Thermodynamics Formula Sheet PDF for JEE (N Awasthi Companion)

Here are the free PDFs for JEE chemical thermodynamics you were looking for: a formula sheet with the condition for every formula, a one-page cheat sheet with exam shortcuts, and a list of the mistakes that cost the most marks. They are original sheets built by us for fast revision. All downloads are free.

What these PDFs are (and what they are not)

These sheets are our own original notes, created by the 1% Syndicate here at AllAboutIIT. They are not a scan of a textbook, a coaching module, or a rip-off from somewhere else. We wrote them in our own words to be a fast, practical companion for your problem-solving sessions, especially if you’re using a book like N Awasthi (N Avasthi) for physical chemistry.

Think of them as the notes a senior who aced this chapter would hand you. They cut straight to the point. They don’t contain any textbook theory, solved examples from the book, or copyrighted material. They are for your personal study only and are not for resale.

Not affiliated with the author or publisher of any book mentioned.

Download the Thermodynamics PDFs

Thermodynamics formula sheet (PDF)

This is your master document. It’s not just a dump of equations; every single formula comes with the specific condition under which it is valid. This is the key to avoiding negative marks. Use this sheet when you’re learning the chapter for the first time or when you’re stuck on a problem and need to find the right tool.

Process types: isothermal, isobaric, isochoric, adiabatic, cyclic
First Law of Thermodynamics: ΔU = q + w with IUPAC sign convention
Pressure-Volume work formulas for different conditions
Enthalpy definition and the relation ΔH = ΔU + Δng RT
Heat capacities Cv and Cp, and the Mayer relation Cp - Cv = R
Kirchhoff’s equation for temperature dependence of enthalpy change
Entropy, the Second Law, and formulas for calculating entropy change
Gibbs free energy, spontaneity, and its link to the equilibrium constant

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Thermodynamics cheat sheet (PDF)

This is your one-page revision weapon for the week before an exam. We’ve compressed the entire chapter into the most critical, high-yield information. It includes a blunt note on sign conventions, a summary of the core ideas, and a list of triggers—phrases you see in a problem that should immediately make you think of a specific concept or formula.

A short, no-nonsense note on getting sign conventions right every time
Core ideas: state vs. path functions, intensive vs. extensive properties
First Law, Enthalpy, Entropy, and Gibbs Free Energy in brief
Exam shortcuts for cyclic processes and comparing work done
The difference between adiabatic and isothermal P-V curve slopes
The spontaneity condition table (based on signs of ΔH and ΔS)
Key points for free expansion and phase changes
An “If you see this, think this” list to map problem language to concepts

Download the Thermodynamics cheat sheet (PDF)

Thermodynamics mistakes and traps (PDF)

Losing marks to silly mistakes is the worst feeling. This document lists the exact traps that make students lose marks in thermodynamics, from sign conventions to unit conversions. We explain why the trap exists and give you a simple rule to avoid it. Review this sheet after every mock test and every time you get a problem wrong.

The classic sign convention mix-up for work done by vs. on the system
Using ΔH = ΔU + PΔV when pressure is not constant
Wrongly assuming ΔU = 0 for all isothermal processes (it’s only for ideal gases)
Forgetting to convert temperature to Kelvin
Counting liquids or solids when calculating Δng
Using the wrong value of R or inconsistent units for energy and R
Confusing system entropy (ΔSsys) with universe entropy (ΔSuniv)
Mixing up standard Gibbs free energy (ΔG°) with ΔG

Download the Thermodynamics mistakes and traps (PDF)

The thermodynamics ideas in plain words

First Law of Thermodynamics

This is just the law of conservation of energy applied to chemical systems. It says the change in a system’s internal energy (ΔU) is the sum of the heat (q) added to it and the work (w) done on it. That’s it. The entire game is about the signs of q and w.

ΔU = q + w
Condition: This is the IUPAC convention. Heat added to the system is positive. Work done on the system is positive.

Pressure-Volume Work

Work, in thermodynamics, is often about expansion or compression of a gas. It’s a path function—the amount depends on how you go from state A to state B. The most common formula is for work done against a constant external pressure.

w = -Pext × ΔV
Condition: This is for an irreversible process happening against a constant external pressure Pext.

Enthalpy (H)

Enthalpy is a concept we invented for convenience. It’s the total heat content of a system at constant pressure. The change in enthalpy (ΔH) is the heat absorbed or released in a reaction at constant pressure, which is how most lab experiments are done.

ΔH = ΔU + Δ(PV)
Condition: This is the fundamental definition. For ideal gases in a chemical reaction, it simplifies to ΔH = ΔU + Δng RT where Δng is the change in moles of gas.

Heat Capacity (C)

Heat capacity is the amount of heat needed to raise the temperature of a substance by one degree. Cv is for constant volume (so no P-V work is done) and Cp is for constant pressure (the system can expand and do work). For an ideal gas, Cp is always greater than Cv.

Cp - Cv = R
Condition: This is Mayer’s relation and it holds true for one mole of an ideal gas.

Entropy (S)

Entropy is a measure of randomness or disorder in a system. The Second Law of Thermodynamics says that for any spontaneous (naturally occurring) process, the total entropy of the universe must increase. Don’t confuse the system’s entropy with the universe’s entropy.

ΔSuniverse = ΔSsystem + ΔSsurroundings > 0
Condition: This is the condition for a process to be spontaneous.

Gibbs Free Energy (G)

Gibbs Free Energy is the ultimate decider of spontaneity for a process at constant temperature and pressure. It combines enthalpy and entropy into a single value. A negative ΔG means the process is spontaneous. A positive ΔG means it’s non-spontaneous. Zero ΔG means the system is at equilibrium.

ΔG = ΔH - TΔS
Condition: This equation holds for a process at constant temperature (T).

Gibbs Energy and Equilibrium

Gibbs Free Energy also connects thermodynamics to chemical equilibrium. The standard free energy change (ΔG°) is related to the equilibrium constant (K) of a reaction. This is a powerful link between how much energy a reaction releases and how far it will proceed.

ΔG° = -RT ln(K)
Condition: This relates the standard free energy change to the equilibrium constant K. For non-standard conditions, the equation is ΔG = ΔG° + RT ln(Q).

Which formula, when

Situation	Formula to use	Watch out for
Find internal energy change from heat and work	`ΔU = q + w`	Use the IUPAC sign convention (work done on system is +ve).
Work in a reversible isothermal expansion of an ideal gas	`w = -nRT ln(V2/V1)`	Temperature T must be in Kelvin. V2 is the final volume.
Work in an irreversible expansion against constant pressure	`w = -Pext × (V2 - V1)`	`Pext` is the external pressure, not the gas’s initial pressure.
Relate enthalpy and internal energy for a reaction	`ΔH = ΔU + Δng RT`	`Δng` is (moles of gas products) – (moles of gas reactants). Ignore solids/liquids.
Process is adiabatic (q=0)	`ΔU = w` and `PV^γ = constant`	`γ` (gamma) is `Cp/Cv`. This is for a reversible adiabatic process.
Check if a process is spontaneous	`ΔG = ΔH - TΔS`	A negative `ΔG` means spontaneous. Ensure units of ΔH and ΔS match (J vs kJ).
Find the equilibrium constant from thermodynamic data	`ΔG° = -RT ln(K)`	This uses `ΔG°` (standard), not `ΔG`. R should be 8.314 J/mol·K.
Calculate entropy change of the surroundings	`ΔSsurr = -qsys / T` or `-ΔHsys / T`	This is valid for constant pressure. `T` is the temperature of the surroundings in Kelvin.
Temperature dependence of reaction enthalpy	`ΔH2 - ΔH1 = ΔCp (T2 - T1)`	This is the Kirchhoff equation. `ΔCp` is `Cp(products) - Cp(reactants)`.

Mistakes that cost the most marks

Mistake: Using the physics sign convention (w = work done by system) in chemistry.
This is the number one source of errors. Chemistry (IUPAC) treats energy entering the system as positive. So, work done on the system (compression) is positive. Fix: Memorize ΔU = q + w and stick to it. Work done by the system is negative.

Mistake: Assuming ΔU = 0 for any isothermal process.
This is only true for an ideal gas undergoing an isothermal process. For real gases, liquids, or solids, intermolecular forces exist, so changing the volume (even at constant T) changes the potential energy, and thus ΔU is not zero.

Mistake: Calculating Δng using moles of liquids or solids.
The formula ΔH = ΔU + Δng RT is derived from Δ(PV) for gases. The volume of solids and liquids is negligible. Fix: When calculating Δng, only count the stoichiometric coefficients of gaseous species.

Mistake: Using Celsius for temperature in any formula.
All major thermodynamics formulas (PV=nRT, ΔG = ΔH - TΔS, w = -nRT ln(V2/V1)) require absolute temperature. Fix: The moment you see a temperature in Celsius, convert it to Kelvin (K = °C + 273.15) before you do anything else.

Mistake: Inconsistent units for R, ΔH, and ΔS.
ΔH is often given in kJ/mol, while ΔS and R (8.314 J/mol·K) are in Joules. This mismatch will kill your ΔG calculation. Fix: Always convert everything to a consistent energy unit, usually Joules, before plugging into ΔG = ΔH - TΔS.

Mistake: Confusing ΔG with ΔG°.
ΔG° is the Gibbs free energy change under standard conditions (1 bar pressure, 1M concentration). ΔG is the free energy change under any non-standard set of conditions. They are related by ΔG = ΔG° + RT ln(Q). Don’t use them interchangeably.

Two worked examples (our own)

Example 1: Irreversible Work Calculation

Problem: 2 moles of an ideal gas expand from a volume of 5 L to 25 L against a constant external pressure of 1 atm at 300 K. Calculate the work done by the gas in Joules.

Given:
n = 2 mol
V1 = 5 L
V2 = 25 L
Pext = 1 atm
T = 300 K

Formula and Logic:
The process is an expansion against a constant external pressure, which means it’s irreversible. We must use the formula for irreversible P-V work: w = -Pext × ΔV. The work will be negative because the system (the gas) is doing work on the surroundings by expanding.

Calculation:

Calculate the change in volume, ΔV:
ΔV = V2 - V1 = 25 L - 5 L = 20 L
Calculate work in L·atm:
w = -Pext × ΔV = -1 atm × 20 L = -20 L·atm
Convert L·atm to Joules. The conversion factor is 1 L·atm = 101.3 J.
w = -20 L·atm × 101.3 J/L·atm
w = -2026 J

Answer: The work is w = -2026 J: the gas does 2026 J of work on the surroundings.

Example 2: Spontaneity Calculation

Problem: For a certain reaction, the change in enthalpy is ΔH = +100 kJ/mol and the change in entropy is ΔS = +400 J/mol·K. Above what temperature will this reaction become spontaneous?

Given:
ΔH = +100 kJ/mol
ΔS = +400 J/mol·K

Formula and Logic:
Spontaneity is determined by the sign of Gibbs Free Energy, ΔG. A reaction becomes spontaneous when ΔG becomes negative. The crossover point from non-spontaneous to spontaneous happens at equilibrium, where ΔG = 0. We will use the formula ΔG = ΔH - TΔS, set ΔG to 0, and solve for T.

Calculation:

First, make the units consistent. Convert ΔH to J/mol.
ΔH = +100 kJ/mol × 1000 J/kJ = +100000 J/mol
Set ΔG = 0 to find the equilibrium temperature.
0 = ΔH - TΔS
TΔS = ΔH
T = ΔH / ΔS
Solve for T:
T = (100000 J/mol) / (400 J/mol·K)
T = 250 K

Answer: The reaction becomes spontaneous at temperatures above 250 K. At T > 250 K, the TΔS term will be larger than the ΔH term, making ΔG negative.

How to use these sheets with a problem book

Don’t just read the PDFs. Integrate them into your workflow. Here’s a simple system:

Before you start: Read our Formula Sheet once. Don’t memorize. Just get familiar with the tools available and the conditions for using them.
During problem-solving: Attempt the problems in your book (like N Awasthi) on your own. When you get stuck, refer back to the formula sheet to find the right equation for the situation described.
After you solve (or fail): For every single question you get wrong, open the Mistakes and Traps sheet. Find the specific trap you fell into. Understand why you made the mistake. This is more important than just seeing the right solution.
For revision: Before a test or a new chapter, use the Cheat Sheet. It’s your high-speed review tool to bring all the key ideas back into your active memory.

A three-session plan for thermodynamics

Break this beast of a chapter down. Don’t try to eat it all in one sitting. Use three focused sessions of 60-90 minutes.

Session 1: The First Law & Energy. Focus on the basics. Master the concepts of system, surroundings, state functions, and path functions. Get the sign convention for q and w locked in. Solve problems related to calculating ΔU and work done in isothermal/isobaric/isochoric processes.
Session 2: Enthalpy, Entropy & The Second Law. Introduce ΔH and its relation to ΔU (Δng RT). Understand what entropy really means (disorder) and how to calculate ΔS for simple processes. Grasp the core idea of the Second Law: ΔSuniverse must increase for a spontaneous process.
Session 3: Gibbs Free Energy & Equilibrium. This is the synthesis session. Understand how ΔG combines ΔH and ΔS to be the final word on spontaneity at constant T and P. Practice problems using ΔG = ΔH - TΔS. Finally, connect ΔG° to the equilibrium constant K.

Quick self-test

A gas is compressed from 10L to 2L by a constant external pressure of 5 atm. Is the work done (w) positive or negative?
For the reaction N2(g) + 3H2(g) -> 2NH3(g), what is the value of Δng?
If a reaction is endothermic (ΔH > 0) and causes an increase in disorder (ΔS > 0), will it be more spontaneous at high or low temperatures?
Calculate the change in internal energy (ΔU) if a system absorbs 500 J of heat and does 200 J of work on the surroundings.
Under what two conditions is ΔH exactly equal to ΔU for a process?

Answers:

Positive. The work is done on the system (compression), so w > 0.
Δng = (moles of gas products) – (moles of gas reactants) = 2 – (1 + 3) = -2.
High temperatures. The -TΔS term in ΔG = ΔH - TΔS will become more negative at high T, eventually overcoming the positive ΔH to make ΔG negative.
q = +500 J (absorbed). w = -200 J (work done by system). ΔU = q + w = 500 + (-200) = 300 J.
(1) For any process involving only solids and liquids where volume change is negligible. (2) For a gaseous reaction where the number of moles of gas does not change (Δng = 0).

FAQ

What is the sign convention for work in JEE chemistry?
The official IUPAC convention is used. Work done on the system (e.g., compression) is positive (+w). Work done by the system (e.g., expansion) is negative (-w). Heat absorbed by the system is positive (+q). Heat released by the system is negative (-q).

Is ΔU zero for every isothermal process?
No. This is a common and costly mistake. ΔU = 0 is only true for an ideal gas undergoing an isothermal process. For real gases, liquids, and solids, intermolecular forces contribute to internal energy, so ΔU is not necessarily zero even if temperature is constant.

When is ΔH equal to ΔU?
From the definition ΔH = ΔU + Δ(PV), ΔH equals ΔU when the Δ(PV) term is zero. This happens in two main cases: (1) For processes involving only solids and liquids, as their volume change is negligible. (2) For gaseous reactions where the total moles of gas on the product side equal the total moles on the reactant side, making Δng = 0.

How is Gibbs free energy related to the equilibrium constant?
The standard Gibbs free energy change (ΔG°) is related to the equilibrium constant (K) by the equation ΔG° = -RT ln(K). This powerful formula links the thermodynamic favorability of a reaction under standard conditions to the position of its equilibrium.

Are these PDFs free to download?
Yes, all the PDFs linked on this page are completely free for your personal, non-commercial use. Our goal is to provide real, useful resources without barriers.

What is the difference between a state function and a path function?
A state function depends only on the initial and final states of the system, not on the path taken to get there. Examples include Internal Energy (U), Enthalpy (H), Entropy (S), and Gibbs Free Energy (G). A path function depends on the specific route taken. The two most important path functions are work (w) and heat (q).

What’s the difference between Cv and Cp?
Cv is the heat capacity at constant volume. When you add heat, all of it goes into increasing the internal energy because no P-V work can be done. Cp is the heat capacity at constant pressure. When you add heat, some of it increases internal energy and some is used to do expansion work, so you need to add more heat to get the same temperature rise. That’s why Cp is always greater than Cv for a gas.

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