First law of thermodynamics

Heat is energy transfer caused by a temperature difference. Work covers other macroscopic transfer mechanisms, such as a shaft or electrical terminals. Internal energy is instead a state property. A body contains internal energy, not stored amounts of “heat” and “work”.[2]

You heat a gas while it moves a piston. Part of the transferred energy can increase its internal energy; another part leaves as work. The first law accounts for both. It answers a balance question, without determining how fast the process occurs or whether it is reversible.[1]

Draw the system boundary before calculating. For the gas alone, piston motion transfers work across the boundary. If the piston is included, the same transfer can initially change its kinetic energy. An energy contribution must not be counted both as exported work and as energy stored inside the same system.[3]

Consider a fixed amount of matter as a closed system: energy may cross its boundary but mass may not. Gas in a cylinder differs from a system containing gas, piston and surroundings together. The simplified balance $\Delta U=Q-W$ requires changes in bulk kinetic and potential energy to vanish or be negligible. If the entire system moves or rises appreciably, first balance total energy E.[3]

U contains microscopic motion and interaction energy according to the material model. Temperature alone does not determine it in general: phase changes and real gases require additional state information. The first law still applies. Flow through turbines and compressors needs a control-volume balance including energy carried by the mass streams. Heat capacity relates heat to a temperature change under the appropriate material and process assumptions; the first law alone does not supply that constitutive relation.

Both forms describe the same balance: use a minus sign for work by the system and a plus sign for work on the system. $Q>0$ means heat enters; heat rejected gives $Q<0$. $W>0$ means work performed by the system on its surroundings. External compression makes system work W negative. Terms such as heat absorbed and work delivered define signs; numbers must not simply be copied as positive.[1]

A negative work sign specifies the direction of energy transfer. With $Q=500\,\mathrm J$ and $W_{\mathrm{by}}=180\,\mathrm J$, internal energy rises by 320 J. If 180 J is instead done on the system, $W_{\mathrm{by}}=-180\,\mathrm J$, and the same absorbed heat gives a 680 J increase. These are two different processes.

The first law is an empirical conservation law, not an algebraic proof specific to ideal gases. A balance equates stored energy after minus before with energy entering minus energy leaving. Heat and work are separate transfers across the chosen boundary. With unchanged bulk energy contributions, this accounting gives $\Delta U=Q-W$.[1]

U has a value for each equilibrium state. Q and W are not stored material properties. Writing U₂ − U₁ is meaningful, whereas a supposed heat content Q₂ − Q₁ is not. Two paths can have different heat and work transfers even though their difference gives the same internal-energy change.[3]

For a quasistatic mechanically balanced path, gas pressure and external loading coincide in the limiting description, allowing integration of p(V). In rapid irreversible expansion, internal gas pressure cannot automatically be treated as a uniform path variable. Free expansion into vacuum delivers no external boundary work despite increasing volume. Friction and piston acceleration require appropriate additional mechanical accounting.

Isothermal means constant T, unlike adiabatic, which means $Q=0$. Ideal-gas U depends only on T, giving $\Delta U=0$ and $Q=W$ for isothermal state changes. Constant volume removes boundary work only: a stirrer can still supply work. A reversible adiabat additionally needs the appropriate gas relation; adiabatic alone does not establish reversibility.

The second path includes isothermal expansion of an ideal gas: its work depends on the path even though both paths reach the same final state. Compare the individual work and heat steps rather than only the initial and final temperatures.[1][2]

No: only pressure–volume work vanishes. A rigid vessel prevents volume work but not energy transfer through a stirrer shaft. The next example accounts for this work mode separately.[1][2]

Zero ΔU does not mean every stage has zero energy change. An engine receives heat and rejects part of it; only the signed sum over the entire cycle equals net work. Irreversible cycles can also restore their initial state. The working medium’s entropy then returns as well, while entropy can be generated in engine plus surroundings.

Conservation alone does not determine spontaneous direction or maximum work obtainable from heat. Complete cyclic conversion of heat from a single reservoir to work could formally balance energy but violates the second law. Carnot efficiency and entropy therefore supplement the calculation. The first law requires neither an ideal gas nor an actually reversible path; these assumptions belong to additional material or work relations.[4]

Isochoric means constant volume and hence no pressure–volume work; isobaric means constant pressure. Adiabatic means $Q=0$. These process conditions do not replace a material equation and do not by themselves determine other work transfers or reversibility.[1][2]

For a closed stationary system with no change in external energies, $\Delta U=Q-W_{\mathrm{by}}$. Establish the boundary and sign convention first, then add all heat and work contributions. If motion or height changes, the corresponding external energy changes also belong on the left-hand side.[1]