Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Physics
Classical Physics
Thermodynamics
External Pressure vs. Internal Pressure when considering work
Reply to thread
Message
[QUOTE="Chestermiller, post: 6547540, member: 345636"] This answer is incorrect. The ideal gas law (or any real gas equation of state) only applies to a gas at thermodynamic equilibrium (or to a gas experiencing a reversible expansion or compression, which consists of a continuous sequence of thermodynamic equilibrium states). In a gas experiencing a rapid irreversible deformation, the properties of density (inertia) and viscosity come into play; both of these cause the force per unit area exerted by the gas on the inside face of the piston to deviate significantly from the pressure calculated from the ideal gas law. The inertia of the gas allows compression and expansion waves to travel through the gas so that the compressive stress in the gas is not even spatially uniform. The viscosity of the gas allows anisotropic viscous tensile or compressive stresses to develop within the gas that are proportional to the derivatives of the gas velocity with respect to spatial position and add to the equilibrium thermodynamic pressure. You will learn about viscous stresses when you study fluid mechanics. The bottom line is that the ideal gas law cannot be used to calculate the force per unit area on the piston face in a rapid irreversible deformation. Incorrect again. By Newton's 3rd law, the force per unit area exerted by the inside face of the piston (which we usually call the "external pressure") on the gas is exactly equal to the force per unit area exerted by the gas on the inside face of the piston. But, as we've already noted, the force per unit area exerted by the gas on the inside face of the piston is not equal to that predicted by the ideal gas law. So to solve problems involving rapid irreversible expansions or compression of ideal gases, we have a dilemma. How are we supposed calculate the work done by the gas in applying the first law of thermodynamics? Thoughts? I hope that this exercise so far has improved your understanding of internal and external "pressure" in a rapid irreversible deformation. [/QUOTE]
Insert quotes…
Post reply
Forums
Physics
Classical Physics
Thermodynamics
External Pressure vs. Internal Pressure when considering work
Back
Top