The First Law of Thermodynamics

Draw the boundary before you write a single term

Thermodynamics starts the way statics starts: you isolate something. In statics that is a free-body diagram; here it is a system, meaning the matter or the region of space whose behavior you intend to track. Everything else is the surroundings. Between them sits the boundary, which is imagined rather than physical, and which is where every transfer you are about to count must cross.

How you draw it determines the classification, and the classification determines which terms appear in your equation. A closed system contains a fixed quantity of matter. Energy may cross the boundary as heat or work, but mass may not. A sealed piston-cylinder is the standard case: the boundary is the cylinder wall plus the moving face of the piston, and it deforms as the piston travels. An open system, also called a control volume, is a region of space that mass flows through. A pipe, a compressor, a turbine, a nozzle, a heat exchanger, and the condensing unit outside a house are all analyzed this way. Mass crossing the boundary is the defining feature; heat and work may or may not also cross. An isolated system exchanges neither mass nor energy. It is a hypothetical construct, useful for arguments, and no real device achieves it.

This is not bookkeeping pedantry. The same physical energy transfer can count as heat for one choice of boundary and as work for another. If you have not said where the boundary is, the terms in your equation do not yet have meaning, and neither do their signs. Almost every sign error a student cannot debug traces back to a boundary that was never drawn, or that quietly moved partway through the problem.

State functions, path functions, and why the notation is asymmetric

Take a gas from one equilibrium state to another by two different routes. Measure the heat that flowed in and the work the gas did on the way. Both numbers depend on the route. Run the experiment again with a slower expansion or a different intermediate pressure and you get different numbers again.

The observation that founds thermodynamics is that one particular combination does not vary: the heat in minus the work out is the same for every route between those two states. A quantity whose change depends only on the endpoints is behaving like a property, so we name it. That property is the internal energy, written U, and it is the total energy stored in the matter of the system by virtue of its microscopic motion and interactions.

This is why the notation is asymmetric, and the asymmetry is worth taking seriously. We write delta-U because a system has an internal energy in each state and you can subtract them. We never write delta-Q or delta-W, because a system does not possess a heat and does not possess a work. Heat and work are transfers, things that happen at the boundary during a process, not contents that sit inside waiting to be counted. A system in a given state has no history of how it got there, so asking how much of its energy is heat is a category error. Quantities like U that depend only on the state are called state functions; quantities like Q and W that depend on the route are path functions.

The equation, and the convention you must declare

For a closed system whose kinetic and potential energy do not change, the first law is

delta-U = Q - W

with Q counted positive when heat enters the system and W counted positive when the system does work on its surroundings. Heat leaving is negative Q; work done on the system, as in a compression, is negative W. That is the convention used in physics and in most mechanical engineering thermodynamics texts, and it is the one this lesson uses throughout.

It is not the only one in print. Chemistry texts, including OpenStax Chemistry, write the first law as delta-U = Q + W and define W as the work done on the system. Both are correct. The second convention flips the sign of the symbol W as well as the operator in front of it, so the two cancel and the physical answer is identical. What is not correct is mixing them, or reading a value out of one book into an equation from another. Before you compute anything, write down which convention you are using and what a positive W means in it. If you inherit a problem statement, find out what its author meant.

The units are unambiguous even when the signs are not. The joule is the SI coherent derived unit of energy, work, and amount of heat alike, so U, Q, and W all carry joules and there is no conversion hiding between the terms. Carry the unit through every line.

Boundary work and the p-V diagram

The work mode that dominates introductory thermodynamics is boundary work: the work done as a system's boundary moves, typically a gas pushing a piston. Suppose the gas is at pressure p and the piston has area A, so the gas pushes with force pA. Let the piston move an infinitesimal distance dx, so the volume increases by dV = A dx. The infinitesimal work is dW = pA dx = p dV, and over a finite change

W = the integral of p dV from V1 to V2.

This expression requires the pressure to be defined throughout the process, which means the system must be in equilibrium at every instant. That is the quasi-static idealization: the change is imagined to happen infinitesimally slowly, so the system is always at equilibrium with itself and with its surroundings. Because no finite change can truly satisfy this, every real process is non-quasi-static to some degree, and for a genuinely fast process you cannot draw a path at all, only the endpoints.

When the path can be drawn, the integral has a picture. On a p-V diagram, with volume on the horizontal axis and pressure on the vertical, W is the area under the process curve. Expansion moves rightward and sweeps positive area; compression moves leftward and gives negative work. Two paths joining the same two states enclose different areas, so they involve different work, which is the path dependence of W made visible. A closed loop encloses a net area, and that area is the net work per cycle.

Boundary work is not the only work mode. Shaft work, electrical work, and spring work all cross boundaries too, and a stirrer driven by a falling weight does work on a fluid in a rigid tank whose volume never changes. Do not assume a rigid boundary means zero total work; it means zero boundary work.

Enthalpy, and the first law for flowing streams

Define a new property by combining ones you already have:

H = U + pV

Enthalpy is a state function because U, p, and V all are. Two things make it worth defining. First, consider a constant-pressure process in a closed system. There W = p(V2 - V1), so substituting into the first law and regrouping gives (U2 + pV2) - (U1 + pV1) = Q, that is, H2 - H1 = Q. At constant pressure the enthalpy change is simply the heat transferred. Second, and this is the reason engineers meet it everywhere, a fluid entering a control volume carries not only its internal energy but also the work required to push it across the inlet against the pressure there. That pushing term is exactly pV per unit mass, so enthalpy is the natural per-unit-mass accounting quantity for anything that flows.

For a control volume operating at steady state, nothing inside changes with time. Mass in equals mass out, and the energy balance at introductory depth reads

Qdot + sum over inlets of mdot(h + V^2/2 + gz) = Wdot + sum over exits of mdot(h + V^2/2 + gz)

where mdot is a mass flow rate, h is specific enthalpy, V is flow speed, and z is elevation. Each stream carries its enthalpy, its kinetic energy, and its potential energy across the boundary; Qdot and Wdot are the heat and shaft-work rates. In most textbook devices the kinetic and potential terms are small enough to drop, leaving a compact statement: for an adiabatic turbine, the shaft power out equals the mass flow times the drop in specific enthalpy. Nozzles, diffusers, throttles, pumps, and heat exchangers are all handled by deleting the terms that do not apply.

What the law reduces to for the standard processes

Most introductory problems are one of five idealized processes, and the value of the first law is that each one deletes a term.

Isochoric, constant volume: the boundary does not move, so boundary work is zero and delta-U = Q. Heat a gas in a rigid sealed tank and all of it goes into internal energy.

Isobaric, constant pressure: W = p(V2 - V1), a rectangle on the p-V diagram, and Q = delta-H. This is the case where enthalpy pays for itself.

Isothermal, constant temperature: for an ideal gas the internal energy depends only on temperature, so delta-U = 0 and Q = W. Every joule of heat that enters leaves again as work. On a p-V diagram the path is a hyperbola, since p is proportional to 1/V at fixed temperature.

Adiabatic: no heat crosses the boundary, so Q = 0 and delta-U = -W. A gas that expands adiabatically does work at the expense of its own internal energy and cools; compressed adiabatically it heats. This is part of why a bicycle pump warms as you work it.

Cyclic: the system returns to its initial state, so delta-U = 0 by definition of a state function, and therefore Q = W over the cycle. Net heat in equals net work out.

Perpetual motion, and what the first law refuses to tell you

The cyclic result is the death of one family of inventions. A perpetual motion machine of the first kind is a device that returns to its initial state each cycle and delivers net work while taking in no energy. But a cycle forces delta-U = 0, hence Q = W; with Q = 0 the net work must be zero as well. No arrangement of gears, magnets, or fluid loops evades this, because the argument never inspects the mechanism, only the boundary. The conclusion is treated as settled well beyond physics: the U.S. Patent and Trademark Office's examining manual notes that, apart from cases involving perpetual motion, a working model is not ordinarily required to show that a device operates.

Now the limit. The first law is a constraint, not a prediction. It says a process must conserve energy; it does not say the process will happen. Heat flowing spontaneously from a cold block into a hot one conserves energy perfectly. So does a shattered mug reassembling while the countertop cools slightly. Neither is observed. The first law permits many states that nature never produces, and identifying which direction a process actually runs requires the second law and the property of entropy. That is a separate lesson; this one does not cover it. If a problem asks which way something goes, or how efficient it can possibly be, the first law alone cannot answer.

Finally, name your idealizations. Ideal gas, quasi-static, adiabatic, negligible kinetic and potential energy, steady state: each is a modeling choice with a domain of validity, and results inherit the limits of the weakest assumption. This lesson is educational material, not engineering design guidance. Real thermal equipment, pressure vessels, and energy systems must be designed by a licensed engineer working to the governing code.