Laws of Thermodynamics – IIT‑JEE (Main + Advanced) Study Notes
Objective – Provide a concise, rigorously‑derived, exam‑oriented treatise on the four laws of thermodynamics with key formulae, subtle points, and a challenging practice problem.
Prerequisites – Familiarity with classical thermodynamics (state functions, processes, ideal‑gas equations), calculus, and basic statistical mechanics.
1. Zeroth Law of Thermodynamics
| Statement | Consequence |
|---|---|
| If two systems $A$ and $B$ are each in thermal equilibrium with a third system $C$, then $A$ and $B$ are in thermal equilibrium with one another. | Thermal equilibrium is an equivalence relation → Temperature is a state function. |
| Mathematical form – When $A \leftrightarrow C$ and $B \leftrightarrow C$ ⇒ $A \leftrightarrow B$. | → Defines a measurable quantity $T$ (temperature). |
| Measurement – Thermometer as a system $C$ in equilibrium with $A$ and $B$. | |
| Important point – Temperature is a property that can be used as an independent variable in thermodynamic equations. |
2. First Law of Thermodynamics (Conservation of Energy)
2.1 General Statement
For a closed system (no mass transfer), the change in internal energy $U$ equals the heat supplied $\delta Q$ minus the work done on the system $\delta W$:
[ \boxed{\displaystyle dU = \delta Q - \delta W} ]
- Sign convention:
- $+\delta Q$ – heat added to the system.
- $+\delta W$ – work done by the system (so $\delta W$ appears with a minus sign).
- Some texts adopt the opposite sign for work; be careful with the convention used in the exam.
2.2 Differential Forms for Simple Compressible Systems
For a simple compressible system (no chemical reactions, no electrical work), the only work is $p!dV$:
[ \boxed{\displaystyle dU = \delta Q - p,dV} ]
If heat transfer is reversible, $\delta Q = T,dS$:
[ \boxed{\displaystyle dU = T,dS - p,dV} \tag{1} ]
Equation (1) is the fundamental thermodynamic identity.
2.3 Internal Energy of an Ideal Gas
For an ideal gas, $U$ depends only on temperature:
[ U = n,u(T) \quad\Rightarrow\quad dU = n,C_V,dT ]
where $C_V$ is the molar heat capacity at constant volume.
2.4 Heat Capacities – $C_P$ vs. $C_V$
From the identity (1) and the ideal‑gas equation $pV = nRT$, we derive the heat capacity relation:
[ C_P - C_V = R \tag{2} ]
-
Derivation (brief):
[ \begin{aligned} dU &= nC_V dT \ \delta Q_{\text{rev}} &= T,dS = dU + p,dV = nC_V dT + p,dV \ \text{But}\quad dS &= \frac{nR}{T},dT + nR,\frac{dV}{V} \ \Rightarrow \delta Q_{\text{rev}} &= nC_P dT \end{aligned} ]
Equating the two expressions for $\delta Q_{\text{rev}}$ gives (2).
2.5 Adiabatic Processes
-
Reversible adiabatic ($\delta Q = 0$):
[ dU = -p,dV \quad \Rightarrow \quad nC_V dT = -p,dV ] Using $pV = nRT$, we get the adiabatic law:
[ TV^{\gamma -1} = \text{constant},\quad pV^\gamma = \text{constant} ] where $\displaystyle \gamma = \frac{C_P}{C_V}$.
-
Isentropic (= reversible adiabatic) → $\Delta S = 0$.