Isothermal compression

Isothermal process

Why must a conducting piston move slowly? A simple model couples the <first law of thermodynamics> to <Newton's law of cooling>. Assume a fixed amount of <ideal gas> with uniform <temperature> $T(t)$ and constant whole-sample <heat capacity at constant volume> $C_V$, a <heat bath> maintained at $T_b$, negligible <friction>, and mechanically slow motion. Let $G>0$ be the effective thermal conductance, the reciprocal of the <thermal resistance>, in $\mathrm{W\,K^{-1}}$. Then
$$
\dot Q=G(T_b-T),\qquad p=\frac{nRT}{V},\qquad
C_V\dot T=G(T_b-T)-p\dot V.
$$
Positive <heat transfer> must offset expansion <work> to keep $T$ near $T_b$. A finite $G$ and nonzero piston speed therefore require a <temperature> difference. This uniform-<temperature> model neglects spatial <temperature gradients>, sensor response, cylinder <heat capacity> and any change of $G$ with piston position; it is an illustrative calculation, not a fit to the <isothermal gas syringe experiment>.

\Image[/isothermal-piston-thermal-bath-temperature-lag.png]
{title=Conducting piston in a thermal bath and modeled temperature lag during expansion}

For a prescribed exponential <volume (thermodynamics)> change $V=V_i e^{\lambda t}$, define $\tau=C_V/G$, $s=t/\tau$ and $\epsilon=nR\lambda/G$. The dimensionless <temperature> $\theta=T/T_b$ obeys a <linear ordinary differential equation> with initial value $\theta(0)=1$:
$$
\frac{d\theta}{ds}=1-(1+\epsilon)\theta,\qquad
\theta(s)=\frac{1}{1+\epsilon}
+\frac{\epsilon}{1+\epsilon}e^{-(1+\epsilon)s}.
$$
The plotted expansion uses stipulated $C_V=(5/2)nR$ and stops at $V_f=2V_i$, at $s_f=\ln2/[(5/2)\epsilon]$. During expansion, the larger $\epsilon$ produces the larger <temperature> deficit; once the piston is held fixed, the deficit undergoes <exponential decay> as $e^{-(s-s_f)}$. Reducing $|\epsilon|$ makes the path more nearly an <isothermal process>, while increasing its duration. Reversing the piston with $\lambda<0$ produces a <temperature> excess and outward <heat transfer>; the small-$|\epsilon|$ regime is the intended near-isothermal approximation. Mechanical slowness alone does not ensure sufficient <heat transfer>.


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 Incoming links: Finite-rate isothermal piston model