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Physics Thermodynamics

Thermodynamics Formula Sheet

All key Thermodynamics formulas for JEE Main & Advanced: laws of thermodynamics, processes, heat capacities, engines, Carnot cycle, and entropy quick revision.

8 min read Updated Jun 2026 #formula sheet#quick revision#jee-main

Everything you need for last-minute Thermodynamics revision: every formula, constant, and must-know result from the chapter, grouped by sub-topic and ready to scan before the exam.

Temperature and Scales

QuantityFormulaNotes
Scale relation$\dfrac{C}{100} = \dfrac{F-32}{180} = \dfrac{K-273}{100}$Master conversion equation
Celsius from Fahrenheit$C = \dfrac{5}{9}(F-32)$
Fahrenheit from Celsius$F = \dfrac{9}{5}C + 32$
Kelvin from Celsius$K = C + 273$Use $273.15$ for ice-point precision
Temperature difference$\Delta C = \Delta K$NOT equal to $\Delta F$
Fahrenheit difference$\Delta F = \dfrac{9}{5}\Delta C = \dfrac{9}{5}\Delta K$
High-Yield Reminder
Always convert temperature to Kelvin before plugging into any thermodynamic formula. Use $\Delta C = \Delta K$ for temperature changes (no $+273$), but $K = C + 273$ for absolute temperature. The C and F scales read equal only at $-40$.

Zeroth Law: If A is in thermal equilibrium with C, and B is in thermal equilibrium with C, then A and B are in thermal equilibrium. This defines temperature and makes thermometers possible.

First Law of Thermodynamics

$$\boxed{\Delta Q = \Delta U + \Delta W \quad\Longleftrightarrow\quad \Delta U = Q - W}$$
QuantityFormulaNotes
Work done by gas$W = \displaystyle\int_{V_1}^{V_2} P\,dV$Area under P-V curve
Internal energy change$\Delta U = nC_V\Delta T$True for all ideal-gas processes
Internal energy (ideal gas)$U = \dfrac{f}{2}nRT = nC_VT$$f$ = degrees of freedom
Elementary work$dW = P\,dV$
Constant-pressure work$W = P\Delta V$

Sign convention: $Q > 0$ heat absorbed; $W > 0$ work done by system (expansion); $\Delta U > 0$ temperature rises.

State vs Path
$\Delta U$ is a state function (path-independent) and equals $nC_V\Delta T$ for an ideal gas in every process. $Q$ and $W$ are path-dependent. For a cyclic process $\Delta U = 0$, so $Q = W$.

Free expansion (into vacuum): $Q = 0$, $W = 0$, so $\Delta U = 0$ and $\Delta T = 0$ for an ideal gas.

Heat, Heat Capacity and Specific Heat

QuantityFormulaNotes
Heat capacity$C = \dfrac{\Delta Q}{\Delta T}$Unit J/K
Specific heat$c = \dfrac{\Delta Q}{m\Delta T} = \dfrac{C}{m}$Unit J/(kg·K)
Molar heat capacity$C_m = \dfrac{\Delta Q}{n\Delta T} = \dfrac{C}{n}$Unit J/(mol·K)
Heat (solid/liquid)$Q = mc\Delta T$Use mass, not moles
Heat–work conversion$1\ \text{cal} = 4.186\ \text{J}$

Reference values: specific heat of water $c = 4186\ \text{J/(kg·K)}$ (highest among common substances).

Molar Heat Capacities of Gases

$$\boxed{C_P - C_V = R \quad\text{(Mayer's relation)}}$$
QuantityFormulaNotes
At constant volume$C_V = \dfrac{f}{2}R = \dfrac{R}{\gamma - 1}$$Q = nC_V\Delta T$
At constant pressure$C_P = \left(\dfrac{f}{2}+1\right)R = \dfrac{\gamma R}{\gamma - 1}$$Q = nC_P\Delta T$
Adiabatic exponent$\gamma = \dfrac{C_P}{C_V} = 1 + \dfrac{2}{f}$Dimensionless, $\gamma > 1$

Gas-Type Reference Table

Gas Type$f$$C_V$$C_P$$\gamma$
Monatomic (He, Ar)3$\tfrac{3}{2}R$$\tfrac{5}{2}R$$\tfrac{5}{3} \approx 1.67$
Diatomic (H₂, N₂, O₂)5$\tfrac{5}{2}R$$\tfrac{7}{2}R$$\tfrac{7}{5} = 1.4$
Polyatomic (CO₂, NH₃)6+$3R$$4R$$\approx 1.33$
Special resultFormulaNotes
Gas mixture $C_V$$C_{V,\text{mix}} = \dfrac{n_1C_{V1} + n_2C_{V2}}{n_1 + n_2}$Mole-weighted average
Dulong–Petit (solids)$C_V \approx 3R \approx 25\ \text{J/(mol·K)}$Room temperature
Polytropic heat capacity$C_n = C_V\dfrac{\gamma - n}{1 - n}$For $PV^n = \text{const}$
3-5-7 Mnemonic
Mono 3-5, Di 5-7 (numerators of $C_V$, $C_P$ with $R/2$). Difference is always $2 \times \tfrac{R}{2} = R$, confirming Mayer’s relation. Polytropic special cases: $n=0 \Rightarrow C_n = C_P$; $n=1 \Rightarrow C_n = \infty$; $n=\gamma \Rightarrow C_n = 0$; $n\to\infty \Rightarrow C_n = C_V$.

The Four Thermodynamic Processes

ProcessConstantRelation$W$$Q$$\Delta U$$C$
Isothermal$T$$PV = \text{const}$$nRT\ln\dfrac{V_2}{V_1}$$W$$0$$\infty$
Adiabatic$Q = 0$$PV^\gamma = \text{const}$$\dfrac{nR(T_1 - T_2)}{\gamma - 1}$$0$$-W$$0$
Isobaric$P$$\dfrac{V}{T} = \text{const}$$P\Delta V = nR\Delta T$$nC_P\Delta T$$nC_V\Delta T$$C_P$
Isochoric$V$$\dfrac{P}{T} = \text{const}$$0$$nC_V\Delta T$$nC_V\Delta T$$C_V$

Isothermal Process

$$P_1V_1 = P_2V_2 \qquad W = Q = nRT\ln\frac{V_2}{V_1} = nRT\ln\frac{P_1}{P_2}$$

Alternative work forms: $W = P_1V_1\ln\dfrac{V_2}{V_1} = P_2V_2\ln\dfrac{V_2}{V_1}$.

Adiabatic Process

$$\boxed{PV^\gamma = \text{const}, \qquad TV^{\gamma-1} = \text{const}, \qquad TP^{(\gamma-1)/\gamma} = \text{const}}$$
QuantityFormula
Between states$P_1V_1^\gamma = P_2V_2^\gamma$; $T_1V_1^{\gamma-1} = T_2V_2^{\gamma-1}$
Temperature–pressure$\dfrac{T_1}{T_2} = \left(\dfrac{P_1}{P_2}\right)^{(\gamma-1)/\gamma}$
Work (temperatures)$W = \dfrac{nR(T_1 - T_2)}{\gamma - 1}$
Work (P, V)$W = \dfrac{P_1V_1 - P_2V_2}{\gamma - 1}$
Work (from $\Delta U$)$W = -\Delta U = nC_V(T_1 - T_2)$

Curve Slopes on a P-V Diagram

ProcessSlope $\dfrac{dP}{dV}$
Isothermal$-P/V$
Adiabatic$-\gamma P/V$ (steeper)
Isobaric$0$ (horizontal)
Isochoric$\infty$ (vertical)

Steepness order: Isochoric $>$ Adiabatic $>$ Isothermal $>$ Isobaric. The adiabatic slope is $\gamma$ times the isothermal slope at any common point.

Work Comparison

For the same expansion from $V_1$ to $V_2$: $W_{\text{isobaric}} > W_{\text{isothermal}} > W_{\text{adiabatic}}$, because a higher curve encloses more area.

Match the heat-capacity subscript to the constant: V constant uses $C_V$, P constant uses $C_P$.

P-V Diagrams and Cyclic Processes

QuantityFormulaNotes
Work by gas$W = \displaystyle\int_{V_1}^{V_2} P\,dV$Area under curve
Cyclic net work$W_{\text{net}} = \displaystyle\oint P\,dV$ = area enclosed$\Delta U = 0$, so $Q_{\text{net}} = W_{\text{net}}$
Rectangular cycle$W = \Delta P \times \Delta V$Area of rectangle
Triangular cycle$W = \tfrac{1}{2} \times \text{base} \times \text{height}$

Cycle direction: Clockwise loop $\Rightarrow W > 0$ (heat engine); counter-clockwise loop $\Rightarrow W < 0$ (refrigerator/heat pump).

graph LR
    A[Cyclic Process] --> B["Clockwise loop
W > 0
Heat Engine"]
    A --> C["Counter-clockwise loop
W < 0
Refrigerator / Heat Pump"]

Second Law of Thermodynamics

  • Kelvin–Planck: No engine working in a cycle can convert heat from a single reservoir entirely into work. Consequently $\eta = \dfrac{W}{Q_H} < 1$ always.
  • Clausius: Heat cannot spontaneously flow from a colder to a hotter body without external work input.

Carnot’s theorem: No engine between two temperatures can exceed the Carnot (reversible) efficiency.

$$\eta_{\text{reversible}} = \eta_{\text{Carnot}} = 1 - \frac{T_C}{T_H} \qquad \eta_{\text{irreversible}} < \eta_{\text{Carnot}}$$

Heat Engines, Carnot Cycle and Refrigerators

$$\boxed{\eta = \frac{W}{Q_H} = \frac{Q_H - Q_C}{Q_H} = 1 - \frac{Q_C}{Q_H}}$$$$\boxed{\eta_{\text{Carnot}} = 1 - \frac{T_C}{T_H}}$$
Device / CycleFormulaNotes
First law (engine)$Q_H = W + Q_C$Per cycle, $\Delta U = 0$
Carnot heat exchange$\dfrac{Q_C}{Q_H} = \dfrac{T_C}{T_H}$Depends only on temperatures
Otto cycle$\eta_{\text{Otto}} = 1 - \dfrac{1}{r^{\gamma-1}}$$r$ = compression ratio
Diesel cycle$\eta_{\text{Diesel}} = 1 - \dfrac{1}{r^{\gamma-1}}\cdot\dfrac{\rho^\gamma - 1}{\gamma(\rho - 1)}$$\rho$ = cutoff ratio
Stirling cycle$\eta_{\text{Stirling}} = \eta_{\text{Carnot}}$Matches Carnot
Brayton cycle$\eta_{\text{Brayton}} = 1 - \dfrac{1}{r_p^{(\gamma-1)/\gamma}}$$r_p$ = pressure ratio
Refrigerator COP$\text{COP}_R = \dfrac{Q_C}{W} = \dfrac{Q_C}{Q_H - Q_C}$Carnot: $\dfrac{T_C}{T_H - T_C}$
Heat pump COP$\text{COP}_{HP} = \dfrac{Q_H}{W} = \dfrac{Q_H}{Q_H - Q_C}$Carnot: $\dfrac{T_H}{T_H - T_C}$
COP relation$\text{COP}_{HP} = \text{COP}_R + 1$Always

Carnot cycle: isothermal expansion at $T_H$ → adiabatic expansion → isothermal compression at $T_C$ → adiabatic compression. Carnot heat absorbed $Q_H = nRT_H\ln\dfrac{V_B}{V_A}$, rejected $Q_C = nRT_C\ln\dfrac{V_C}{V_D}$.

Series enginesResult
Two Carnot engines in seriesOverall $\eta$ = single Carnot from $T_{\text{highest}}$ to $T_{\text{lowest}}$
$n$ engines, equal efficiencyTemperatures in GP: $T_i = T_H\left(\dfrac{T_C}{T_H}\right)^{i/n}$
Two-stage equal $\eta$Intermediate $T = \sqrt{T_H T_C}$ (geometric mean)
Engine vs Refrigerator
For engines $\eta < 1$ always; for refrigerators and heat pumps the COP can exceed 1 because they move existing heat rather than create work. Increase efficiency by widening the gap: $\eta = \dfrac{T_H - T_C}{T_H}$. Raising $T_H$ and $T_C$ by the same $\Delta T$ decreases efficiency (same numerator, larger denominator).

Entropy

$$\boxed{dS = \frac{dQ_{\text{rev}}}{T}, \qquad \Delta S = \int \frac{dQ_{\text{rev}}}{T}}$$

Second Law (entropy form): for an isolated system $\Delta S_{\text{universe}} \geq 0$ — equality for reversible, strict inequality ($>0$) for irreversible processes. Units: J/K.

ProcessEntropy change $\Delta S$
Isothermal$\dfrac{Q}{T} = nR\ln\dfrac{V_2}{V_1} = nR\ln\dfrac{P_1}{P_2}$
Isochoric$nC_V\ln\dfrac{T_2}{T_1}$
Isobaric$nC_P\ln\dfrac{T_2}{T_1}$
Reversible adiabatic$0$ (isentropic)
Phase change$\dfrac{Q}{T} = \dfrac{mL}{T}$
General ideal gas$nC_V\ln\dfrac{T_2}{T_1} + nR\ln\dfrac{V_2}{V_1}$
General (alt. forms)$nC_P\ln\dfrac{T_2}{T_1} - nR\ln\dfrac{P_2}{P_1} = nC_V\ln\dfrac{P_2}{P_1} + nC_P\ln\dfrac{V_2}{V_1}$
Special caseEntropy result
Boltzmann (microscopic)$S = k_B\ln\Omega$, $\ k_B = 1.38\times10^{-23}\ \text{J/K}$
Free expansion$\Delta S_{\text{gas}} = nR\ln\dfrac{V_2}{V_1} > 0$ (even though $Q=0$)
Heat across finite $\Delta T$$\Delta S = Q\left(\dfrac{1}{T_C} - \dfrac{1}{T_H}\right) > 0$
Mixing of ideal gases$\Delta S_{\text{mix}} = -nR(x_1\ln x_1 + x_2\ln x_2) > 0$
Two equal bodies mixing$\Delta S = C\ln\dfrac{(T_1+T_2)^2}{4T_1T_2} > 0$ for $T_1 \neq T_2$
Clausius inequality$\displaystyle\oint \dfrac{dQ}{T} \leq 0$ ($=0$ reversible, $<0$ irreversible)

Carnot cycle entropy: $\Delta S_{\text{gas}} = 0$ and $\Delta S_{\text{universe}} = 0$ (reversible). On a T-S diagram the Carnot cycle is a rectangle with area $W = (T_H - T_C)(S_2 - S_1)$.

Entropy Traps

A reversible adiabatic has $\Delta S = 0$, but an irreversible adiabatic (free expansion) has $\Delta S > 0$. Local entropy can decrease (e.g. freezing water), but $\Delta S_{\text{universe}}$ never does — always check system + surroundings, and always use Kelvin.

Useful Constants

ConstantValue
Gas constant$R = 8.314\ \text{J/(mol·K)}$
Boltzmann constant$k_B = 1.38\times10^{-23}\ \text{J/K}$
Mechanical equivalent of heat$1\ \text{cal} = 4.186\ \text{J}$
Ice point$273.15\ \text{K} \approx 273\ \text{K}$
Specific heat of water$4186\ \text{J/(kg·K)}$

Keep going in this chapter

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