Suppose we have an ideal gas filled inside a container fitted with a piston. The external pressure is equal to internal pressure.

If we suddenly decrease the external pressure, we say that the gas will expand until the internal pressure and external pressure become equal.

Considering an irreversible process, shouldn't the gas expand beyond that point because by that time the piston will have gained some velocity?


You're right. The piston can gain kinetic energy and overshoot its equilibrium position, but then, the force of the gas on the piston will drop until the piston comes back the other way. The piston will undergo an oscillation back and forth, but this oscillation will be damped as a result of viscous dissipation stresses in the gas. Eventually, the oscillation will decay until the piston stops. At this point, the gas pressure will be in equilibrium with the lower pressure outside.

If I do a force balance (Newton's 2nd law) on the piston, I get $$F_\mathrm{g} - p_0 A = m \frac{\mathrm{d}v}{\mathrm{d}t},$$ where $F_\mathrm{g}$ is the force the gas exerts on the piston, $p_0$ is the outside pressure, $A$ is the piston cross sectional area, $m$ is the mass of the piston, and $v$ is the piston velocity. If I multiply this equation by the velocity of the piston $v = \mathrm{d}x/\mathrm{d}t$, and integrate with respect to time, I obtain:
$$ W_\mathrm{g}(t) = \int\limits_0^{x(t)} F_\mathrm{g}\,\mathrm{d}x = p_0(V(t) - V(0)) + m\frac{v(t)^2}{2}, $$ where $W_\mathrm{g}(t)$ is the work done by the gas on the piston (its surroundings) up to time $t$, $V(t)$ is the gas volume at time $t$, and $m\frac{v(t)^2}{2}$ is the kinetic energy of the piston at time $t$. At infinite time, the piston is no longer moving, so its kinetic energy is zero at final equilibrium. Therefore, at final equilibrium, the work done by the gas is just $$W_\mathrm{g}(\infty) = p_0\Delta V, \text{ where } \Delta V = V(\infty)-V_0.$$

  • $\begingroup$ If we use damping factors to explain these points, then shouldn't we count them too while calculating work? We only count external pressure. $\endgroup$
    – Amogh
    May 19 at 19:23
  • $\begingroup$ No. That is internal. The net work done on the surroundings when the final equilibrium state is attained is just the outside pressure times the volume change. $\endgroup$ May 19 at 19:33
  • $\begingroup$ BTW, I can prove that if you want to see it. $\endgroup$ May 19 at 19:42
  • $\begingroup$ Please show it, if it can be done using only high school calculus. $\endgroup$
    – Amogh
    May 19 at 19:59
  • 1
    $\begingroup$ @ado sar Sure, as long as you recognize that, for the reversible case, the external pressure cannot be held constant, and has to be decreased gradually. $\endgroup$ Jul 18 at 13:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.