Thermodynamics — Physics Class 11 Notes (CBSE & HBSE)
Free NCERT Physics notes for Thermodynamics (Class 11) on Siksha Sarovar, aligned to CBSE and Haryana Board (HBSE). This chapter is broken into 3 topics with clear explanations, formulas, solved examples and board-pattern practice — free to read, no sign-up required.
Board exam focus — Thermodynamics (CBSE & HBSE)
Thermodynamics studies energy transformations between heat and work in macroscopic systems. This chapter introduces thermal equilibrium and the zeroth law, internal energy, the first law (energy conservation for heat and work), the main thermodynamic processes (isothermal, adiabatic, isobaric, isochoric), and the second law with heat engines, the Carnot cycle and refrigerators. CBSE and HBSE emphasise sign conventions, P-V work and efficiency calculations.
Zeroth Law, Internal Energy and First Law
Thermal Equilibrium and the Zeroth Law
Two systems are in thermal equilibrium when there is no net heat flow between them; they then share a common temperature. The Zeroth law states:
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 each other.
This law justifies temperature as a measurable property and the use of thermometers.
Internal Energy
The internal energy (U) of a system is the sum of the kinetic and potential energies of its molecules. It is a state function — it depends only on the state (P, V, T), not on the path taken to reach it. For an ideal gas, U depends only on temperature.
First Law of Thermodynamics
The first law is the law of conservation of energy applied to heat:
$$\Delta Q = \Delta U + \Delta W$$
where $\Delta Q$ is heat supplied to the system, $\Delta W = P\,\Delta V$ is work done by the system, and $\Delta U$ is the change in internal energy.
Sign convention (CBSE/HBSE):
| Quantity | Positive when |
|---|---|
| ΔQ | Heat is added to the system |
| ΔW | Work is done by the system (expansion) |
| ΔU | Internal energy increases (temperature rises) |
Trap: ΔU and ΔQ are path independent? Only ΔU is a state function; both ΔQ and ΔW depend on the path. Their difference ΔQ − ΔW = ΔU is path independent.
Thermodynamic Processes
Four Key Processes
| Process | Condition | First law result |
|---|---|---|
| Isothermal | T constant, ΔU = 0 | ΔQ = ΔW |
| Adiabatic | ΔQ = 0 | ΔW = −ΔU |
| Isobaric | P constant | ΔQ = ΔU + PΔV |
| Isochoric | V constant, ΔW = 0 | ΔQ = ΔU |
Isothermal Process
The gas is kept at constant temperature (slow process, in contact with a reservoir). Since $U$ depends only on $T$, $\Delta U = 0$. Work done:
$$W = nRT \ln\!\left(\frac{V_2}{V_1}\right)$$
The relation $PV = \text{constant}$ holds (Boyle's law).
Adiabatic Process
No heat exchange ($\Delta Q = 0$); the process is fast or well insulated. Here $PV^\gamma = \text{constant}$, where $\gamma = C_P/C_V$. Work done:
$$W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1} = \frac{nR(T_1 - T_2)}{\gamma - 1}$$
Trap: In adiabatic expansion the gas does work at the cost of its internal energy, so the gas cools. In adiabatic compression the gas heats up — this is why a bicycle pump warms.
Isobaric and Isochoric
- Isobaric (constant P): $W = P\Delta V = nR\Delta T$.
- Isochoric (constant V): no work is done, all heat changes internal energy.
Work done = area under the P-V curve. A cyclic process returns to its start, so ΔU = 0 over a complete cycle and the net heat absorbed equals the net work done (area enclosed).
Second Law, Heat Engines and Carnot Cycle
Second Law of Thermodynamics
The first law allows energy conservation but says nothing about direction. The second law fixes direction:
- Kelvin-Planck statement: No process is possible whose sole result is the absorption of heat from a reservoir and its complete conversion into work.
- Clausius statement: No process is possible whose sole result is the transfer of heat from a colder to a hotter body.
Heat Engine
A heat engine absorbs heat $Q_1$ from a hot source, does work $W$, and rejects heat $Q_2$ to a cold sink. Its efficiency is:
$$\eta = \frac{W}{Q_1} = 1 - \frac{Q_2}{Q_1}$$
Carnot Engine
The Carnot cycle (two isothermals + two adiabatics) is the most efficient engine working between temperatures $T_1$ (source) and $T_2$ (sink):
$$\eta_{\text{Carnot}} = 1 - \frac{T_2}{T_1}$$
(temperatures in kelvin). No engine can exceed Carnot efficiency.
Trap: Efficiency is 100% only if $T_2 = 0$ K, which is unattainable. Efficiency improves more by lowering $T_2$ than by raising $T_1$ for the same change.
Refrigerator / Heat Pump
A refrigerator is a heat engine run in reverse: it extracts $Q_2$ from a cold body and rejects $Q_1$ to a hot body using work $W$. Its coefficient of performance:
$$\beta = \frac{Q_2}{W} = \frac{Q_2}{Q_1 - Q_2} = \frac{T_2}{T_1 - T_2}$$
Frequently asked questions
Are these Thermodynamics notes free?
Yes — the Thermodynamics notes for Physics (Class 11) on Siksha Sarovar are completely free to read, with no account required.
Do these notes follow CBSE and HBSE?
Yes. The Thermodynamics notes are NCERT-aligned and include guidance for both CBSE and Haryana Board (HBSE), with important questions and MCQs for revision.
What does the Thermodynamics chapter cover?
Concept explanations, key formulas and definitions, fully solved examples and board-pattern practice questions for Thermodynamics.