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PhysicsThermal PhysicsTop Priority (4-5 Qs in JEE)

Heat & Thermodynamics Formulas

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

10 Formulas•1,411+ Questions in Bank

Concept Diagram

Carnot Engine P-V Cycle
PV1 (T₁)2 (T₁)3 (T₂)4 (T₂)Isothermal Exp (Q₁ in)Adiabatic ExpIsothermal Comp (Q₂ out)Work (W)η = 1 - T₂/T₁ = W/Q₁
Reversible thermodynamic cycle: 1→2 Isothermal expansion, 2→3 Adiabatic expansion, 3→4 Isothermal compression, 4→1 Adiabatic compression.

Core Formulas & Mathematical Identities

10
01

First Law of Thermodynamics (Physics)

ΔQ=ΔU+W,ΔU=nCvΔT\Delta Q = \Delta U + W, \quad \Delta U = n C_v \Delta T
Diagram Reference:First Law of Thermodynamics (Physics)

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

Examiner Tip: Internal energy ΔU depends ONLY on initial and final temperatures, independent of process path.
02

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.

Examiner Tip: In adiabatic process, PV^γ = constant and TV^(γ-1) = constant.
03

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).

Examiner Tip: At high temperatures, vibrational degrees of freedom add 2 extra degrees.
04

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.

Examiner Tip: COP of refrigerator relates to engine efficiency: β = (1 - η) / η.
05

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).

Examiner Tip: Thermal resistance R_th = L / (kA). Series thermal resistance R_s = R₁ + R₂. Parallel: 1/R_p = 1/R₁ + 1/R₂.
06

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).

Examiner Tip: 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.
07

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.

Examiner Tip: 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.
08

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.

Examiner Tip: Total translational kinetic energy of n moles of any gas is E = (3/2) nRT = (3/2) PV.
09

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.

Examiner Tip: Mean free path λ is directly proportional to temperature T and inversely proportional to pressure P.
10

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.

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

Top Examiner Traps & Pitfalls

Frequently tested misconceptions in Heat & Thermodynamics

  • •Sign convention for work done: In Physics, W = ∫ P dV (work done BY system is positive). In Chemistry, W = -∫ P_ext dV.
  • •Temperatures in Carnot efficiency MUST be converted to absolute Kelvin (K), never Celsius.
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