= Isothermal process
{wiki}
An <isothermal process> changes a <thermodynamic system> while keeping its <temperature> constant. It need not keep its <pressure>, <volume (thermodynamics)> or <internal energy> constant, and it does not generally have zero <heat> transfer. An <adiabatic process> specifies zero <heat> transfer instead. For example, a slowly expanding <ideal gas> can absorb <heat> from a <heat bath> and deliver the same amount of <work> while its <temperature> stays fixed.
For a fixed <amount of substance> and composition, <internal energy of an ideal gas> depends only on <temperature>. With $Q>0$ for <heat> entering and $W_{\rm by}>0$ for <work> delivered to the surroundings, the <first law of thermodynamics> therefore gives
$$
\Delta U=Q-W_{\rm by}=0,\qquad Q=W_{\rm by}.
$$
This equality does not determine either transfer until the path is specified. <Isothermal ideal-gas work> gives the reversible value; an insulated <free expansion> has a different value even when its equilibrium endpoints have the same <temperature>. The <closed system> convention here excludes changes in bulk <kinetic energy> and external <potential energy>.
A conducting cylinder in a large, stirred <heat bath> approximates an <isothermal process> when the piston moves slowly enough for <thermal conduction> to keep up. A <quasistatic process> keeps the <gas> close to internal <thermodynamic equilibrium>; a <reversible thermodynamic process> additionally requires negligible <friction> and vanishing driving differences in <pressure> and <temperature>. Exact equality of the <gas> and <heat bath> <temperatures> gives no finite <heat transfer> rate through a finite <thermal resistance>. The reversible description is a limit of arbitrarily small differences and arbitrarily slow motion, illustrated by the <finite-rate isothermal piston model>.<University Physics Volume 2 by Samuel J. Ling and others/3.4 Thermodynamic Processes>[§3.4, “Isothermal Processes”, Figure 3.9 and the discussion of removing piston weights in infinitesimal steps]{ref}
For a <real gas>, <internal energy at constant temperature> can change with <volume (thermodynamics)> because molecular interaction <potential energy> changes. In one smooth <thermodynamic phase> of a fixed-composition <simple compressible system>,
$$
\left(\frac{\partial U}{\partial V}\right)_{T,n}
=T\left(\frac{\partial p}{\partial T}\right)_{V,n}-p.
$$
Thus an <isothermal process> generally has $Q=\Delta U+W_{\rm by}$, with no reason to set $\Delta U=0$. The right-hand side vanishes for the <ideal gas law>, but not generally for another <equation of state>.<Thermodynamics and Statistical Mechanics by Richard Fitzpatrick/Free Expansion of Gas>[Equations 6.151–6.154: the volume contribution to internal-energy changes]{ref}
Back to article page
