First law of thermodynamics

Pump air into a bicycle tyre by hand, fast and hard, and the pump barrel gets noticeably warm. Nobody lit a flame under it. No heat was poured in from outside. The warmth came entirely from the work your arm did squeezing the air — and that single observation, generalised, is the whole first law of thermodynamics. Energy supplied to a system as heat, plus energy supplied to it as work, has nowhere to hide: it must show up as a change in the system's stored internal energy. Nothing is created, nothing is destroyed, only converted and moved around between three accountable quantities — heat $\Delta Q$ , work $\Delta W$ , and internal energy $\Delta U$ . This sounds almost too simple to be a "law," and that is exactly its power. It is the law of conservation of energy, translated into the language of gases, pistons and cylinders, and it is the single equation that every later question in this chapter — isothermal expansions, adiabatic compressions, cyclic engines-in-waiting — reduces to. Get the bookkeeping right, in particular the sign convention for who is doing what to whom, and the rest of thermodynamics becomes an exercise in careful accounting rather than new physics. This article builds the first law from the ground up: the exact CBSE form $\Delta Q = \Delta U + \Delta W$ , why $\Delta U$ alone is path-independent while $Q$ and $W$ separately are not, how the law simplifies for isochoric, isothermal, adiabatic and cyclic processes, and a full derivation of the work done by an isothermal gas — the calculation examiners ask for most often. Two worked examples, one board-level and one competitive-level, carry the arithmetic all the way through, extending the Pass-1 numbers you may already have seen.

Key takeaways

The first law is energy conservation for heat and work: $\Delta Q = \Delta U + \Delta W$ , where $\Delta W = P\,\Delta V$ for a gas.
Sign convention decides everything. Heat absorbed and work done by the system are positive; heat released and work done on the system are negative.
Internal energy is a state variable. $\Delta U$ depends only on initial and final states, so it is path-independent even though $Q$ and $W$ individually are not.
Four special cases recur constantly: isochoric ( $\Delta W = 0$ ), isothermal ideal gas ( $\Delta U = 0$ ), adiabatic ( $\Delta Q = 0$ ), and a full cycle ( $\Delta U_{cycle} = 0$ ).
Isothermal work has a clean formula: $\Delta W = nRT\ln\!\left(\dfrac{V_2}{V_1}\right)$ , derived directly from $\Delta W = \int P\,dV$ using $PV = nRT$ .
Adiabatic changes of state obey $PV^\gamma = \text{const}$ , quoted here and used to relate pressure and volume when no heat is exchanged.
Key idea: heat supplied to a system is conserved as it splits between raising internal energy and doing work, $\Delta Q = \Delta U + \Delta W$ , with the split depending on the path even though $\Delta U$ itself does not.

The core insight

Energy given to a system as heat cannot vanish — the first law $\Delta Q = \Delta U + \Delta W$ says it must appear either as a rise in the system's internal energy or as work done by the system on its surroundings, with $\Delta U$ fixed by the endpoints alone while $Q$ and $W$ separately depend on the path taken between them.

Where it fits in the syllabus

The first law is the centrepiece of the CBSE Class 11 Thermodynamics unit, sitting right after heat, internal energy and work are introduced and right before the specific-process language (isothermal, adiabatic, isochoric, cyclic) that dominates the rest of the chapter. You are expected to state it, apply the exact form $\Delta Q = \Delta U + \Delta W$ with $\Delta W = P\,\Delta V$ , use the sign convention without error, and derive the work done in an isothermal change of state. Heat engines, their efficiency, and the second law's quantitative machinery are out of scope for the current 2025-26 prescription — the second law appears here only as the qualitative Kelvin-Planck and Clausius statements, with heat-engine efficiency left as a one-line forward pointer to what you will meet in a higher class.

The physical picture

Picture a gas sealed in a cylinder by a frictionless piston. Heat a system a little, and that energy has exactly two places to go: it can shake the molecules harder, raising the internal energy stored in their random kinetic and potential motion, or it can push the piston outward, doing mechanical work on whatever lies beyond it. The first law simply insists that the books balance between these two destinations.

\Delta Q = \Delta U + \Delta W

where: $\Delta Q$ is the heat supplied to the system, $\Delta U$ is the change in the system's internal energy, and $\Delta W$ is the work done by the system on its surroundings, with $\Delta W = P\,\Delta V$ for a gas held at pressure $P$ as its volume changes by $\Delta V$ . Read left to right, it says: whatever heat you put in comes back out as some combination of a hotter gas and a pushed piston, in a ratio fixed by the process, never more and never less. The sign convention is what turns this equation from a slogan into a working tool. $\Delta Q$ is positive when heat flows into the system and negative when it flows out. $\Delta W$ is positive when the system expands and does work on the surroundings, negative when the surroundings compress the system and do work on it. $\Delta U$ is positive when the internal energy — and for an ideal gas, therefore the temperature — rises. Every worked problem in this chapter is really just a careful application of this convention to a specific scenario.

Figure

A cylinder with a piston showing heat entering from below split into two arrows, one raising the internal energy of the gas and one pushing the piston outward to do work, with the first law equation written beneath

Because internal energy is a state variable, $\Delta U = U_{final} - U_{initial}$ depends only on where the system started and ended, never on the road taken between them. Take two different paths from the same initial state $A$ to the same final state $B$ — one that dumps a lot of heat in and does a lot of work, another that dumps in less heat and does correspondingly less work — and $\Delta U$ comes out identical both times, while $Q$ and $W$ individually can be completely different numbers on the two paths. That is the deep content of the first law: it is not merely "heat splits into internal energy and work," it is the guarantee that however you split it, the internal-energy share is fixed by the endpoints alone. Four special cases recur so often that it is worth fixing them before touching a single numerical problem. In an isochoric process the volume is held fixed, so $\Delta W = P\,\Delta V = 0$ and the entire heat supplied goes into internal energy: $\Delta Q = \Delta U$ . In an isothermal process on an ideal gas the temperature does not change, and since the internal energy of an ideal gas depends only on temperature, $\Delta U = 0$ , so all the heat supplied is converted straight into work: $\Delta Q = \Delta W$ . In an adiabatic process no heat is exchanged with the surroundings at all, $\Delta Q = 0$ , so $\Delta U = -\Delta W$ — any work done by the gas comes entirely out of its own internal energy, and any work done on it goes entirely into raising it. And over one complete cycle, the system returns to its starting state, so $\Delta U_{cycle} = 0$ and $\Delta Q_{cycle} = \Delta W_{cycle}$ — the net heat absorbed over a full cycle equals the net work delivered, the relation on which every heat engine (a topic you will meet formally in a later class) ultimately rests. The Class 11 syllabus also asks you to recognise isothermal ( $PV = \text{const}$ ) and adiabatic ( $PV^\gamma = \text{const}$ ) changes of state by their $P$ - $V$ curves: an adiabatic curve is always steeper than an isothermal one through the same point, because compressing a gas adiabatically also heats it, which pushes the pressure up faster than temperature-constant compression alone would. The second law of thermodynamics, at this level, adds only the qualitative statements — the Kelvin-Planck statement (no engine can convert heat completely into work with no other effect) and the Clausius statement (heat cannot spontaneously flow from a colder to a hotter body) — which set a direction to natural processes without yet giving you a formula to compute. Quantitative heat-engine efficiency is studied in a higher class.

Derivation AWork done by an ideal gas in an isothermal expansion

The goal is to find an exact formula for the work done by an ideal gas as it expands from volume $V_1$ to volume $V_2$ at constant temperature $T$ , starting from nothing more than the definition of work for a gas, $\Delta W = \int P\,dV$ , and the ideal gas equation of state $PV = nRT$ . This single result is the one examiners ask for most often when a first-law question specifies an isothermal process, because it converts the abstract $\Delta W = P\,\Delta V$ of the first law into a number you can actually compute for a continuously changing pressure.

Step 1. Write the work done for a small volume change. As the gas pushes the piston out by an infinitesimal $dV$ at whatever pressure $P$ it currently has, it does work $dW = P\,dV$ . Because the pressure itself falls as the gas expands, this cannot be replaced by a single constant $P$ times the total $\Delta V$ — it must be summed, that is, integrated, over the whole expansion from $V_1$ to $V_2$ : $\Delta W = \displaystyle\int_{V_1}^{V_2} P\,dV$ .

Step 2. Substitute the equation of state. Since the gas is ideal, $PV = nRT$ at every instant, so $P = \dfrac{nRT}{V}$ . Along an isothermal process $T$ is fixed, so $n$ , $R$ and $T$ can all be pulled outside the integral, leaving only the $V$ -dependence inside: $\Delta W = nRT\displaystyle\int_{V_1}^{V_2}\dfrac{dV}{V}$ .

Step 3. Evaluate the integral. The standard result $\displaystyle\int \dfrac{dV}{V} = \ln V$ gives $\Delta W = nRT\left[\ln V\right]_{V_1}^{V_2} = nRT\left(\ln V_2 - \ln V_1\right)$ , which combines into a single logarithm of a ratio: \Delta W = nRT\,\ln\!\left(\frac{V_2}{V_1}\right).

Step 4. Read the sign and connect back to the first law. If the gas expands, $V_2 > V_1$ , so the logarithm is positive and $\Delta W > 0$ — the gas does positive work on the surroundings, matching the sign convention. Since the process is isothermal for an ideal gas, $\Delta U = 0$ , so the first law collapses to $\Delta Q = \Delta W = nRT\ln(V_2/V_1)$ : every joule of work done by the expanding gas is supplied instant-by-instant as heat flowing in from a reservoir holding the temperature fixed, exactly the balance the first law demands. For an adiabatic change of state, by contrast, the corresponding pressure-volume relation is the quoted result $PV^\gamma = \text{const}$ , with $\gamma = C_p/C_v$ ; deriving that relation needs the specific-heat and equipartition ideas of the next article and is not required here.

Worked examples

Board level - heat, work and the rise in internal energy

A gas absorbs 850 J of heat while expanding, doing 320 J of work on the surroundings. Find the change in the internal energy of the gas, and state whether its temperature rises or falls.

Step 1. Identify the signs. Heat is absorbed by the system, so $\Delta Q = +850\ \text{J}$ . The gas does work on the surroundings (expansion), so $\Delta W = +320\ \text{J}$ .

Step 2. Apply the first law, solved for $\Delta U$ : $\Delta U = \Delta Q - \Delta W$ .

Step 3. Substitute the numbers: $\Delta U = 850 - 320 = 530\ \text{J}$ .

Step 4. Interpret the result. The internal energy increases, so — since the internal energy of an ideal gas tracks its temperature directly — the temperature of the gas rises. Answer: $\Delta U = +530\ \text{J}$ , temperature rises. Even though the gas is doing work while expanding, it is absorbing more heat than it is giving out as work, so on balance it still gets hotter, not cooler.

Competitive level - compression, heat release, and the implied temperature change

A gas is compressed from state A to state B, during which 250 J of work is done on the gas by an external agent, and the gas releases 90 J of heat to the surroundings. (a) Find the change in internal energy. (b) State whether the temperature rises or falls. (c) If this sample is 0.10 mol of a monatomic ideal gas, estimate the temperature change using the quoted relation $\Delta U = \tfrac{3}{2}nR\,\Delta T$ (derived from the equipartition theorem in the kinetic theory chapter).

Step 1. Fix the signs. Work is done on the gas, so by the "work done by the system" convention $\Delta W = -250\ \text{J}$ . Heat is released by the gas, so $\Delta Q = -90\ \text{J}$ .

Step 2. Apply the first law: $\Delta U = \Delta Q - \Delta W = (-90) - (-250) = -90 + 250 = 160\ \text{J}$ .

Step 3. Interpret part (b). The internal energy rises by 160 J, so the gas's temperature rises — physically, the 250 J of compression work fed in outweighs the 90 J of heat that leaked out, leaving a net energy gain, the same effect that warms a bicycle pump under fast strokes.

Step 4. Solve part (c) for $\Delta T$ : rearranging $\Delta U = \tfrac{3}{2}nR\,\Delta T$ gives $\Delta T = \dfrac{2\,\Delta U}{3nR} = \dfrac{2 \times 160}{3 \times 0.10 \times 8.314} = \dfrac{320}{2.494} \approx 128\ \text{K}$ . Answer: $\Delta U = +160\ \text{J}$ ; temperature rises; for 0.10 mol of a monatomic ideal gas this corresponds to a rise of about $128\ \text{K}$ . Timing note: under ninety seconds for parts (a) and (b) once the sign convention is fixed in your head first — write down the sign of $\Delta Q$ and $\Delta W$ before touching the formula, never while substituting. Part (c) borrows a single quoted formula from kinetic theory ( $\Delta U = \tfrac32 nR\Delta T$ for a monatomic gas); do not attempt to derive it here, just plug in the $\Delta U$ you already found in part (a).

Common traps

Sign errors on $\Delta W$ or $\Delta Q$ . Decide first whether heat is absorbed or released, and whether work is done by or on the system, and write the sign down explicitly before substituting — this single habit removes most first-law mistakes.
Forgetting $\Delta U = 0$ over a full cycle. Because internal energy is a state variable, a system returning to its starting point after one complete cycle has $\Delta U_{cycle} = 0$ , so $\Delta Q_{cycle} = \Delta W_{cycle}$ — students sometimes try to compute a nonzero $\Delta U$ for a closed cycle by mistake.
Applying $\Delta U = 0$ outside an isothermal ideal-gas process or a full cycle. It does not hold for isochoric, adiabatic or general processes — check which special case actually applies before dropping a term.
Using $\Delta W = P\Delta V$ when pressure is not constant. That shortcut only works for an isobaric step; for an isothermal or adiabatic change, $\Delta W$ must come from the integral $\int P\,dV$ , giving $nRT\ln(V_2/V_1)$ or the corresponding adiabatic result, not a simple $P\Delta V$ .
Confusing the first law with the zeroth or second law. The zeroth law is about the existence of thermal equilibrium and a well-defined temperature; the first law is energy conservation among $Q$ , $U$ and $W$ ; the second law (Kelvin-Planck, Clausius, qualitative here) is about the direction natural processes can run. They answer three different questions and are not interchangeable on an exam.

Common trapQ and W are not separately conserved — only their difference, delta U, is

The single biggest misconception is treating heat and work as though each is individually a fixed, path-independent quantity like internal energy. They are not. Take a gas from the same initial state to the same final state along two different routes — one mostly by heating at constant volume then letting it expand, another by a slow isothermal expansion — and you will find different values of $Q$ and different values of $W$ on the two paths, even though $\Delta U$ comes out exactly the same both times. Heat and work are process quantities, meaningful only for a specific path; internal energy is a state quantity, meaningful for a specific state. Writing $Q_{final} - Q_{initial}$ or treating "the heat in the gas" as something the gas simply possesses is a category error — a system has internal energy, not "heat" or "work" stored inside it.

You should be able to…

State the first law of thermodynamics and explain it as a statement of energy conservation among heat, work and internal energy.
Apply the correct sign convention to $\Delta Q$ , $\Delta W$ and $\Delta U$ for expansion, compression, heat absorption and heat release.
Solve for any one of $\Delta Q$ , $\Delta U$ , $\Delta W$ given the other two, in both board-style and multi-part competitive problems.
Explain why $\Delta U$ is path-independent while $Q$ and $W$ individually are not, using the state-variable versus process-quantity distinction.
Derive the work done by an ideal gas in an isothermal expansion, $\Delta W = nRT\ln(V_2/V_1)$ , from $\Delta W = \int P\,dV$ and $PV = nRT$ .
Apply the first law correctly to isochoric, isothermal, adiabatic and cyclic special cases, and recognise isothermal ( $PV=\text{const}$ ) versus adiabatic ( $PV^\gamma=\text{const}$ ) curves on a P-V diagram.

Ask the tutor: Want to work through a multi-part first-law problem with your own numbers, or see the adiabatic-work formula derived the same way as the isothermal one above — ask the AI tutor for guided practice.

Why is delta W positive when the gas expands and not when it is compressed?

This is purely a convention, chosen so that the first law reads naturally: heat in becomes internal energy plus work delivered outward. When a gas expands it pushes its surroundings — piston, atmosphere, whatever is outside — so it is doing positive work on them, and that is defined as positive $\Delta W$ . When a gas is compressed, the surroundings are pushing on the gas instead, doing work on it, so from the gas's point of view $\Delta W$ is negative (equivalently, the external agent's work on the gas is $+250\ \text{J}$ but the gas's own $\Delta W$ is $-250\ \text{J}$ ). As long as you stay consistent with one convention throughout a problem, the arithmetic comes out right either way — the trap is switching conventions halfway through.

If internal energy only depends on temperature, does that mean heat and work don't matter at all for an ideal gas?

Heat and work absolutely matter — they are exactly what changes the temperature and hence the internal energy in the first place. The point is subtler: for an ideal gas, internal energy depends on temperature alone, not on pressure or volume separately. So two very different processes — one supplying a lot of heat while doing a lot of work, another supplying little heat while doing little work — can land the gas at the same final temperature and therefore the same $\Delta U$ , even though the heat and work involved were completely different along the way. Heat and work are how you get there; internal energy (through temperature) is where you end up.

Why do adiabatic and isothermal curves on a P-V graph look different, and which is steeper?

An isothermal curve follows $PV = \text{const}$ , so $P \propto 1/V$ — a single, gentle hyperbola. An adiabatic curve follows $PV^\gamma = \text{const}$ with $\gamma > 1$ , which falls off faster with increasing $V$ , so the adiabatic curve is always steeper than the isothermal curve through the same starting point. Physically this makes sense from the first law: in an adiabatic compression, none of the compression work can escape as heat, so all of it drives the temperature (and hence the pressure) up harder than an isothermal compression at the same volume change would, where the extra heat is instead allowed to leave and keep the temperature fixed.

Is the second law, or heat engines, going to show up in my Class 11 exam?

The 2025-26 CBSE Class 11 syllabus keeps the second law to its qualitative statements only — the Kelvin-Planck statement that no engine converts heat completely into work with no other effect, and the Clausius statement that heat does not spontaneously flow from cold to hot. You are not expected to compute engine efficiency, Carnot-cycle numbers, or entropy changes at this stage; those quantitative heat-engine and entropy topics are studied in a higher class. If a question mentions "efficiency of a heat engine," treat it as outside this chapter's numerical scope unless your teacher has specifically extended the syllabus.