Thermodynamics Short Notes for Class 11, JEE & NEET

This page is about Thermodynamics in Chemistry. It is Chapter 5 of the current NCERT Class 11 Chemistry book (Chapter 6 in older books). Class 11 Physics also has a chapter with the same name, which covers heat engines, the laws in terms of gases and the Carnot cycle, so do not mix the two. These Thermodynamics Short Notes follow the NCERT chemistry chapter, and they link it to equilibrium and electrochemistry.

Thermodynamics Short Notes: Chapter Overview and Syllabus

Check the latest syllabi of CBSE, NTA (JEE Main) and NMC (NEET) for the exact lists.

💡 Expert Tip by Prateek Gupta, IIT Bombay: "In JEE Main, Thermodynamics questions are almost always formula-application based — not conceptual traps. If you know the formulas and the sign conventions cold, you pick up these marks in under 2 minutes per question. That is free time for harder questions."

Thermodynamics Short Notes 

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Basic Terms in Thermodynamics Short Notes

• System: the part of the universe that is under study. Surroundings: everything else. Boundary: the real or imaginary surface between them.
• A state function depends only on the initial and the final state: $U$, $H$, $S$ and $G$. A path function depends on the path: $q$ and $w$.
• Extensive properties depend on the amount of matter ($U$, $H$, $S$, $G$, $V$), and intensive properties do not ($T$, $P$, density).

First Law of Thermodynamics in Thermodynamics Short Notes

Zeroth law: if A and B are each in thermal equilibrium with C, then A and B are in thermal equilibrium with each other. This is the basis of temperature measurement.

First law: the internal energy of an isolated system is constant, and energy is conserved:

$$\Delta U = q + w$$

Sign convention (IUPAC): $q > 0$ when heat is absorbed by the system, and $w > 0$ when work is done on the system.

For an isothermal change of an ideal gas, $\Delta U = 0$, so $q = -w$. (1 L bar $= 100$ J.)

Enthalpy and Heat Capacity in Thermodynamics Short Notes

$$H = U + PV, \qquad \Delta H = \Delta U + \Delta n_g\,RT$$

$\Delta n_g$ is the change in the number of moles of gaseous substances only, which is the moles of gaseous products minus the moles of gaseous reactants. At constant pressure, $q_p = \Delta H$, and at constant volume, $q_V = \Delta U$.

Heat capacity: $q = C\,\Delta T = m\,c\,\Delta T$. For one mole of an ideal gas, $C_p - C_V = R$. A bomb calorimeter measures $\Delta U$ at constant volume, and a coffee-cup calorimeter measures $\Delta H$ at constant pressure.

Types of Enthalpy Change

Hess’s Law and Bond Enthalpy in Thermodynamics Short Notes

Hess’s law: the enthalpy change of a reaction is the same whether it takes place in one step or in several steps. To use it:

1. Write the target reaction.
2. Reverse a given reaction when its substance is on the wrong side, and change the sign of $\Delta H$.
3. Multiply a reaction by a factor, and multiply its $\Delta H$ by the same factor.
4. Add the reactions, and add the $\Delta H$ values.

$$\Delta H^\circ_{rxn} = \sum \Delta_f H^\circ(\text{products}) - \sum \Delta_f H^\circ(\text{reactants})$$

Bond enthalpy method (gaseous species only):

$$\Delta H^\circ_{rxn} = \sum (\text{bond enthalpies of bonds broken}) - \sum (\text{bond enthalpies of bonds formed})$$

Breaking bonds takes energy (positive), and forming bonds releases energy (negative).

Entropy, Second Law and Third Law in Thermodynamics Short Notes

Entropy $S$ measures the disorder of a system. For a reversible change at a temperature $T$:

$$\Delta S = \frac{q_{rev}}{T}$$

Second law: in a spontaneous process, the entropy of the universe increases:

$$\Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings} > 0$$

Third law: the entropy of a perfectly crystalline substance is zero at $0$ K.

Gibbs Energy and Spontaneity in Thermodynamics Short Notes

$$\Delta G = \Delta H - T\,\Delta S$$

A process is spontaneous at constant $T$ and $P$ if $\Delta G < 0$, it is at equilibrium if $\Delta G = 0$, and it is non-spontaneous if $\Delta G > 0$.

At equilibrium, $\Delta G = 0$, so $T_{eq} = \dfrac{\Delta H}{\Delta S}$.

Links to equilibrium and electrochemistry:

$$\Delta G^\circ = -RT\ln K = -2.303\,RT\log K, \qquad \Delta G = \Delta G^\circ + RT\ln Q, \qquad \Delta G^\circ = -nFE^\circ$$

The first two are used in eSaral’s Chemical Equilibrium Class 11 notes, and the third in the Electrochemistry Class 12 notes.

Key Formulas in Thermodynamics Short Notes

Solved Examples from Thermodynamics Short Notes

Example 1 (work of expansion): A gas expands from 2 L to 5 L against a constant external pressure of 1 bar. Find the work done.

$w = -P_{ext}\Delta V = -1 \times (5 - 2) = -3$ L bar $= -300$ J. The negative sign shows that the system does the work.

Example 2 (first law): A system absorbs 150 J of heat and does 60 J of work. Find $\Delta U$.

$q = +150$ J and $w = -60$ J, so $\Delta U = 150 - 60 = +90$ J.

Example 3 ($\Delta H$ and $\Delta U$): For $\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}$, $\Delta H = -92.4$ kJ at 298 K. Find $\Delta U$.

$\Delta n_g = 2 - 4 = -2$, so $\Delta U = \Delta H - \Delta n_g RT = -92.4 - (-2)(8.314 \times 10^{-3})(298) \approx -92.4 + 4.96 = -87.4$ kJ.

Example 4 (Hess’s law): Given $\mathrm{C(s) + O_2(g) \rightarrow CO_2(g)}$, $\Delta H = -393.5$ kJ, and $\mathrm{CO(g) + \tfrac{1}{2}O_2(g) \rightarrow CO_2(g)}$, $\Delta H = -283.0$ kJ, find $\Delta H$ for $\mathrm{C(s) + \tfrac{1}{2}O_2(g) \rightarrow CO(g)}$.

Reverse the second reaction (so $\Delta H = +283.0$ kJ), and add it to the first: $\Delta H = -393.5 + 283.0 = -110.5$ kJ.

Example 5 (bond enthalpy): Find $\Delta H$ for $\mathrm{H_2 + Cl_2 \rightarrow 2HCl}$, given the bond enthalpies H–H $= 436$, Cl–Cl $= 242$ and H–Cl $= 431$ kJ mol⁻¹.

$\Delta H = (436 + 242) - 2(431) = 678 - 862 = -184$ kJ.

Example 6 (spontaneity): A reaction has $\Delta H = +30$ kJ mol⁻¹ and $\Delta S = +100$ J K⁻¹ mol⁻¹. Above which temperature is it spontaneous?

Both are positive, so it is spontaneous at high temperature. $T_{eq} = \dfrac{30\,000}{100} = 300$ K, so it is spontaneous above 300 K.

Example 7 ($\Delta G^\circ$ and $K$): Find $\Delta G^\circ$ for a reaction with $K = 10$ at 300 K.

$\Delta G^\circ = -2.303\,RT\log K = -2.303 \times 8.314 \times 300 \times 1 \approx -5744$ J, or $-5.74$ kJ mol⁻¹.

Example 8 (entropy of vaporisation): The enthalpy of vaporisation of water is 40.7 kJ mol⁻¹ at 373 K. Find $\Delta S_{vap}$.

$\Delta S_{vap} = \dfrac{\Delta H_{vap}}{T} = \dfrac{40\,700}{373} \approx 109$ J K⁻¹ mol⁻¹.

Example 9 (reversible isothermal expansion): One mole of an ideal gas at 300 K expands reversibly and isothermally to twice its volume. Find $w$, $q$ and $\Delta S$.

$w = -2.303 \times 8.314 \times 300 \times \log 2 \approx -1729$ J. $\Delta U = 0$, so $q = +1729$ J. $\Delta S = R\ln 2 \approx 5.76$ J K⁻¹.

Common Mistakes in Thermodynamics Short Notes Problems

How to Study Thermodynamics Short Notes

1. Read the NCERT Chemistry chapter once fully, and download the textbook from eSaral’s NCERT books for Class 11 page. Understand the reason for each law before you learn the formulas.
2. Write all the formulas on one page, with the sign convention at the top.
3. Practise numericals in this order: work and $\Delta U$, then $\Delta H$ and Hess’s law, then bond enthalpy, and then entropy and Gibbs energy.
4. Revise with eSaral’s Class 11 Chemistry notes and JEE Main Chemistry revision notes.
5. Then solve previous years’ questions with eSaral’s JEE Main chapter-wise PYQs and NEET chapter-wise PYQs.