Heat internal energy and work

Put a bicycle pump in your hand, block its outlet, and push the piston down fast. The barrel gets warm — not because you rubbed it, not because heat leaked in from outside, but because you did work on the trapped air and that work turned into internal energy. Run the same pump the other way — let compressed air escape and expand — and it cools. No flame touched it either time. Something inside the gas itself is keeping score of every push and every escape, and thermodynamics is the bookkeeping system built to track it. That bookkeeping rests on three quantities that sound like everyday synonyms but are, in physics, sharply different animals. Heat and work are both energy caught crossing a boundary — a temperature difference driving one, a moving piston driving the other — while internal energy is what the system actually holds, a running total set entirely by its current state. Confuse "in transit" with "stored" and every thermodynamics problem from here onward becomes guesswork; keep them straight and the rest of the chapter — starting with the first law in the very next topic — falls into place almost mechanically. This article builds that distinction from scratch: what each quantity physically means, the sign convention CBSE expects you to apply without hesitation, the formula $\Delta W = P\,\Delta V$ that lets you actually compute work done by an expanding or compressing gas, and why heat and work depend on the path taken while internal energy does not. Two worked examples — one pump-style constant-pressure expansion, one compression with heat removed — carry the arithmetic all the way through, using the same numbers a board exam would set.

Key takeaways

Internal energy ( $U$ ) is the total microscopic kinetic and potential energy stored in a system's molecules — a state variable fixed by the current state ( $P$ , $V$ , $T$ ), not by history.
Heat ( $Q$ ) is energy crossing a system's boundary because of a temperature difference — energy in transit, never something a system "contains."
Work ( $W$ ) is energy crossing the boundary through a force acting over a displacement — for a gas, usually a moving piston.
Sign convention (CBSE/NCERT): heat absorbed by the system is positive, heat released is negative; work done by the system (expansion) is positive, work done on it (compression) is negative.
The core formula for a gas is $\Delta W = P\,\Delta V$ for a small change, and $W = \int_{V_1}^{V_2} P\,dV$ (the area under the curve on a $P$ – $V$ diagram) for a finite one.
$Q$ and $W$ are path-dependent — their values depend on exactly how a process unfolds — while $\Delta U$ depends only on the initial and final states.
Key idea: heat and work are energy caught crossing a system's boundary and depend on the path taken, while internal energy is a stored state variable computed from the work formula ΔW = PΔV together with the bookkeeping equation ΔQ = ΔU + ΔW.

The core insight

Heat and work are two different ways energy crosses the boundary of a system — heat driven by a temperature difference, work by a piston moving through $\Delta W = P\,\Delta V$ — and both are path-dependent, unlike the internal energy $U$ they add up to change, which is a state variable fixed only by the system's current $P$ , $V$ , $T$ .

Where it fits in the syllabus

This topic sits inside the CBSE Class 11 Thermodynamics unit, positioned right before the first law of thermodynamics is formally stated and used. The prescribed scope asks you to know heat, work and internal energy as distinct concepts, the sign convention used throughout the chapter, and the formula $\Delta W = P\,\Delta V$ for work done during expansion or compression of a gas — exactly the ground covered here. It deliberately stops short of the first law equation itself (the very next topic in the syllabus) and stops well short of heat engines, Carnot efficiency, or entropy, none of which are in the 2025-26 Class 11 prescription; those are flagged only as one-line forward pointers where they naturally arise.

The physical picture

Internal energy ( $U$ ) is the sum of the kinetic energy of random molecular motion and the potential energy of intermolecular interactions, all bundled inside the system. For an ideal gas, there is no intermolecular potential energy at all (molecules interact only during instantaneous collisions), so internal energy depends on temperature alone: raise $T$ and $U$ rises; hold $T$ fixed and $U$ does not change, no matter what happens to $P$ or $V$ individually. This is what makes $U$ a state variable — its value is fixed the moment you specify the state ( $P$ , $V$ , $T$ ), completely independent of the sequence of steps that got the gas there.

Heat ( $Q$ ) and work ( $W$ ), by contrast, are not properties a system holds — they are the two channels through which energy crosses its boundary. Heat flows because of a temperature difference between the system and its surroundings; work flows because a force acts through a displacement, which for a gas usually means a piston sweeping through a volume change. Once either has crossed the boundary and been absorbed into the system, it stops being "heat" or "work" and simply becomes part of $U$ — you cannot open up a gas afterwards and separate out "the heat part" from "the work part" of its internal energy. For a gas at pressure $P$ undergoing a small volume change $\Delta V$ , the work done by the gas on its surroundings is

\Delta W = P\,\Delta V

where: $P$ is the gas pressure (Pa) at the moment the volume changes, and $\Delta V = V_2 - V_1$ is the change in volume (m³). If the pressure is not constant over the whole process, this relation applies only to an infinitesimally small step, and the total work for a finite change from $V_1$ to $V_2$ is the sum (integral) of all those small steps — the area swept out under the curve on a $P$ – $V$ diagram:

W = \int_{V_1}^{V_2} P\,dV

where: the integral adds up $P\,dV$ over every sliver of volume change between $V_1$ and $V_2$ ; graphically, $W$ is exactly the shaded area between the process curve and the volume axis on a $P$ – $V$ plot. This is why work is called path-dependent — two different curves joining the same two points $(P_1,V_1)$ and $(P_2,V_2)$ enclose different areas, hence different $W$ , even though the gas starts and ends in identical states.

Sign convention (CBSE/NCERT), used throughout this chapter: heat absorbed by the system is taken positive, heat released is negative; work done by the system — the gas expands, $\Delta V>0$ — is positive, and work done on the system — the gas is compressed, $\Delta V<0$ — is negative. This convention feeds directly into the bookkeeping relation of the first law, previewed here only to show why the definitions matter:

\Delta Q = \Delta U + \Delta W

where: $\Delta Q$ is heat absorbed, $\Delta U$ is the resulting change in internal energy, and $\Delta W$ is work done by the gas — the full derivation and use of this equation is the next topic, but notice already that it is exactly a statement that energy in ( $Q$ ) either raises the stored energy ( $U$ ) or leaves as work ( $W$ ) done on the surroundings, or some split of both.

Figure

A pressure volume diagram showing a shaded region under a curved process line between an initial point and a final point, with the shaded area labelled as the work done by the gas equal to the integral of pressure with respect to volume

The physical picture continued: path dependence

Because $W$ is literally the area under a curve, and because many different curves can connect the same two endpoints on a $P$ – $V$ diagram, the same gas going from state 1 to state 2 can do very different amounts of work depending on the road taken — a slow constant-pressure expansion sweeps a different shaped area than a curved, varying-pressure expansion between the identical endpoints. Heat behaves the same way, because it is precisely whatever $Q$ the first law demands to balance a given $\Delta U$ and $\Delta W$ pair; change the path, change $W$ , and $Q$ must change to match, even with $\Delta U$ fixed by the endpoints alone. Internal energy alone escapes this: since $U$ is a state variable, $\Delta U = U_2 - U_1$ is exactly the same number no matter which path was followed to get from state 1 to state 2.

Derivation AWork done by an ideal gas at constant pressure

The goal is to derive, from the basic definition of mechanical work as force times displacement, the formula $W = P\,\Delta V$ that CBSE expects you to apply directly to a gas expanding or being compressed inside a cylinder fitted with a movable, frictionless piston — showing exactly why pressure times volume change gives an energy, and why the sign works out correctly for both expansion and compression.

Step 1. Set up the cylinder. Let the gas occupy a cylinder of cross-sectional area $A$ , closed by a piston that can slide without friction. The gas pushes outward on the piston with a force $F = P\,A$ , where $P$ is the gas pressure at that instant — this is simply the definition of pressure as force per unit area, rearranged.

Step 2. Let the piston move outward by a small distance $dx$ while the pressure stays essentially constant over that tiny step. The elementary work done by the gas on the piston, by the ordinary mechanics definition of work as force times displacement in the direction of the force, is $dW = F\,dx = P\,A\,dx$ .

Step 3. Recognise $A\,dx$ as the small extra volume swept out by the piston, $dV$ . Substituting, $dW = P\,dV$ . This is the elementary work for one infinitesimal step; it already shows the sign works correctly, since an outward motion ( $dV>0$ ) gives positive work done by the gas, and an inward motion ( $dV<0$ , compression) gives negative work done by the gas — matching the CBSE/NCERT sign convention exactly.

Step 4. For a process at constant pressure $P$ from volume $V_1$ to $V_2$ , add up (integrate) every elementary step: $W = \int_{V_1}^{V_2} P\,dV = P\int_{V_1}^{V_2} dV = P(V_2 - V_1) = P\,\Delta V$ , since $P$ can be pulled outside a constant-pressure integral. For a general process where $P$ varies with $V$ , the same integral $W = \int_{V_1}^{V_2} P\,dV$ still holds, but it must be evaluated along the actual $P(V)$ path — which is exactly why $W$ depends on the path and not just the endpoints.

Worked examples

Board level - work done and internal energy change during expansion

An ideal gas expands from a volume of $2\times10^{-3}\,\text{m}^3$ to $5\times10^{-3}\,\text{m}^3$ at a constant pressure of $1.5\times10^5\,\text{Pa}$ , absorbing $700\ \text{J}$ of heat during the process. Find the work done by the gas and the resulting change in its internal energy.

Step 1. Find the volume change. $\Delta V = V_2 - V_1 = 5\times10^{-3} - 2\times10^{-3} = 3\times10^{-3}\,\text{m}^3$ . Since $\Delta V > 0$ , the gas expands, so we expect positive work done by the gas.

Step 2. Apply $W = P\,\Delta V$ at constant pressure. $W = 1.5\times10^5 \times 3\times10^{-3} = 450\ \text{J}$ . The gas does 450 J of work on its surroundings.

Step 3. Apply the sign convention to the given heat. The gas absorbs 700 J, so $\Delta Q = +700\ \text{J}$ directly, no sign flip needed.

Step 4. Use the first-law bookkeeping relation (previewed here, developed fully in the next topic) to find the internal energy change: $\Delta U = \Delta Q - \Delta W = 700 - 450 = 250\ \text{J}$ . Answer: the gas does 450 J of work on its surroundings, and its internal energy increases by 250 J.

Competitive level - compression with heat removed

A fixed mass of gas is compressed from a volume of $6\times10^{-3}\,\text{m}^3$ to $4\times10^{-3}\,\text{m}^3$ at a constant pressure of $2\times10^5\,\text{Pa}$ , while $300\ \text{J}$ of heat is removed from the gas. Find the work done on the gas and the change in its internal energy, and state whether the gas heats up or cools down.

Step 1. Find the volume change and apply the sign convention up front. $\Delta V = 4\times10^{-3} - 6\times10^{-3} = -2\times10^{-3}\,\text{m}^3$ , negative because the gas is compressed — so work done by the gas must come out negative too.

Step 2. Compute work done by the gas: $W = P\,\Delta V = 2\times10^5 \times (-2\times10^{-3}) = -400\ \text{J}$ . The negative sign means the gas does $-400\ \text{J}$ , equivalently 400 J of work is done on the gas by whatever is compressing it.

Step 3. Apply the sign convention to the heat: heat is removed, so $\Delta Q = -300\ \text{J}$ .

Step 4. Apply $\Delta U = \Delta Q - \Delta W = (-300) - (-400) = 100\ \text{J}$ . The internal energy increases by 100 J even though heat left the gas, because the surroundings did more work compressing it (400 J in) than the heat that escaped (300 J out); the net 100 J stays trapped as internal energy, so the gas's temperature rises — it heats up, not cools down, despite losing heat. Answer: 400 J of work is done on the gas, its internal energy rises by 100 J, and it gets hotter. Timing note: under ninety seconds once the sign convention is fixed first. The one trap that costs marks under time pressure is applying $\Delta U = \Delta Q - \Delta W$ before converting the given heat and volume-change directions into signed numbers — decide the sign of $\Delta V$ , then $W$ , then $\Delta Q$ , before touching the first-law arithmetic, and the rest is a single subtraction.

Common traps

Treating heat as something a gas "has." It is wrong to say a system "contains" a certain amount of heat — heat exists only as energy in transit across a boundary. Only internal energy is stored and state-dependent; once heat or work crosses the boundary, it simply becomes part of $U$ and can no longer be labelled "heat" or "work."
Mixing up the sign convention mid-problem. Fix the convention (CBSE/NCERT: $Q$ absorbed positive, $W$ done by the gas positive) before touching any numbers, and apply it to every quantity in the same problem — switching conventions halfway through is the single most common source of wrong answers.
Assuming $Q$ and $W$ depend only on the initial and final states. Only $\Delta U$ is path-independent; both heat and work depend on exactly how the process unfolds between the two states, which is why the same $\Delta V$ at different pressures, or a curved versus a straight path on a $P$ – $V$ diagram, gives different $W$ .
Forgetting that for an ideal gas $U$ depends only on $T$ . Two different processes sharing the same initial and final temperature give exactly the same $\Delta U$ , even when $Q$ and $W$ differ wildly between the two processes — this shortcut is heavily tested.
Confusing "gas loses heat" with "gas cools down." As the competitive example shows, a gas can lose heat and still get hotter if enough work is done on it — always finish the full $\Delta U$ calculation before concluding whether temperature rises or falls.

Common trapHeat is not a substance stored inside a system

The single biggest misconception in this topic is imagining heat as a fluid a body "holds," the way water fills a tank — an idea called the caloric theory, abandoned by physics well over a century ago but still the intuitive default for most students. Heat only exists while it is crossing a boundary, driven by a temperature difference; the instant it arrives, it stops being "heat" and simply adds to internal energy. You cannot ask "how much heat does this gas contain right now" any more than you can ask how much "work" a parked car contains — both questions mistake a transfer for a possession. Only $Q$ (transiting energy from a temperature difference), $W$ (transiting energy from a force through a displacement), and their net effect $U$ (the only one actually stored) are meaningful, and only $U$ is a state variable you can meaningfully speak of "having."

You should be able to…

Define heat, work and internal energy, and explain the physical difference between energy "in transit" and energy "stored."
State and correctly apply the CBSE/NCERT sign convention for $Q$ and $W$ to any expansion or compression problem.
Derive and compute work done by a gas at constant pressure using $W = P\,\Delta V$ , from the basic force-times-displacement definition of work.
Recognise work as the area under a $P$ – $V$ curve, and explain why that makes $W$ (and hence $Q$ ) path-dependent while $\Delta U$ is not.
Explain, using an ideal gas's $U$ depending only on $T$ , why two different processes between the same endpoint temperatures give identical $\Delta U$ despite different $Q$ and $W$ .
Combine the sign convention with $\Delta U = \Delta Q - \Delta W$ to determine whether a gas heats up or cools down in a given process.

Ask the tutor: Want to practise reading work done straight off a $P$ – $V$ diagram, or drill the sign convention with a few more mixed expansion-and-compression problems before the first law arrives? Ask the AI tutor for guided practice on heat, work, and internal energy.

If heat is not stored in a gas, why do we say a hot object radiates heat?

That everyday phrase is really shorthand for "this object is losing energy by heat transfer because it is hotter than its surroundings" — it describes the transfer happening right now, not a stock the object is carrying around. Once that energy has left the hot object and crossed into the surroundings, it stops being called "heat" and becomes part of the surroundings' internal energy instead. Physics keeps this distinction sharp precisely because it matters for calculations: you track $Q$ only as it crosses a boundary, and everywhere else you track $U$ .

Why does work depend on the path when it is basically just force times distance?

Because the force itself changes as the gas expands or compresses — pressure typically falls as volume grows and rises as volume shrinks, so $P$ in $W = \int P\,dV$ is not a constant number but a function that can follow many different curves between the same two endpoints on a $P$ – $V$ diagram. A slow, gradual expansion sweeps a different-shaped area under its curve than a sudden, jagged one, even between identical start and end states, so the two give genuinely different total work — that is the whole meaning of "path-dependent."

In the second worked example, how can the gas end up hotter if we removed heat from it?

Because internal energy responds to the net of heat and work together, not to heat alone. Here 300 J of heat left the gas, but 400 J of work was done on it by compression — more energy came in through the work channel than left through the heat channel, leaving a net gain of 100 J trapped as internal energy. Since internal energy of an ideal gas tracks temperature directly, that net gain means the gas actually warms up, even though it simultaneously lost heat. This is exactly the situation inside a bicycle pump getting warm as you compress it, even with no flame anywhere near it.

Is the equation delta Q equals delta U plus delta W something I need to derive here, or just use?

Just use it as a preview here — it is the first law of thermodynamics, and its full statement, derivation-level treatment and applications to specific processes (isothermal, adiabatic, and so on) are the very next topic in your syllabus. This article's job is to make sure you understand what each of the three symbols in that equation actually means and how to get their signs right, so that when the first law itself arrives, the equation is doing work you already understand rather than one more thing to memorise blindly.