= Quasistatic process
= Quasi-static process
{synonym}
A quasistatic process is an ideal limit in which a changing <thermodynamic system> passes through a sequence of states arbitrarily close to <thermodynamic equilibrium>. Its intermediate <pressure>, <temperature> and <volume (thermodynamics)> are therefore well defined and obey its <equation of state>. Driving must be slow compared with the system's own equilibration, rather than merely taking a long time on a clock. A <reversible thermodynamic process> additionally allows both system and surroundings to be restored without net change; <friction> can produce <entropy> even in a <quasistatic process>.<Statistical Physics by David Tong/4 Classical Thermodynamics>[§§4.2–4.3: quasi-static and reversible processes]{ref}
For a computed example, take a fixed <amount of substance> $n$ of <ideal gas> in a sealed cylinder with a frictionless piston, in contact with a <heat bath> at absolute <temperature> $T$. Reduce the external load in infinitesimal steps so that the piston expands from $V_0$ to $2V_0$. In the reversible limit, the gas <pressure> exceeds the external <pressure> by an infinitesimal amount during expansion, and the <heat transfer> occurs across an infinitesimal <temperature> difference. Exact <mechanical equilibrium> would give no driving <force>; the limiting imbalance supplies motion while preserving equilibrium intermediate states.<University Physics Volume 2 by Samuel J. Ling and others/3.4 Thermodynamic Processes>[§3.4: a piston unloaded in infinitesimal steps]{ref}
Along this reversible <isothermal expansion>, the <ideal gas law> gives
$$
p(V)=\frac{nRT}{V},\qquad nRT=1\,\mathrm{kJ},\qquad V_0=1\,\mathrm L=10^{-3}\,\mathrm{m^3},\qquad
p_0=\frac{nRT}{V_0}=1\,\mathrm{MPa}.
$$
Here $R$ is the <molar gas constant>; specifying $nRT$ fixes an <energy> scale, without choosing $n$ and $T$ separately. The initial and final equilibrium states are $(V_0,p_0,T)$ and $(2V_0,p_0/2,T)$. Count <work> delivered to the surroundings as positive. The reversible <work> is the shaded <area> beneath the <pressure>-<volume (thermodynamics)> curve:
$$
W_{\rm out,rev}=\int_{V_0}^{2V_0}p(V)\,dV
=nRT\ln2\approx0.693147\,\mathrm{kJ}.
$$
The <integral> of $1/V$ supplies the <natural logarithm>. Since <internal energy of an ideal gas> depends only on <temperature>, $\Delta U=0$; the <first law of thermodynamics>, $\Delta U=Q_{\rm in}-W_{\rm out}$, gives $Q_{\rm in,rev}=0.693147\,\mathrm{kJ}$ to the displayed precision. The <heat bath> replaces the <energy> delivered as <work>.<University Physics Volume 2 by Samuel J. Ling and others/3.2 Work, Heat, and Internal Energy>[§3.2: pressure-volume work and the isothermal integral]{ref}
\Image[/quasistatic-isothermal-pv-work.png]
{title=Reversible <isothermal expansion>: <work> is the <area> beneath the equilibrium <pressure> curve}
{description=Fixed <amount of substance> of <ideal gas>, with $nRT=1\,\mathrm{kJ}$ and $V_0=1\,\mathrm L$. The blue curve gives every intermediate equilibrium state; its shaded <area> is $0.693147\,\mathrm{kJ}$. The arrow indicates expansion, without specifying a time dependence.}
Now start from the same initial state, but abruptly lower the external <pressure> to the constant value $p_{\rm ext}=p_0/2=0.5\,\mathrm{MPa}$. There is a finite initial <pressure> imbalance, so this is an irreversible expansion rather than a <quasistatic process>. Assume the piston begins and ends at rest, has no net change in stored <energy>, and has no external <friction>; relaxation occurs within the <gas> and its <heat bath>. After motion has settled and the <heat bath> has restored <temperature> $T$, <mechanical equilibrium> and the <ideal gas law> give $V_f=nRT/p_{\rm ext}=2V_0$. The <work> delivered to the constant load is
$$
W_{\rm out,load}=p_{\rm ext}(V_f-V_0)
=(0.5\,\mathrm{MPa})(1\,\mathrm L)=0.5\,\mathrm{kJ},\qquad
\Delta U=0,\qquad Q_{\rm in,load}=0.5\,\mathrm{kJ}.
$$
This <work> follows directly from load <force> times piston <displacement>, using $dV=A\,dx$ for piston <area> $A$. Intermediate gas states need not have a single equilibrium <pressure> or <temperature>. In particular, neither $p(V)=nRT/V$ at the bath <temperature> nor $p(V)=p_{\rm ext}$ describes the actual gas during this irreversible motion. The rectangle below plots the known external load; its upper edge is not an equilibrium gas path.
\Image[/reversible-versus-constant-load-work.png]
{title=The same equilibrium endpoints give different <work> for reversible and constant-load expansion}
{description=The blue curve is the reversible <ideal gas> path. The dashed orange line is constant external <pressure>, whose rectangle gives $0.5\,\mathrm{kJ}$ of load <work>. The additional blue shading is $0.193147\,\mathrm{kJ}$ available in the reversible case. Black points mark the common gas endpoints; the open orange point marks the initial external load, and the dotted gap indicates the initial imbalance. No irreversible internal gas path is drawn.}
The <entropy> accounting makes the difference precise. Because <entropy> is a <state function>, both expansions have the same gas <entropy> change, computed along the reversible <isothermal expansion>:
$$
\Delta S_{\rm gas}=\int_{V_0}^{2V_0}\frac{\delta Q_{\rm rev}}{T}=nR\ln2.
$$
The <thermal reservoir> stays at $T$ and loses <heat> $Q_{\rm in}=W_{\rm out}$, so $\Delta S_{\rm bath}=-W_{\rm out}/T$. Assume the ideal load stores <work> without an <entropy> change. The combined <entropy production> is therefore
$$
\Delta S_{\rm total}=nR\ln2-\frac{W_{\rm out}}{T}
=\begin{cases}
0,&\text{reversible expansion},\\
nR(\ln2-1/2)>0,&\text{constant external load}.
\end{cases}
$$
No numerical <entropy> unit is implied by $nRT=1\,\mathrm{kJ}$ alone: $nR$ or $T$ must also be specified. The lost <work> opportunity is $T\Delta S_{\rm total}=0.193147\,\mathrm{kJ}$ for this comparison. This calculation uses equilibrium endpoints and the reservoir's <temperature>, without assigning equilibrium intermediate states to the irreversible gas. https://openstax.org/books/university-physics-volume-2/pages/4-6-entropy[OpenStax, §4.6: entropy as a state function and reversible reference paths]{ref}
For another boundary choice, let the same <ideal gas> expand into an initially evacuated compartment of <volume (thermodynamics)> $V_0$ inside a rigid, insulated vessel of total <volume (thermodynamics)> $2V_0$. Removing the partition causes <free expansion>: no load is moved and no <heat> crosses the outer boundary, so $W_{\rm out}=Q_{\rm in}=0$ and $\Delta U=0$. After equilibration the <temperature> is again $T$, the <pressure> is $p_0/2$, and the gas <entropy> has increased by $nR\ln2$. The outer vessel has fixed <volume (thermodynamics)>; the gas's accessible <volume (thermodynamics)> doubles. Restoring the initial gas state requires compression and removal of <heat>, changing the surroundings. Equal endpoints do not make <free expansion> reversible. https://openstax.org/books/university-physics-volume-2/pages/4-1-reversible-and-irreversible-processes[OpenStax, §4.1: insulated free expansion and restoration]{ref}
A slowly moving piston with dry <friction> supplies a different caution: its gas can remain arbitrarily close to <thermodynamic equilibrium>, yet rubbing still converts <work> into <heat> and produces <entropy>. If the frictional <force> has a nonzero slow-motion limit, its dissipated <energy> over a fixed travel does not vanish with speed. Reversing the motion reverses <friction>, so infinitesimally changing the external load cannot retrace the same operation and restore the surroundings. The <heat capacity> calculations additionally exclude <friction> and other dissipative <work>.<University Physics Volume 2 by Samuel J. Ling and others/3.4 Thermodynamic Processes>[§3.4: a quasi-static piston with friction]{ref}
The usefulness of a <quasistatic process> is that it replaces detailed motion by a readable sequence of equilibrium states. An <equation of state> and an <integral> then turn a <pressure>-<volume (thermodynamics)> drawing into an <energy> budget. The beautiful distinction is that <internal energy> and <entropy> depend on the endpoints, while <heat> and <work> record how the change was performed. Reversible benchmarks expose how much <work> an irreversible device sacrifices, while <entropy production> explains why simply making it slower need not recover that opportunity.
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