0
$\begingroup$

Assuming we have an ideal gas expanding against a pistom in a cylinder isothermally, then the pressure drops due to viscous strain within the gas and we can extract less work than for an reversible expansion. But if we want to compress the gas irreversibel, back to the initial state the pressure must be higher than for the expansion to overcome the viscous strain. What I do not understand is that the integrated PV work is higher for.the compression than for the expansion though we have as far as I know the same work loss as friction and the same Volume change. Could you please help me?-

$\endgroup$
0
$\begingroup$

If the process is irreversible and the gas is undergoing viscous strain, then temperature and pressure gradients will exist throughout the volume in the gas during the process. So the process can’t be considered isothermal as there would be no defined temperature (or pressure) throughout the volume of the gas until the end of the process when thermal and mechanical equilibrium is attained.

Hope this helps.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you. It helps a bit. Do you have further tips? $\endgroup$ – Martin sssssssss Jan 8 at 8:01
0
$\begingroup$

The work done by the gas on the piston during irreversible expansion is less than the work done by the piston on the gas during irreversible compression. In either expansion or compression, a rough approximation to the force F on the piston in the irreversible behavior of the gas is given by: $$\frac{F}{A}=\frac{nRT}{V}-\frac{\mu}{V}\frac{dV}{dt}$$where the first term represents the equilibrium equation of state behavior of the gas, and the second term represents the viscous damping. The parameter $\mu$ is proportional either to the actual viscosity or to the turbulent viscosity (depending on whether the flow is turbulent). Note that, in expansion, the viscous term makes a negative contribution to the force on the piston, while, in compression, the viscous term makes a positive contribution to the force on the piston.

For more discussion of the mechanistic picture of irreversible expansion and compression of gases (and the analogy to spring-damper systems), see the following link: https://www.physicsforums.com/insights/reversible-vs-irreversible-gas-compressionexpansion-work/

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ My question was upon reading your text because I did not understand how this works. How can viscosity reduce the pressure for expansion but increases on the other hand the compressive force? $\endgroup$ – Martin sssssssss Jan 9 at 19:35
  • $\begingroup$ In expansion, dV/dt is positive and, in compression, dV/dt is negative. To understand how viscosity can give rise to both compressive and tensile contributions to normal stress, you need to familiarize yourself with the derivation of Newton's law of viscosity in 3D. For this, I refer you to Chapter 1 of Transport Phenomena by Bird, Stewart, and Lightfoot. $\endgroup$ – Chet Miller Jan 9 at 21:55

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.