JEE Mains Physics · Thermodynamics

First Law and Energy Bookkeeping

The heat given to a gas goes into its internal energy and into the work it does, ΔQ = ΔU + W, with W counted positive when the gas expands.

All Thermodynamics notes JEE Mains Physics strategy

Why this matters

Fifteen PYQs, all multiple choice, and four from 2026. Six test the law's statements: the sign of work, which quantities depend on the path, and which are intensive. Nine are energy balances: a liquid turning to vapour, water warming, heat and work given as rates, or a gas pushing a piston. Each needs only the first law and the work done against a steady pressure.

Concept 1 of 2: The first law and its sign convention

Heat given to a gas does two jobs: it raises the internal energy and it lets the gas do work by expanding. The first law is this energy account, ΔQ = ΔU + W. These notes use the physics convention throughout: heat INTO the gas is positive, and W is the work done BY the gas, positive when it expands. Some books count the work done ON the gas instead; then the same law reads ΔQ = ΔU − W.

Definition

$\Delta Q = \Delta U + W$ , with W the work done by the gas: positive when V increases, negative when V decreases.
With $W'$ the work done on the gas, $W' = -W$ and the law reads $\Delta Q = \Delta U - W'$ .
ΔU depends only on the start and end states (a state function). Heat and work depend on the path (path functions).
Heat in need not raise the temperature: if the gas does more work than the heat it gets, ΔU < 0 and it cools.
Positive work by the gas means its volume grew, since $W = \int P\,dV$ and P > 0.
The zeroth law gives the idea of temperature; the first law gives the idea of internal energy.
Extensive quantities grow with the amount of gas (U, V, mass); intensive ones do not (P, T, density). Extensive ÷ extensive, such as energy per kilogram, is intensive; intensive × extensive is extensive.

First law (work done by the gas positive)

\Delta Q = \Delta U + W \qquad W = \int_{V_1}^{V_2} P\,dV

Worked example

A gas absorbs

300\ \text{J}

of heat while it expands and does

450\ \text{J}

of work. Find ΔU. Does its temperature rise or fall?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 8 Apr 2023 · Q6Moderate

Example 1 · Thermodynamics · First Law and Energy Bookkeeping

Given below are two statements: Statement I: If heat is added to a system, its temperature must increase. Statement II: If positive work is done by a system in a thermodynamic process, its volume must increase. In the light of the above statements, choose the correct answer from the options given below

Two sign conventions for work

Physics counts the work done BY the gas as positive and writes ΔQ = ΔU + W. Chemistry counts the work done ON the gas and writes ΔQ = ΔU − W. Both describe the same energy; mixing them flips the sign of W.

Heat added does not mean hotter

If a gas does more work than the heat it receives, its internal energy falls and so does its temperature. Heat in only fixes ΔU = Q − W, not the sign of ΔT.

Drill 5 more on the first law and its sign convention

Concept 2 of 2: Energy balances: phase changes, rates and pistons

Many questions give two of Q, ΔU and W and ask for the third. The work usually comes from a steady outside pressure: W = PΔV for boiling at atmospheric pressure, W = PAΔx for a piston of area A. In a phase change the heat is mL, and not all of it stays as internal energy, because the vapour pushes back the atmosphere. When the data are rates, the same law holds for each second.

Definition

Phase change at constant pressure: $Q = mL$ , $W = P\Delta V$ , $\Delta U = mL - P\Delta V$ .
Boiling: the volume grows a lot, so W is large enough to matter. Melting ice: the volume falls, so W < 0, and the atmosphere does work on the ice–water system.
A liquid or solid warmed: ΔV is tiny, so $\Delta U = mc\Delta T - P\Delta V$ is very close to $mc\Delta T$ .
Rate form: $\dfrac{dQ}{dt} = \dfrac{dU}{dt} + \dfrac{dW}{dt}$ . Heat supplied per second minus work done per second is the rate at which U rises.
Piston of area A moving out a distance Δx at pressure P: $W = PA\,\Delta x$ .
If a stated fraction of the heat goes into work, the rest is ΔU.
Units: $1\ \text{cm}^{3} = 10^{-6}\ \text{m}^{3}$ , $1\ \text{L} = 10^{-3}\ \text{m}^{3}$ , and Pa × m³ = J.

Energy balance at a steady pressure

\Delta U = Q - P\Delta V \qquad \frac{dU}{dt} = \frac{dQ}{dt} - \frac{dW}{dt}

Worked example

10 g of a liquid boils at a steady pressure of

2 \times 10^{5}\ \text{Pa}

, and its volume grows by

5000\ \text{cm}^{3}

. Its latent heat of vaporisation is

1.2 \times 10^{6}\ \text{J/kg}

. Find the heat taken, the work done by the vapour and ΔU.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 11 April 2023 · Q10Moderate

Example 2 · Thermodynamics · First Law and Energy Bookkeeping

1\text{ }kg

of water at

100^{\circ}C

is converted into steam at

100^{\circ}C

by boiling at atmospheric pressure. The volume of water changes from

1.00 \times10^{- 3}{\text{ }m}^{3}

as a liquid to

1.671{\text{ }m}^{3}

as steam. The change in internal energy of the system during the process will be (Given latent heat of vaporisation

= 2257\text{ }kJ/kg

. Atmospheric pressure

= 1 \times10^{5}\text{ }Pa

)

Subtract the work against the atmosphere

When a liquid boils, ΔU is the latent heat minus PΔV, not the latent heat itself. The PΔV term is often what separates two of the options.

Convert volumes before multiplying

cm³ to m³ is a factor 10⁻⁶ and litres to m³ is 10⁻³. Pressure in Pa times volume in m³ gives joules; so does kPa times litres. Any other pair needs converting first.

Drill 8 more on energy balances: phase changes, rates and pistons

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

The first law and its sign convention
First law (work done by the gas positive)
$\Delta Q = \Delta U + W \qquad W = \int_{V_1}^{V_2} P\,dV$
Energy balances: phase changes, rates and pistons
Energy balance at a steady pressure
$\Delta U = Q - P\Delta V \qquad \frac{dU}{dt} = \frac{dQ}{dt} - \frac{dW}{dt}$

Watch out for (4)

Two sign conventions for work
→ The first law and its sign convention
Heat added does not mean hotter
→ The first law and its sign convention
Subtract the work against the atmosphere
→ Energy balances: phase changes, rates and pistons
Convert volumes before multiplying
→ Energy balances: phase changes, rates and pistons

Test yourself on Thermodynamics

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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