# Work done by irreversible expansion of gas against piston of fixed pressure

this has been puzzling me for a while. An ideal gas expands against a piston with mass M and area A, so that pressure external Pex= Mg/A. The initial internal pressure of the gas is Pino > Pex. Because the gas is expanding irreversibly, Pin > Pex, until equilibrium is reached, where Pinf = Pex.

The physical chemistry texts I've read, including Atkins, seem to characterize the work done by the expanding gas as Pex * change in volume. I recognize this as the equation to calculate work for lifting an object at constant velocity over a given distance.

However, since the initial internal pressure of the gas is greater than the external pressure, there is an imbalance in forces, and the mass will accelerate upwards, causing it to acquire kinetic energy that was derived from the original energy of the gas. A rocket is an example of gas causing a mass to accelerate upwards.

My question is why the standard formula of work done by an irreversibly expanding gas is Pex * change in volume, and doesn't include a term for the kinetic energy from acceleration.

## 3 Answers

In an irreversible expansion like this, the aveage pressure of the gas within the cylinder is greater than the pressure acting at the internal piston face $P_{ext}$ (where the work is being performed). Of course, at the internal piston face, by Newton's 3rd law, the pressure exerted by the gas on the piston face is equal to the pressure exerted by the piston face on the gas.

The reason that, in an irreversible expansion, the average pressure of the gas (averaged over the volume) is greater than the value at the piston faces is that the gas pressure is non-uniform spatially within the cylinder. This is the result both of mass times acceleration effects within the gas itself (yes, the gas has mass) as well as viscous stresses present within the gas that are proportional, not to the gas volume, but to the rate at which the gas is deforming locally within the cylinder. So the mechanical behavior of a gas experiencing an irreversible expansion is very different from that of a gas experiencing a reversible (quasi-static) expansion.

Now let's reconsider the force balance on the piston. If $P_{gas}(t)$ represents the instantaneous pressure exerted by the gas on the piston at time t, then the force balance on the piston becomes: $$M\frac{dv}{dt}=P_{gas}(t)A-Mg$$If we multiply both sides of this equation by v=dx/dt and integrate with respect to t (from time = 0 when the piston is released), we obtain: $$\int_0^{x(t)}{P_{gas}Adx}=W(t)=\frac{1}{2}Mv^2-\frac{Mg}{A}[V(t)-V(0)]$$where W(t) is the work done by the gas on the piston up to time t, v is the velocity of the piston at time t, and V(t) is the volume of the gas at time t.

Basically what will happen here at short times is that the piston will rise and gain velocity, but it will overshoot its equilibrium position (like a mass on a spring experiencing simple harmonic motion). It will then slow down, until it begins moving downward. But it will again overshoot the equilibrium position.

If the piston were frictionless and the gas were inviscid, this oscillation would continue forever. However, even if the piston were frictionless, the gas is not inviscid. So the viscosity of the gas will act as a damping factor, and the piston, rather than oscillating forever, will experience a damped oscillation until if finally comes to a stop at the equilibrium position (i.e., when v = 0). So the equation for the work will become: $$W(\infty)=\frac{Mg}{A}(V_{\infty}-V(0))$$ So, even for a piston with mass, the kinetic energy effect will die out and the work will be the same as just the change in potential energy of the piston.

• Thank you for the explanation - I had an intuition that there might be an oscillation. Can I ask a followup? – lamplamp Jul 21 '17 at 19:15
• Is the energy of the oscillation lost as heat due to dampening? Is this heat loss the difference in work output between an irreversible and reversible expansion? – lamplamp Jul 21 '17 at 19:24
• As a reply to your first question, I suggest reading the following analysis I did to get an understanding of reversible vs irreversible work using a spring-damper analogy: physicsforums.com/insights/… – Chet Miller Jul 21 '17 at 19:39
• In the reversible expansion, of course, there is no overshoot. – Chet Miller Jul 21 '17 at 19:40
• I meant to mention in the article that the "piston" is supposed to be massless. Also, in this system, there is no terminal velocity because, as time progresses, the spring is carrying more of the load and the damper is carrying less. – Chet Miller Aug 3 '17 at 12:29

(I'm solving this assuming the container and piston to be adiabatic).

Now, Force (on piston)= (Pint-Pext)A - Mg = M(accl.) and as no net energy loss, we can say that,

Work Done (by gas)= (Pint)A(dx)= [(Pext)A(dx) + Mg(dx) + M(accl.)(dx)]. This is the actual equation of the work done by the gas. Now, as there is no theoretical way of deducing the accelaration of the piston without involving more than one variable in the integral(thus making the mathematics lengthy), we have to assume that the piston is either massless, or its accelaration negligible(which it practically is) to ignore M(accl.)(dx) term.

Otherwise, theoretically, you are correct in saying that the change in Kinetic Energy of piston should be included in the work equation, we just choose to ignore it for calculation purposes.

The point that is being missed here is that the Pressure to be used is neither Pint nor Pext, but Poperating, which depends on the velocity of the piston. Gas molecules striking the moving piston gain or lose speed depending on direction, affecting the energy of the gas. An approximation is

Poperating = Pint(1 +/- u/v)^2 where u is the velocity of the piston and v is the average velocity of the gas molecules in one direction.

"Thermodynamic Calculation of Work for Some Irreversible Processes" Gary L Bertrand, J of Chemical Education, Vol 82, pp 874-877, June 2005

The lifting of a mass by a gas is usually slow compared to molecular velocities, so Poperating is essentially the sane as Pint. For a gas expanding into a vacuum, the imaginary piston moves with the average velocity of the molecules so Poperating is zero. This relationship is expected to fail for compressions approaching the speed of sound.

• Without formatting of equations this doesn't look good... – Mithoron Mar 7 '18 at 20:31