The first law in chemistry

Result

The first law in chemistry

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Statement

Internal energy \(U\) — the sum of every particle's kinetic and potential energy within a closed system — changes only through heat exchanged with the surroundings and work done on the system: \(\Delta U=q+w\). For expansion or compression against a possibly time-varying external pressure, \(w=-\int P_{\text{ext}}\,dV\); a reversible process (external pressure matching the system's own internal pressure at every instant) extracts the maximum possible work from an expansion, while any real, finite-rate (irreversible) process extracts strictly less — demonstrating that \(q\) and \(w\) individually depend on the specific path taken, even though their sum, \(\Delta U\), does not.

Why it matters

calorimetry already used \(\Delta U=q+w\) in its two simplest special cases (constant volume, constant pressure); this result establishes the fully general law and its molecular meaning, extending kinetic-theory-pressure's purely translational-kinetic-energy picture of an ideal gas to the complete internal energy of any system, including intermolecular and intramolecular potential energy. It also introduces the reversible/irreversible work distinction that directly motivates entropy-second-law, the next result in this unit: understanding why a slow, reversible expansion extracts more useful work than a sudden, irreversible one is the essential physical intuition entropy formalises.

Hypotheses

The system is closed: energy, but not matter, is exchanged with the surroundings.Open systems (allowing mass exchange as well) are a further generalisation beyond the scope of this result; ordinary sealed reaction vessels and calorimeters, as already used throughout this unit and calorimetry, satisfy the closed-system assumption. Only expansion/compression (\(PV\)) work is considered, with no other work type (electrical, magnetic, etc.) present.This matches the same assumption already used implicitly in calorimetry, appropriate for ordinary benchtop chemical reactions and phase changes; systems involving electrochemical work (electrochemistry, later in this curriculum) require an additional work term beyond \(PV\) work alone.

Proof

\Delta U = q + w

The first law itself: energy conservation applied to a closed system, with every possible energy change accounted for entirely by heat and work exchanged with the surroundings, and nothing left over. A

w = -\int_{V_1}^{V_2} P_{\text{ext}}\,dV

Work done on the system during a volume change depends, in general, on how the external pressure \(P_{\text{ext}}\) varies throughout the process, not merely on the initial and final volumes — a first hint that \(w\) alone is not a state function. A

\text{Reversible: } P_{\text{ext}}=P_{\text{gas}}\text{ at every instant} \Rightarrow w_{\text{rev}} = -\int_{V_1}^{V_2}P_{\text{gas}}\,dV = -nRT\ln\frac{V_2}{V_1}\ (\text{isothermal, ideal gas})

A reversible process is an idealised limit in which the external pressure is adjusted infinitesimally slowly to always match the system's own internal pressure, keeping the system in equilibrium throughout; for an isothermal ideal gas, substituting \(P_{\text{gas}}=nRT/V\) (ideal-gas-law) and integrating gives the explicit logarithmic work formula. A

\text{Irreversible (constant external pressure } P_{\text{ext}}\text{): } w_{\text{irrev}} = -P_{\text{ext}}(V_2-V_1)

A sudden expansion against a fixed, lower external pressure extracts strictly less work in magnitude than the reversible case between the identical initial and final volumes (Worked examples), since the gas is not held at equilibrium throughout the process. A

\text{For an isothermal ideal gas, } \Delta U=0 \Rightarrow q = -w

An ideal gas's internal energy depends only on temperature (it has no intermolecular potential energy, per van-der-waals-equation's Hypotheses, and its kinetic energy depends only on \(T\), per kinetic-theory-pressure); at constant \(T\), \(\Delta U=0\) exactly, so any difference between the reversible and irreversible work (Steps 3–4) must be exactly compensated by an equal and opposite difference in heat absorbed, \(q=-w\) — the direct, quantitative demonstration that \(q\) and \(w\) individually are path-dependent even though \(\Delta U\) (their sum) is identical in both cases. A

Result

\Delta U = q+w, \qquad w_{\text{rev}}=-nRT\ln\frac{V_2}{V_1}, \qquad |w_{\text{rev}}| \ge |w_{\text{irrev}}|\ (\text{expansion, same }V_1,V_2)

Reading. Internal energy is a state function (fixed by initial and final states alone), but heat and work individually are path functions (depending on exactly how the process was carried out) — the reversible-expansion work formula gives the largest magnitude either quantity can take between two given states.

Scope. The explicit logarithmic work formula (Step 3) assumes an ideal gas at constant temperature; the general path-dependence conclusion (Step 5) holds far more broadly, for any process where reversible and irreversible paths connect the identical initial and final states.

Corollaries & converses

The reversible work magnitude is always the theoretical maximum obtainable from a given expansion (or the theoretical minimum required for a given compression) between two fixed states; any real, finite-rate process necessarily does worse, extracting less useful work or requiring more work input — a direct preview of the "wasted" energy concept entropy-second-law formalises rigorously.
Since \(\Delta U\) is identical regardless of path, Hess's law (hess-law) applies equally validly to internal energy changes, not only to enthalpy changes — both are state functions, for exactly the same underlying reason.
Converse: if two different processes connecting the same initial and final states are observed to have different \(q\) and \(w\) individually but the same total \(\Delta U\), this is not a contradiction but the expected, guaranteed behaviour of any state function decomposed into path-dependent contributions (Worked examples).

Fails without

Assume \(q\) alone, or \(w\) alone, must be identical between any two processes connecting the same states (treat them as if they were state functions): the Worked example's explicit numeric comparison directly contradicts this — \(q_{\text{rev}}\ne q_{\text{irrev}}\) and \(w_{\text{rev}}\ne w_{\text{irrev}}\) even though \(\Delta U=0\) identically in both cases, since only their sum, not either individually, is guaranteed path-independent.
Use the constant-external-pressure work formula (\(w=-P_{\text{ext}}\Delta V\)) for a genuinely reversible process: this systematically understates the true magnitude of reversible work, since a reversible process's external pressure is not constant but continuously tracks the changing internal pressure throughout — only the integral form (Step 3) correctly captures reversible work.

Common errors

Treating \(q\) or \(w\) individually as state functions, rather than only their sum \(\Delta U\) (Fails without, first bullet).
Using the sign convention \(\Delta U=q-w\) (work done by the system, an older physics-textbook convention) while mixing in formulas or tabulated values that assume the IUPAC convention \(\Delta U=q+w\) (work done on the system, used throughout this result) — the two conventions are both used across different textbooks, and mixing them within a single calculation silently flips the sign of every work term.
Applying the constant-pressure work formula to a process that is actually reversible, or vice versa (Fails without, second bullet).

Discussion

The first law of thermodynamics was established through the 1840s by James Joule, Julius Robert von Mayer, and Hermann von Helmholtz, working largely independently, cementing energy conservation as a universal physical principle — work that, as already noted in hess-law and calorimetry, came after Hess's own 1840 empirically discovered chemical law, which the first law would soon be recognised as directly explaining.

The reversible-versus-irreversible work distinction, while introduced here purely through the mechanical example of gas expansion, is one of the most consequential ideas in all of thermodynamics: it is the direct conceptual seed of entropy (entropy-second-law), and more broadly of the second law's statement that no real, finite-rate process can be perfectly efficient — some capacity to do useful work is always, unavoidably, lost to irreversibility in any real process, a conclusion whose quantitative, general form is developed in the very next result of this unit.

Common misconception: that a "reversible process" describes something that literally happens in the real world, perhaps very slowly. A reversible process is a strict mathematical idealisation — infinitely slow, remaining in equilibrium at every instant — that no real, finite-rate physical process can ever fully achieve; it serves as a theoretical limiting case (the best possible work extraction or the least possible work input) against which real, always-somewhat-irreversible processes are compared and evaluated.

Worked examples

1.00\,\text{mol ideal gas, isothermal at } 298\,\text{K, expanding from } 10.0\,\text{L to } 20.0\,\text{L, reversibly}: \quad w_{\text{rev}}=-(1.00)(8.314)(298)\ln(2)\approx-1717\,\text{J}

A direct application of Step 3's logarithmic formula; the negative sign indicates work is done by the gas on the surroundings during this expansion (energy leaves the system as work). A

\text{Same expansion, irreversibly, against a constant external pressure } P_{\text{ext}}=P_2=\frac{nRT}{V_2}: \quad w_{\text{irrev}}=-P_2(V_2-V_1)\approx-1239\,\text{J}

Since \(\Delta U=0\) for both processes (identical initial and final temperature, ideal gas), \(q_{\text{rev}}=+1717\,\text{J}\) and \(q_{\text{irrev}}=+1239\,\text{J}\) — different heat absorbed in each case, despite \(\Delta U\) being exactly zero both times, directly confirming Step 5's central claim. A

|w_{\text{rev}}|\approx1717\,\text{J} > |w_{\text{irrev}}|\approx1239\,\text{J}; \qquad q_{\text{rev}}\ne q_{\text{irrev}}\text{ despite identical }\Delta U=0

Reading. The reversible pathway between two identical states extracts noticeably more work (and correspondingly absorbs more heat) than the irreversible pathway, a concrete, fully numeric demonstration of \(q\) and \(w\)'s path-dependence.

Scope. The magnitude comparison \(|w_{\text{rev}}|>|w_{\text{irrev}}|\) holds generally for any expansion compared against a sudden, constant-external-pressure alternative between the same two volumes.

Problems

Repeat the reversible work calculation from Worked example 1 for \(2.00\,\text{mol}\) of ideal gas (instead of \(1.00\,\text{mol}\)) expanding isothermally at \(298\,\text{K}\) from \(10.0\,\text{L}\) to \(20.0\,\text{L}\). Compare with the original \(1.00\,\text{mol}\) result and explain the relationship.
Solution
\(w_{\text{rev}}=-nRT\ln(V_2/V_1)=-(2.00)(8.314)(298)\ln(2)\approx-3435\,\text{J}\), exactly double the \(1.00\,\text{mol}\) result (\(\approx-1717\,\text{J}\)), since \(w_{\text{rev}}\) is directly proportional to \(n\) with every other quantity held fixed.
Now consider the reverse process: compressing the same \(1.00\,\text{mol}\) ideal gas isothermally at \(298\,\text{K}\) from \(20.0\,\text{L}\) back to \(10.0\,\text{L}\), first reversibly, then irreversibly against a constant external pressure equal to the final pressure \(P_1=nRT/V_1\). Compute both work values and compare their magnitudes.
Solution
Reversible: \(w_{\text{rev}}=-nRT\ln(V_1/V_2)=-(1.00)(8.314)(298)\ln(0.5)\approx+1717\,\text{J}\) (positive: work is done on the gas during compression). Irreversible (constant \(P_{\text{ext}}=P_1\)): \(w_{\text{irrev}}=-P_1(V_1-V_2)=+nRT\left(1-\dfrac{V_2}{V_1}\right)\times(-1)\cdot(-1)\approx+2478\,\text{J}\). Here \(|w_{\text{irrev}}|>|w_{\text{rev}}|\) — the opposite-direction version of the Result's inequality: irreversible compression requires more work input than reversible compression between the same two states, exactly complementary to Worked example 2's finding that irreversible expansion extracts less work than reversible expansion.
Explain, using only the first law and the state-function property of \(\Delta U\), why it is not a contradiction for \(q_{\text{rev}}\ne q_{\text{irrev}}\) in Worked example 2, even though both processes start and end at identical states.
Solution
The first law states \(\Delta U=q+w\); \(\Delta U\) is fixed once the initial and final states are fixed (a state function), but this only constrains the sum \(q+w\), not \(q\) and \(w\) individually. Since \(w_{\text{rev}}\ne w_{\text{irrev}}\) (Step 4, Worked example), and \(\Delta U\) is the same (\(=0\)) in both cases, \(q\) must differ by exactly the same amount that \(w\) differs, in the opposite direction, to keep their sum constant — \(q_{\text{rev}}-q_{\text{irrev}}=-(w_{\text{rev}}-w_{\text{irrev}})\) exactly. This is expected, not contradictory, behaviour for path functions summing to a state function.
A student claims that since \(w_{\text{rev}}\) always has the largest magnitude of any possible pathway between two states, a reversible process is always the "best" real-world choice for running an industrial gas expansion. Explain what is misleading about this claim, using the Common misconception discussion.
Solution
A truly reversible process requires infinitely slow, quasi-static change, remaining in equilibrium at every instant — no real, finite-rate industrial process can actually achieve this. While the reversible limit does correctly identify the theoretical maximum extractable work between two states, it is not itself a practically achievable process; real industrial processes must trade off closer-to-reversible operation (extracting more useful work, but requiring impractically slow operation) against practical considerations of speed, cost, and finite operating time, and always fall somewhat short of the true reversible-limit work value.