamp black and platinum black absorb about 97% of incident radiant heat and so closely resemble a perfect blackbody.
  • Ferry's blackbody: a double-walled hollow sphere with a tiny aperture; space between the walls evacuated, outer surface silvered, inner surface coated with lamp black, with a conical projection opposite the aperture. Radiation entering suffers multiple reflections and is almost completely absorbed. The aperture behaves like a perfect blackbody, and its area is the effective area of the blackbody.
  • Good absorbers are good emitters; poor absorbers are poor emitters. A poor absorber is a good reflector, so a good reflector is a poor emitter — the reason for silvering the walls of a thermos flask.
  • Cavity radiator: a block with an internal cavity connected to the outside by a small hole. Radiation entering cannot escape, so the cavity acts as a blackbody. When heated, it emits cavity radiation whose nature depends only on the temperature of the walls, not on the shape, size or material of the cavity.

3.12 Emission of Heat Radiation

  • Prevost's theory of exchange of heat (1792): all bodies at all temperatures above 0 K radiate thermal energy and simultaneously absorb radiation from their surroundings.
  • If a body radiates more than it absorbs its temperature falls, and vice versa. When the rate of emission equals the rate of absorption, the temperature is constant and the body is in thermal equilibrium with its surroundings.
  • Amount of heat radiated depends on: absolute temperature TT; the nature of the body (material, polished or not); surface area AA; and the time tt for which it radiates.
  • Emissive power (radiant power) R=QAtR = \dfrac{Q}{At}: heat radiated per unit area per unit time. SI unit W m−2^{-2}, dimensions [L0M1T−3][L^{0}M^{1}T^{-3}].
  • Coefficient of emission (emissivity) e=RRBe = \dfrac{R}{R_{B}}, the ratio of the emissive power of the surface to that of a perfect black surface at the same temperature. e=1e = 1 for a perfect blackbody, e=0e = 0 for a perfect reflector; for copper e=0.3e = 0.3. Emissivity is larger for rough surfaces and smaller for smooth, polished ones, and varies somewhat with temperature and wavelength.
  • Colour with temperature: below about 800 ∘800\,^{\circ}C the radiation is longer than visible wavelengths (infrared, invisible); around 800 ∘800\,^{\circ}C the body appears red; around 3000 ∘3000\,^{\circ}C it looks white hot, like a tungsten filament.

3.13 Kirchhoff's Law of Heat Radiation

  • Statement: at a given temperature, the ratio of the emissive power to the coefficient of absorption of a body is equal to the emissive power of a perfect blackbody at the same temperature, for all wavelengths.
  • Equivalent statement: for a body emitting and absorbing thermal radiation in thermal equilibrium, the emissivity is equal to the absorptivity. Symbolically a=ea = e, or more specifically a(λ)=e(λ)a(\lambda) = e(\lambda).
  • Consequence: a body with high emissive power also has high absorptive power, and vice versa.
  • Everyday applications: the bottom of a cooking utensil is blackened so that it absorbs heat well, while the top is polished so that it emits (and so loses) little; a good reflector such as the silvered wall of a thermos is a poor emitter.

3.14 Spectral Distribution of Blackbody Radiation

  • Lummer and Pringsheim measured the radiant power of a cavity radiator at different wavelengths, keeping the source at different fixed temperatures, and plotted RλR_{\lambda} against λ\lambda. The distribution depends only on the absolute temperature and not on the material.
  • Diagram in words: a family of humped curves, RλR_{\lambda} on the yy-axis against λ\lambda on the xx-axis, one curve per temperature. Each rises from zero, peaks at λmax\lambda_{max}, then falls asymptotically. Higher-temperature curves lie entirely above lower-temperature ones and their peaks are shifted to shorter wavelengths (to the left).
  • Observations from the curves: (i) the energy is not uniformly distributed over the spectrum; (ii) with increasing λ\lambda the radiant power first increases, reaches a maximum and then decreases; (iii) the area under a curve gives the total energy emitted per unit time per unit area at all wavelengths; (iv) λmax\lambda_{max} decreases as TT increases; (v) the total radiant power increases with temperature; (vi) at 300 K the most intense wave has λ≈5×10−5\lambda \approx 5 \times 10^{-5} m, in the infrared.
  • Wien's displacement law: the wavelength for which the emissive power of a blackbody is maximum is inversely proportional to the absolute temperature: λmaxT=b\lambda_{max}T = b, with b=2.897×10−3b = 2.897 \times 10^{-3} m K. Used to find the temperature of distant stars, the Sun and the Moon. White dwarfs have surface temperature about 10000 K, red giants about 3000 K.
  • Theories: Wien's expression from thermodynamics fitted only short wavelengths; the Rayleigh-Jeans formula from equipartition fitted long wavelengths but diverged at short wavelengths. Planck combined the two with the hypothesis of quantised oscillator energy E=nhνE = nh\nu, with h=6.626×10−34h = 6.626 \times 10^{-34} J s and nn a positive integer, and described the whole spectrum.

3.15 Stefan-Boltzmann Law of Radiation

  • Statement: the rate of emission of radiant energy per unit area, or the power radiated per unit area, of a perfect blackbody is directly proportional to the fourth power of its absolute temperature.

R=σT4,σ=5.67×10−8 W m−2K−4R = \sigma T^{4}, \qquad \sigma = 5.67 \times 10^{-8}\ \text{W m}^{-2}\text{K}^{-4}

  • Power radiated by a perfect blackbody depends only on its temperature, not on colour, material or nature of surface. For an ordinary body R=eσT4R = e\sigma T^{4}.
  • Body at TT in surroundings at T0T_{0}: net loss per unit area per unit time =eσ(T4−T04)= e\sigma\left(T^{4} - T_{0}^{4}\right); if T<T0T < T_{0} the body gains eσ(T04−T4)e\sigma\left(T_{0}^{4} - T^{4}\right) instead.
  • Doubling the absolute temperature increases the radiated power by 24=162^{4} = 16 times.

Traps that cost marks

  • Always convert ∘^{\circ}C to K before using any of vrmsv_{rms}, K.E., Wien's law or Stefan's law. 127 ∘127\,^{\circ}C is 400 K, not 127 K.
  • λmax\lambda_{max} is the wavelength of maximum intensity, not the longest wavelength emitted.
  • In the mean free path formula the diameter is squared and there is a factor 2\sqrt{2}: λ=12πd2(N/V)\lambda = \dfrac{1}{\sqrt{2}\pi d^{2}(N/V)}, not 12πd(N/V)\dfrac{1}{\sqrt{2}\pi d(N/V)}.
  • Mean speed vˉ\bar{v} and rms speed vrmsv_{rms} are different quantities; do not square the mean, average the squares.
  • For a net rate of loss of heat use (T4−T04)\left(T^{4} - T_{0}^{4}\right), not (T−T0)4\left(T - T_{0}\right)^{4}.
  • σ=5.67×10−8\sigma = 5.67 \times 10^{-8} carries a negative exponent; a positive exponent is a very common slip.
  • Mayer's relation CP−CV=RC_{P} - C_{V} = R is per mole. Use SP−SV=R/(M0J)S_{P} - S_{V} = R/(M_{0}J) when the data are per unit mass in calories.
  • Degrees of freedom for a diatomic molecule are 5 only if it is treated as a rigid rotator; if vibration is included they are 7 quadratic terms.
  • aa, rr and trt_{r} have no units and no dimensions; do not attach J or W to them.
  • In PV=NkBTPV = Nk_{B}T, NN is the number of molecules; in PV=nRTPV = nRT, nn is the number of moles. Mixing them up changes the answer by NAN_{A}.
Solved Numerical ExamplesWorked numericals to attempt first, each with a hint before the full solution.Coming soon

These are being added chapter by chapter. They are solved numericals from the prescribed book and the reference guides — not previous board questions, which live under Practice — so that you can work through the method before meeting it in a real paper.

Important DerivationsEvery derive / obtain / prove question in this chapter. 1 have come up in two or more papers.7
  1. 1
    Pressure exerted by a gas (kinetic theory of gases) Repeats
    Derive an expression for the pressure exerted by a gas on the basis of the kinetic theory of gases.

    Asked in: October 2009; March 2010; March 2022 (Q.27, 4 marks); February 2023 (Q.27, 4 marks)

    Show the derivation
    Suppose, P = pressure exerted by the gas, V = volume of the gas, N = number of molecules of the gas, m = mass of each molecule of the gas
    ∴\therefore Total mass of the gas, M = mN
    Step 1
    From kinetic theory of gases, P=13mNVv2‾P=\dfrac{1}{3}\dfrac{mN}{V}\overline{v^2}
    Step 2
    Pressure exerted by a gas in an enclosed vessel, P=23NV(12mv2‾)P=\dfrac{2}{3}\dfrac{N}{V}\left(\dfrac{1}{2}m\overline{v^2}\right)
    Step 3
    But, 12mv2‾=\dfrac{1}{2}m\overline{v^2}= K.E. = constant for all the gases at a given temperature. N = number of molecules which is constant for a given mass of the gas.
    ∴P=constantV\therefore P=\dfrac{\text{constant}}{V}
    ∴P∝1V\therefore P\propto\dfrac{1}{V}
    Hence, at constant temperature, pressure of the gas is increased if its volume is reduced.
  2. 2
    Average kinetic energy per molecule is directly proportional to absolute temperature
    Show that the average energy per molecule is directly proportional to the absolute temperature 'T' of the gas.

    Asked in: July 2024 (Q.3, 2 marks)

    Show the derivation

    The question is a real board question. This working is ours — no source printed a derivation for it.

    What is assumed

    The expression for the pressure of an ideal gas, obtained from kinetic theory,

    P=13 Nmv2‾VP = \dfrac{1}{3}\,\dfrac{N m \overline{v^{2}}}{V}

    where NN is the number of molecules in volume VV, mm the mass of one molecule and v2‾\overline{v^{2}} the mean square speed.

    Derivation

    Multiply and divide the right-hand side by 2 so that the kinetic energy appears explicitly:

    PV=13Nmv2‾=23N(12mv2‾)PV = \dfrac{1}{3} N m \overline{v^{2}} = \dfrac{2}{3} N \left(\dfrac{1}{2} m \overline{v^{2}}\right)

    The bracket is exactly the average translational kinetic energy of one molecule. Writing it as E‾\overline{E},

    PV=23NE‾…(1)PV = \dfrac{2}{3} N \overline{E} \qquad \dots (1)

    For nn moles of an ideal gas the equation of state is

    PV=nRT…(2)PV = nRT \qquad \dots (2)

    Comparing (1) and (2),

    23NE‾=nRT⇒E‾=32 nRTN\dfrac{2}{3} N \overline{E} = nRT \quad \Rightarrow \quad \overline{E} = \dfrac{3}{2}\,\dfrac{nRT}{N}

    Now N=nNAN = n N_A, where NAN_A is Avogadro's number, so n/N=1/NAn/N = 1/N_A, and R/NA=kBR/N_A = k_B, the Boltzmann constant. Hence

     E‾=32kBT \boxed{\,\overline{E} = \dfrac{3}{2} k_B T\,}

    Conclusion

    Since kBk_B is a universal constant, E‾∝T\overline{E} \propto T: the average translational kinetic energy of a molecule of an ideal gas depends only on the absolute temperature and not at all on the nature of the gas or on its pressure. Two gases at the same temperature therefore have molecules of the same average kinetic energy, although the heavier molecules move more slowly.

    Absolute zero is the temperature at which this average energy would be zero.

  3. 3
    Mayer's relation, Cp−Cv=RJC_p - C_v = \dfrac{R}{J}
    Prove Mayer's relation: Cp−Cv=RJC_p - C_v = \dfrac{R}{J}

    Asked in: February 2024 (Q.25, 3 marks)

    Show the derivation
    Step 1
    Consider one mole of an ideal gas that is enclosed in a cylinder by light, frictionless airtight piston.
    Step 2
    Let P, V and T be the pressure, volume and temperature respectively of the gas.
    Step 3
    If the gas is heated so that its temperature rises by dT, but the volume remains constant, then the amount of heat supplied to the gas(dQ1dQ_1) is used to increase the internal energy of the gas (dE). Since volume of the gas is constant, no work is done in moving the piston.
    ∴dQ1=dE=CVdT....(1)\therefore dQ_1=dE=C_VdT\qquad....(1)
    where CVC_V is the molar specific heat of the gas at constant volume.
    Step 4
    On the other hand, if the gas is heated to the same temperature, at constant pressure, volume of the gas increases by an amount say dV. The amount of heat supplied to the gas is used to increase the internal energy of the gas as well as to move the piston backwards to allow expansion of gas. The work done to move the piston, dW=PdVdW=PdV.
    ∴dQ2=dE+dW=CPdT....(2)\therefore dQ_2=dE+dW=C_PdT\qquad....(2)
    Where, CPC_P is the molar specific heat of the gas at constant pressure.
    Step 5
    From equations (1) and (2), CPdT=CVdT+dWC_PdT=C_VdT+dW
    ∴(CP−CV)dT=PdV....(3)\therefore (C_P-C_V)dT=PdV\qquad....(3)
    Step 6
    For one mole of gas, PV = RT
    ∴PdV=RdT\therefore PdV=RdT, since pressure is constant.
    Substituting equation (3), we get (CP−CV)dT=RdT(C_P-C_V)dT=RdT
    ∴CP−CV=R\therefore C_P-C_V=R
    In mechanical units, i.e., joules, CP−CV=RJC_P-C_V=\dfrac{R}{J}
    Where, J is mechanical equivalent of heat.
    This is known as Mayer's relation between CPC_P and CVC_V.
  4. 4
    Adiabatic constant for a rigid diatomic gas molecule
    Explain the thermodynamics of an adiabatic process to obtain the adiabatic constant for a rigid diatomic molecule of a gas.

    Asked in: June 2025 (Q.5, 2 marks)

    Show the derivation

    The question is a real board question. This working is ours — no source printed a derivation for it.

    The adiabatic constant

    The adiabatic constant (or ratio of specific heats) is

    γ=CPCV\gamma = \dfrac{C_P}{C_V}

    where CPC_P and CVC_V are the molar specific heats at constant pressure and at constant volume. In an adiabatic process no heat is exchanged, ΔQ=0\Delta Q = 0, so the first law of thermodynamics gives ΔU=−W\Delta U = -W: the work done by the gas comes entirely out of its internal energy. The relation between pressure and volume in such a process is PVγ=constantPV^{\gamma} = \text{constant}, which is why γ\gamma is called the adiabatic constant, and its value is fixed by how many ways the molecule can store energy.

    Degrees of freedom of a rigid diatomic molecule

    A rigid diatomic molecule (the bond does not stretch, so there is no vibrational contribution) has

    • 3 translational degrees of freedom, and
    • 2 rotational degrees of freedom — rotation about the two axes perpendicular

    to the line joining the atoms. Rotation about the bond axis itself carries negligible moment of inertia and is not counted.

    f=3+2=5f = 3 + 2 = 5

    Applying the law of equipartition of energy

    Each degree of freedom carries an average energy 12kBT\tfrac{1}{2} k_B T per molecule, so for one mole the internal energy is

    U=f×12kBT×NA=52RTU = f \times \dfrac{1}{2} k_B T \times N_A = \dfrac{5}{2} RT

    At constant volume no work is done, so all the heat supplied raises the internal energy:

    CV=dUdT=52RC_V = \dfrac{dU}{dT} = \dfrac{5}{2} R

    From Mayer's relation for one mole, CP−CV=RC_P - C_V = R,

    CP=CV+R=52R+R=72RC_P = C_V + R = \dfrac{5}{2}R + R = \dfrac{7}{2} R

    Result

    γ=CPCV=72R52R=75\gamma = \dfrac{C_P}{C_V} = \dfrac{\tfrac{7}{2}R}{\tfrac{5}{2}R} = \dfrac{7}{5}

     γ=1.4 \boxed{\,\gamma = 1.4\,}

    In general, for a molecule with ff degrees of freedom, γ=1+2/f\gamma = 1 + 2/f; putting f=5f = 5 gives 1.41.4, and putting f=3f = 3 recovers 1.671.67 for a monoatomic gas.

  5. 5
    Molar specific heat of a monoatomic gas at constant volume
    Deduce an expression for molar specific heat of a monoatomic gas at constant volume.

    Asked in: July 2024 (Q.29(i), 4 marks)

    Show the derivation

    The question is a real board question. This working is ours — no source printed a derivation for it.

    Definition

    The molar specific heat at constant volume is the heat needed to raise the temperature of one mole of the gas by one kelvin while its volume is held fixed:

    CV=(dQdT)VC_V = \left(\dfrac{dQ}{dT}\right)_{V}

    Internal energy of one mole of a monoatomic gas

    A monoatomic molecule is a single atom. It can only translate, so it has f=3f = 3 degrees of freedom and no rotational or vibrational contribution.

    By the law of equipartition of energy, each degree of freedom carries an average energy 12kBT\tfrac{1}{2} k_B T per molecule, so the average energy of one molecule is

    E‾=32kBT\overline{E} = \dfrac{3}{2} k_B T

    For one mole there are NAN_A molecules, so the internal energy of one mole is

    U=NA×32kBT=32(NAkB)T=32RTU = N_A \times \dfrac{3}{2} k_B T = \dfrac{3}{2} \left(N_A k_B\right) T = \dfrac{3}{2} RT

    using NAkB=RN_A k_B = R.

    Heating at constant volume

    At constant volume the gas does no work, dW=P dV=0dW = P\,dV = 0, so the first law dQ=dU+dWdQ = dU + dW reduces to

    dQ=dUdQ = dU

    Therefore

    CV=(dQdT)V=dUdT=ddT(32RT)C_V = \left(\dfrac{dQ}{dT}\right)_{V} = \dfrac{dU}{dT} = \dfrac{d}{dT}\left(\dfrac{3}{2} RT\right)

     CV=32R≈12.47 J mol−1 K−1 \boxed{\,C_V = \dfrac{3}{2} R \approx 12.47\ \text{J mol}^{-1}\,\text{K}^{-1}\,}

    Note

    The result is independent of temperature and the same for every monoatomic gas — helium, neon, argon — which is one of the successes of kinetic theory. With Mayer's relation it also gives CP=52RC_P = \tfrac{5}{2}R and γ=5/3≈1.67\gamma = 5/3 \approx 1.67.

  6. 6
    Relation between coefficients of absorption, reflection and transmission
    Derive the relation between the coefficient of absorption, coefficient of reflection and coefficient of transmission.

    Asked in: February 2026 (Q.29(i), 4 marks)

    Show the derivation
    Step 1
    Coefficient of absorption: The ratio of amount of heat absorbed to total quantity of heat incident is called the coefficient of absorption.
    Step 2
    Coefficient of reflection: The ratio of amount of radiant energy reflected to the total energy incident is called the coefficient of reflection.
    Step 3
    Coefficient of transmission: The ratio of amount of radiant energy transmitted to total energy incident is called the coefficient of transmission.
  7. 7
    Kirchhoff's law of radiation (theoretical proof)
    Prove Kirchhoff's law of radiation theoretically.

    Asked in: July 2017

    Show the derivation
    Theoretical proof:
    Step 1
    Consider an ordinary body A and a perfect blackbody B of identical geometric shapes placed in an enclosure. In thermal equilibrium, both bodies will be at same temperature as that of the enclosure.
    Step 2
    Let R = emissive power of body A, RBR_B = emissive power of blackbody B, a = coefficient of absorption of body A. Q = quantity of radiant heat incident on each body in unit time and QaQ_a = quantity of radiant heat absorbed by the body A,
    then Qa=aQQ_a=aQ.
    Step 3
    As the temperatures of the body A and blackbody B remain the same, both must emit the same amount as they absorb in unit time.
    ∴\therefore Quantity of radiant heat absorbed by body A = Quantity of heat emitted by body A
    ∴aQ=R....(1)\therefore aQ=R\qquad....(1)
    Step 4
    For the perfect blackbody B, Q=RB....(2)Q=R_B\qquad....(2)
    Step 5
    Dividing equation (1) by equation (2), we get, a=RRBa=\dfrac{R}{R_B}
    Step 6
    But by definition of emissivity, e=RRBe=\dfrac{R}{R_B}
    ∴a=e\therefore a=e