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Thermal Physics
PhysicsThermal PhysicsJEE Main & Advanced

Thermal Physics Formulas

Complete chapter-wise formula sheet for Thermal Physics (Physics) covering key equations, parameters, traps, and concept diagrams for JEE Main & Advanced.

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Formulas Reference (Thermal Physics)

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Thermal PhysicsTop Priority (4-5 Qs in JEE)1,411 PYQs

Heat & Thermodynamics

First law of thermodynamics, work done in isobaric/isothermal/adiabatic processes, Carnot engine, and heat transfer laws.

Chapter

First Law of Thermodynamics (Physics)

ΔQ=ΔU+W,ΔU=nCvΔT\Delta Q = \Delta U + W, \quad \Delta U = n C_v \Delta T

Energy conservation: heat supplied ΔQ equals increase in internal energy plus work done by the system.

Internal energy ΔU depends ONLY on initial and final temperatures, independent of process path.

Work Done in Thermodynamic Processes

Wisochoric=0,Wisobaric=PΔV=nRΔT,Wisothermal=nRTln⁡V2V1,Wadiabatic=P1V1−P2V2γ−1W_{\text{isochoric}} = 0, \quad W_{\text{isobaric}} = P\Delta V = nR\Delta T, \quad W_{\text{isothermal}} = nRT\ln\frac{V_2}{V_1}, \quad W_{\text{adiabatic}} = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}

Formulas for area under P-V indicator diagram.

In adiabatic process, PV^γ = constant and TV^(γ-1) = constant.

Molar Heat Capacities & Mayer's Relation

Cp−Cv=R,γ=CpCv=1+2f,Cv=f2RC_p - C_v = R, \quad \gamma = \frac{C_p}{C_v} = 1 + \frac{2}{f}, \quad C_v = \frac{f}{2}R

Degrees of freedom f: Monoatomic = 3 (γ = 5/3), Diatomic = 5 (γ = 7/5), Polyatomic non-linear = 6 (γ = 4/3).

At high temperatures, vibrational degrees of freedom add 2 extra degrees.

Carnot Engine Efficiency & Refrigerator COP

η=1−TCTH=WQH,βrefrigerator=QCW=TCTH−TC\eta = 1 - \frac{T_C}{T_H} = \frac{W}{Q_H}, \quad \beta_{\text{refrigerator}} = \frac{Q_C}{W} = \frac{T_C}{T_H - T_C}

Maximum theoretical efficiency of reversible heat engine operating between reservoirs T_H and T_C.

COP of refrigerator relates to engine efficiency: β = (1 - η) / η.

Heat Transfer Laws (Conduction & Radiation)

dQdt=kA(T1−T2)L,Pradiation=σeA(T4−T04),σ=5.67×10−8 W m−2K−4\frac{dQ}{dt} = \frac{kA(T_1 - T_2)}{L}, \quad P_{\text{radiation}} = \sigma e A (T^4 - T_0^4), \quad \sigma = 5.67 \times 10^{-8} \text{ W m}^{-2}\text{K}^{-4}

Fourier conduction law, Stefan-Boltzmann radiation law for net radiation from body at T in surrounding T₀. ε = emissivity (0 ≤ ε ≤ 1; black body ε = 1).

Thermal resistance R_th = L / (kA). Series thermal resistance R_s = R₁ + R₂. Parallel: 1/R_p = 1/R₁ + 1/R₂.

Wien's Displacement Law & Black Body Radiation

λmT=b=2.898×10−3 m⋅K,Eλ∝1λ5⋅1ehc/λkBT−1  (Planck’s Law)\lambda_m T = b = 2.898 \times 10^{-3} \text{ m·K}, \quad E_\lambda \propto \frac{1}{\lambda^5} \cdot \frac{1}{e^{hc/\lambda k_B T} - 1} \; (\text{Planck's Law})

Wien's law: wavelength of peak emission intensity is inversely proportional to absolute temperature. As temperature increases, peak shifts to shorter wavelengths (blue shift).

Sun (T ≈ 5800 K): λ_m ≈ 500 nm (visible green). Human body (T ≈ 310 K): λ_m ≈ 9.4 μm (infrared). Used in NEET & JEE to identify stellar temperatures from emission spectra.

Van der Waals Gas & Critical Constants

(P+aVm2)(Vm−b)=RT,Tc=8a27Rb,Pc=a27b2,Vc=3b,Zc=PcVcRTc=38\left(P + \frac{a}{V_m^2}\right)(V_m - b) = RT, \quad T_c = \frac{8a}{27Rb}, \quad P_c = \frac{a}{27b^2}, \quad V_c = 3b, \quad Z_c = \frac{P_c V_c}{RT_c} = \frac{3}{8}

Van der Waals constants: a accounts for intermolecular attraction (pressure correction), b accounts for finite molecular volume (volume correction). Critical compressibility factor Z_c = 3/8 for any VdW gas.

Boyle temperature T_B = a/(Rb) where the gas behaves most ideally over a wide pressure range. T_B = (27/8) T_c. At T < T_c, the gas can be liquefied by pressure alone.

Kinetic Theory of Gases (KTG) Speeds

vrms=3RTM,vavg=8RTπM,vmp=2RTM,vrms:vavg:vmp=3:8/π:2v_{\text{rms}} = \sqrt{\frac{3RT}{M}}, \quad v_{\text{avg}} = \sqrt{\frac{8RT}{\pi M}}, \quad v_{\text{mp}} = \sqrt{\frac{2RT}{M}}, \quad v_{\text{rms}} : v_{\text{avg}} : v_{\text{mp}} = \sqrt{3} : \sqrt{8/\pi} : \sqrt{2}

Root-mean-square, average, and most probable speeds of ideal gas molecules of molar mass M.

Total translational kinetic energy of n moles of any gas is E = (3/2) nRT = (3/2) PV.

Mean Free Path & Pressure of Ideal Gas

λ=12πnd2=kBT2πd2P,P=13ρvrms2=23Ev\lambda = \frac{1}{\sqrt{2}\pi n d^2} = \frac{k_B T}{\sqrt{2}\pi d^2 P}, \quad P = \frac{1}{3}\rho v_{\text{rms}}^2 = \frac{2}{3}E_v

Average distance traveled by a molecule of diameter d between successive collisions in molecular density n.

Mean free path λ is directly proportional to temperature T and inversely proportional to pressure P.

Newton's Law of Cooling

dTdt=−K(T−T0)  ⟹  T1−T2t=K(T1+T22−T0)\frac{dT}{dt} = -K(T - T_0) \implies \frac{T_1 - T_2}{t} = K\left(\frac{T_1 + T_2}{2} - T_0\right)

Rate of loss of heat by a body is directly proportional to temperature difference between body and surroundings.

Valid strictly when (T - T₀) is small (typically ≤ 30 °C). Exact solution is T(t) = T₀ + (T_initial - T₀) e^(-Kt).