Table of Contents
How to Use This Page
Read each concept carefully, then check the formula, common mistake, and exam tip before moving to the next. This page completely covers Thermodynamics for CBSE Class 11 Physics, bridging the microscopic world of heat with macroscopic mechanical work.
Key Concepts
Thermodynamics
The rules that run the universe
1
Zeroth Law of Thermodynamics
Definition
If 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 defines Temperature.
$$T_A = T_C \text{ and } T_B = T_C \implies T_A = T_B$$
2
First Law of Thermodynamics
Formula
The law of conservation of energy. The heat supplied to a system ($\Delta Q$) is equal to the increase in its internal energy ($\Delta U$) plus the work done by the system on its surroundings ($\Delta W$).
$$\Delta Q = \Delta U + \Delta W$$
3
Work Done by a Gas
Formula
When a gas expands, it pushes against the surroundings, doing positive work. On a P-V (Pressure-Volume) diagram, work done is the Area under the curve.
$$W = \int_{V_i}^{V_f} P \, dV$$
4
Specific Heats of Gases ($C_p$ and $C_v$)
Formula
Gases have two principal specific heats: at constant volume ($C_v$) and at constant pressure ($C_p$). $C_p$ is always greater than $C_v$ because, at constant pressure, extra heat is needed to do expansion work.
$$C_p – C_v = R \text{ (Mayer’s Relation)}$$
5
Isothermal Process
Formula
A process occurring at constant temperature ($T = \text{const}$). The system must exchange heat with the surroundings slowly. Since $\Delta T = 0$, the change in internal energy $\Delta U = 0$.
$$PV = \text{constant} \quad | \quad W = nRT \ln\left(\frac{V_f}{V_i}\right)$$
6
Adiabatic Process
Formula
A rapid process where NO heat is exchanged with the surroundings ($\Delta Q = 0$). The system is perfectly insulated. Expansion causes cooling; compression causes heating.
$$PV^\gamma = \text{constant} \quad | \quad W = \frac{P_1 V_1 – P_2 V_2}{\gamma – 1}$$
$$\text{Where } \gamma = \frac{C_p}{C_v}$$
7
Isochoric & Isobaric Processes
Case Formulas
Isochoric: Volume is constant ($\Delta V = 0$). Therefore, Work Done $W = 0$, and $\Delta Q = \Delta U$.
Isobaric: Pressure is constant. Work Done $W = P(V_f – V_i)$.
Isobaric: Pressure is constant. Work Done $W = P(V_f – V_i)$.
$$\text{Isochoric: } \Delta Q = n C_v \Delta T \quad | \quad \text{Isobaric: } \Delta Q = n C_p \Delta T$$
8
Second Law of Thermodynamics
Definition
Kelvin-Planck Statement: It is impossible to construct an engine that extracts heat from a hot reservoir and converts it completely into work (100% efficiency is impossible).
Clausius Statement: Heat cannot spontaneously flow from a colder body to a hotter body.
Clausius Statement: Heat cannot spontaneously flow from a colder body to a hotter body.
$$\eta < 1 \text{ (Always)}$$
9
Heat Engines & Efficiency ($\eta$)
Formula
A device that takes heat ($Q_1$) from a hot reservoir, does some work ($W$), and dumps the remaining heat ($Q_2$) into a cold sink. Efficiency is Output Work / Input Heat.
$$\eta = \frac{W}{Q_1} = \frac{Q_1 – Q_2}{Q_1} = 1 – \frac{Q_2}{Q_1}$$
10
Carnot Engine
Formula
An ideal, reversible heat engine with the absolute maximum possible efficiency operating between two temperatures ($T_1$ for Source, $T_2$ for Sink).
$$\eta_{\text{carnot}} = 1 – \frac{T_2}{T_1}$$
Concept Deep Dive
01
The Great Sign Convention War
Physics vs. ChemistryIf you take both Physics and Chemistry, Thermodynamics can be incredibly confusing because the sign conventions are reversed!
In Physics: We study engines. We *want* the gas to do work for us. So, Work done BY the gas (expansion) is Positive ($+W$). Work done ON the gas (compression) is Negative. The First Law is written as $\Delta Q = \Delta U + \Delta W$.
In Chemistry: They study the system itself. If the gas expands, it loses energy to do work, so Work done BY the gas is Negative ($-W$). Their First Law is written as $\Delta U = q + w$.
Stay entirely in the “Physics Mindset” for your Physics exam: Expansion = $+W$, Compression = $-W$.
In Physics: We study engines. We *want* the gas to do work for us. So, Work done BY the gas (expansion) is Positive ($+W$). Work done ON the gas (compression) is Negative. The First Law is written as $\Delta Q = \Delta U + \Delta W$.
In Chemistry: They study the system itself. If the gas expands, it loses energy to do work, so Work done BY the gas is Negative ($-W$). Their First Law is written as $\Delta U = q + w$.
Stay entirely in the “Physics Mindset” for your Physics exam: Expansion = $+W$, Compression = $-W$.
02
Reading Cyclic P-V Diagrams
Clockwise vs Counter-ClockwiseIn a cyclic process, the gas undergoes a series of changes and returns to its initial state. The net work done is the Area enclosed inside the loop.
If the cycle runs Clockwise, the expansion phase happens at a higher pressure than the compression phase. Thus, the net work done BY the gas is Positive (it acts as a Heat Engine).
If the cycle runs Counter-Clockwise, compression happens at a higher pressure. The net work done is Negative (it acts as a Refrigerator or Heat Pump).
If the cycle runs Clockwise, the expansion phase happens at a higher pressure than the compression phase. Thus, the net work done BY the gas is Positive (it acts as a Heat Engine).
If the cycle runs Counter-Clockwise, compression happens at a higher pressure. The net work done is Negative (it acts as a Refrigerator or Heat Pump).
Compare & Contrast
✗ Isothermal Expansion
- Temperature remains constant ($\Delta T = 0$).
- Requires highly conducting walls and a very slow process.
- Since $\Delta U = 0$, all heat added converts to work ($\Delta Q = \Delta W$).
- Equation: $PV = \text{const}$.
- Slope on P-V diagram is $-P/V$ (flatter).
✓ Adiabatic Expansion
- Heat exchange is zero ($\Delta Q = 0$).
- Requires perfectly insulating walls and a very fast process.
- Gas does work at the expense of its internal energy, causing temperature to drop ($\Delta W = -\Delta U$).
- Equation: $PV^\gamma = \text{const}$.
- Slope on P-V diagram is $-\gamma(P/V)$ (steeper).
Common Mistakes to Avoid
Mistake 1
Using Celsius in Carnot Efficiency: The formula $\eta = 1 – (T_2/T_1)$ ONLY works if temperatures are in Kelvin. If a question gives Source at $100^\circ\text{C}$ and Sink at $0^\circ\text{C}$, the efficiency is NOT $1 – (0/100) = 100\%$. It is $1 – (273/373) = 26.8\%$.
Mistake 2
Confusing Heat ($Q$) with Internal Energy ($U$): Heat is energy in transit due to a temperature difference. Internal energy is the energy stored in the gas due to molecular motion. A system does not “contain” heat; it contains internal energy.
Mistake 3
Assuming $\Delta Q = 0$ means $T$ is constant: This is entirely wrong! In an adiabatic process ($\Delta Q = 0$), expanding the gas forces it to do work using its own internal energy, which drastically drops its temperature. The sudden chilling of air when you release it from a highly pressurized tire is an adiabatic expansion.
Exam Tips
Tip 1
Finding $\Delta U$ from any path: Because Internal Energy is a state function, $\Delta U$ between state A and state B is exactly the same regardless of what weird path the PV diagram shows. If you know the temperatures, $\Delta U = n C_v (T_B – T_A)$ always works, even if the process is isobaric or random!
Tip 2
Refrigerator CoP: For a refrigerator, we don’t use “efficiency”, we use Coefficient of Performance ($\beta$ or $\alpha$). $\beta = \frac{\text{Heat extracted from cold}}{\text{Work input}} = \frac{Q_2}{W} = \frac{T_2}{T_1 – T_2}$. Unlike efficiency, $\beta$ can be greater than $1$.
Expected Exam Questions
SQ
Board Pattern Questions
Class 11 · Thermodynamics · CBSE Exam1
A Carnot engine operates between a source at $500 \text{ K}$ and a sink at $300 \text{ K}$. If it absorbs $1000 \text{ J}$ of heat per cycle from the source, calculate the work done per cycle and the heat rejected to the sink. [3 marks]
Answer
Work = $400 \text{ J}$, Heat Rejected = $600 \text{ J}$
📝
Explanation
1. Calculate Carnot Efficiency: $\eta = 1 – \frac{T_2}{T_1} = 1 – \frac{300}{500} = 1 – 0.6 = 0.4$ (or $40\%$).
2. Work Done: $\eta = \frac{W}{Q_1} \implies 0.4 = \frac{W}{1000} \implies W = 400 \text{ J}$.
3. Heat Rejected ($Q_2$): By conservation of energy, $Q_1 = W + Q_2$.
$1000 = 400 + Q_2 \implies Q_2 = 600 \text{ J}$.
2
A thermodynamic system undergoes a process in which its internal energy decreases by $500 \text{ J}$ and it does $200 \text{ J}$ of work on its surroundings. Calculate the heat transferred to or from the system. [2 marks]
Answer
$\Delta Q = -300 \text{ J}$ (Heat is lost)
📝
Explanation
Use the First Law of Thermodynamics: $\Delta Q = \Delta U + \Delta W$.
Apply proper sign conventions:
Internal energy decreases $\implies \Delta U = -500 \text{ J}$.
System does work on surroundings (expansion) $\implies \Delta W = +200 \text{ J}$.
$\Delta Q = (-500) + (200) = -300 \text{ J}$.
The negative sign indicates that $300 \text{ J}$ of heat was rejected/lost from the system to the surroundings.
3
Why is $C_p$ strictly greater than $C_v$ for a gas? [1 mark]
Answer
Constant pressure requires extra heat to do expansion work.
📝
Explanation
When heat is supplied at constant volume ($C_v$), all the heat goes directly into increasing the internal energy (raising the temperature). When heat is supplied at constant pressure ($C_p$), the gas expands. Some of the supplied heat is used to do mechanical work against the surroundings ($P\Delta V$), and only the remaining heat raises the temperature. Hence, more total heat is required at constant pressure.
Concept Map
Thermodynamics connects to →
Heat Physics
Thermal Properties (Calorimetry bridge)
Kinetic Theory of Gases (Microscopic derivation of U)
Chemistry (Physical Chemistry crossover)
Mechanics (Work done by force)
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