1
$\begingroup$

Say that I have a vertical thermally insulated cylinder which contains an ideal monoatomic gas under a weightless piston and the system is initially in equilibrium.

Now if a load of weight $W$ is kept on the piston,the piston would fall by some distance and the system would attain a new equilibrium. It is intuitive that the sum of the work done by gravity and the work done by atmospheric pressure would change into internal energy of the gas.

However, my book mentions that only the work done by gravity goes into the increase in internal energy of gas. It does not say anything about where the atmospheric work would go. Why is it so?

$\endgroup$
7
  • 1
    $\begingroup$ What is holding the system in equilibrium initially? Is the gas pressure initially equal to the external atmospheric pressure? $\endgroup$ – Chet Miller Apr 25 '20 at 11:47
  • $\begingroup$ Yes it is initially equal to external atmospheric pressure. $\endgroup$ – Dylan Rodrigues Apr 25 '20 at 12:17
  • $\begingroup$ Then the work done by the atmospheric pressure should be included. $\endgroup$ – Chet Miller Apr 25 '20 at 13:55
  • $\begingroup$ Actually the situation I have described is a problem of the book SS Krotov. In the solution they only included the work done by gravity which they equated with the change in internal energy of gas $\endgroup$ – Dylan Rodrigues Apr 25 '20 at 13:57
  • $\begingroup$ With all due respect to your book, that is incorrect. $\endgroup$ – Chet Miller Apr 25 '20 at 14:51
1
$\begingroup$

If we carry out a force balance on the piston in the initial equilibrium state of the system, we have $$F_{gi}=P_{atm}A$$where $F_{gi}$ is the force the gas exerts on the piston in the initial state and A is the area of the piston. Similarly, a force balance on the piston during the irreversible deformation that results when we suddenly apply a mass m to the piston reads: $$F_g(t)-mg-P_{atm}A=m\frac{dv}{dt}$$where $F_g(t)$ is the force exerted by the gas on the piston at time t during the deformation and v is the piston velocity. If we multiply this equation by the piston velocity v = dz/dt and then integrate up to time t, we obtain: $$W_g(t)=\int_0^t{F_g(t)dt}=mg\Delta z+P_{atm}\Delta V+m\frac{v^2}{2}$$where $W_g{t}$ is the work done by the gas on the piston up to time t and the 3rd term on the right hand side of the equation is the kinetic energy of the mass.

After infinite time passes, the system equilibrates, and the piston and mass come to rest. At this final steady state, the work done by the gas on its surroundings (the piston) is$$W_g=mg\Delta z+P_{atm}\Delta V$$So there is no doubt that the work done by the gas must include the work done against the external atmosphere. And this is the work that causes the internal energy to change.

$\endgroup$
3
  • $\begingroup$ You've assumed that process is reversible but what if it is irreversible? $\endgroup$ – Dylan Rodrigues Apr 26 '20 at 18:47
  • $\begingroup$ I've assumed that the process is irreversible, not reversible. One aspect of this is the damping of the mass kinetic energy. There is nothing I have written to suggest that I have treated the process as reversible. $\endgroup$ – Chet Miller Apr 26 '20 at 19:14
  • 1
    $\begingroup$ I'm curious. What is it about my analysis that made you think it assumes a reversible process; certainly not the use of Newton's 2nd law, since we use that all the time to analyze kinetic friction processes (which are obviously irreversible). $\endgroup$ – Chet Miller Apr 26 '20 at 21:24

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.