How Slow is “sufficiently slow” ?

A  quasi-static process that is reversible would be one which takes an infinitely long time to complete and during which the system and the surroundings are infinitessimally close to equilibrium. Sadly both are ideal situations and are not really achievable in practice

However, if I can execute the process sufficiently slow, I should be able to get close to reversibility. So the question is:

How slow do we have to go to get the reversible quasi-static assumption to be valid?

It depends…… Depends on what???

Let’s see…

Consider a system(piston) that is to be led along the quasi-static adiabatic process.

The constraint are to be moved slowly step by step, the system being permitted at each step to come to a new equilibrium state while fulfilling the condition PV^\gamma=constant. After a slight outward motion of the piston we must wait until the system fully achieves equilibrium, then we proceed with the next movement and we wait again, and so forth. Although this is the theoretical prescribed procedure, the practical realization of the process seldom follows this prescription. In practice the constraints usually are, relaxed continuously, at some “sufficiently slow” rate.

The characteristic time tau it takes for a system to reach equilibrium called the relaxation time. Although thermodynamics is a powerful theory that has been successfully predicts the fate of so many physical systems with some environment, strictly speaking thermodynamics does not deal with what is happening during relaxation. It describes the initial and final states only.

So, if the adiabatic expansion of the gas in the cylinder is performed in times much longer than this relaxation time the expansion occurs reversibly and isentropically. If the expansion is performed in times comparable to or shorter than the relaxation time there is an irreversible increase in entropy within the system and the expansion. The external pressure is decreased too rapidly that the resulting rapid motion of the piston is accompanied by turbulence and inhomogeneous flow within the cylinder. The process is then neither quasi-static nor reversible.

To estimate the relaxation time we first recognize that a slight outward motion of the piston reduces the density of the gas immediately adjacent to the piston. If the expansion is to be reversible this local “rarefaction” in the gas must be homogenized before the piston again moves appreciably. The rarefaction itself propagates through the gas with the velocity of sound, reflects from the walls of the cylinder and gradually dissipates. The mechanism of dissipation involves both diffusive reflection from the walls and viscous damping within the gas which is the damping due to imaginary part of the refractive index of sound wave in air.

Let’s assume that the second one is more dominant

Divide the air inside the piston into thin slices of thickness dx that are perpendicular to the direction of sound propagation.The intensity of sound wave emerges from the layer immediately adjacent to the piston will decay as it travel through distance because some of the them have run into particles and did not make it to the next layer or end up having lower average velocity. The change in intensity due to a slice of air is


To determine k, we need to know about the cross section of the particles . We assume that dx is sufficiently small that one particle in the slab cannot obscure another particle in the slab when viewed along the x direction. It follows that the fraction of particles collided when passing through this slab is equal to the total opaque area of the particles in the slab, \sigma nAdx, divided by the area of the slab A, which yields on \sigma ndx. Thus the fraction of particles collided when passing through the slab is given by:

\frac{dI}{I}=-\sigma ndx

Integrating both sides we get

I=I_{0}e^{-\sigma nx}=I_{0}e^{-\sigma nv_{s}t}

Where $v_{s}$  is the velocity of sound. Which is easily derived and given by

v_{s}=\sqrt{\frac{\gamma kT}{\mu}}

Thus the characteristic relaxation time or the time it takes for the the rarefaction pulse to be effectively dissipated is given by

\tau=\frac{1}{\sigma nv_{s}}=\frac{1}{\sigma n}\sqrt{\frac{\mu}{\gamma kT}}

So, if the adiabatic expansion of the gas in the cylinder is performed in times much longer than this relaxation time, the  quasi-static assumption is a good one.

Assume the following numerical values:

v_s=340 m/s

\sigma=3 \times 10^{-19} m^2 (hydrogen atom)

n=10^{23} m^{-3}

* the numerical value of the cross section of hydrogen is “stolen” from this paper

Then we get

\tau=9.8\times 10^{-8} s

Which is surprisingly  small… Therefore we are safe to assume the adiabatic condition in real life