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The his thermodynamics book, Zemansky refers to an example in which a gas in a piston is expanding for which the piston is accelerating.

It is obvious that no equation of state exists for the states traversed by a system that is not in mechanical and thermal equilibrium, since such states cannot be described in terms of thermodynamic coordinates referring to the system as a whole. For example, if a gas cylinder were to expand and to impart to a piston an accelerated motion, the gas might have, at any moment, a definite volume and temperature, but the corresponding pressure, calculated from an equation of state, would not apply to the system as a whole. The pressure would not be a thermodynamic coordinate, because it would depend not only on the velocity and acceleration of the piston but would also vary from point to point.

Extracted from "Heat and Thermodynamics" by Zemansky and Dittman, page 32.

He goes on to state that the pressure is not a thermodynamic coordinate in this case.

How is it possible to state that pressure is not a thermodynamic variable in case of an accelerating expansion of a gas ?

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  • $\begingroup$ Are you familiar with the property of viscosity, and the behavior of viscous fluids (which behave in a way that the stresses within the fluid are functions not only of the amount of deformation, but also on the rate of deformation)? $\endgroup$ – Chet Miller May 20 '18 at 21:57
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The behavior of a gas experiencing a rapid expansion (or compression) is fundamentally very different from that of the same gas under static conditions (i.e., at thermodynamic equilibrium), even if the piston velocity is not high enough to be comparable to the speed of sound. Because of the viscous character of gases, the behavior depends not only on amount of the volume change that occurs, but also on the rate at which the volume is changing (i.e., dV/dt). For example, a crude approximation to the force F exerted by the gas on the piston face (where the gas is doing work) is given by the equation: $$\frac{F}{A}=\frac{nRT}{V}-\frac{k(T)}{V}\frac{dV}{dt}\tag{1}$$where k(T) is a temperature-dependent gas physical property proportional to the gas viscosity. Note that, if the rate of change of volume approaches zero, the thermodynamic equilibrium behavior of the gas is recovered. Note also that, although Eqn. 1 has been written for the specific case of an ideal gas, the same basic principle applies for non-ideal gas equations of state.

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  • $\begingroup$ "... even if the piston velocity is not high enough to be comparable to the speed of sound ...". Why did you refer to the speed of sound ? $\endgroup$ – The Monk May 21 '18 at 7:55
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    $\begingroup$ The purpose of this aside was to address a factor alluded to in your original post. A gas, in addition to being viscous, also possesses (distributed) mass (inertia). If the piston is moving extremely rapidly (as, say, if its mass was very low), this would allow compression and expansion waves to travel down the length of the cylinder at the speed of sound and would result in substantial pressure variations with spatial position, such that a single unique value of the pressure could not be assigned to the gas. This would likewise contribute to the irreversible deformation behavior. $\endgroup$ – Chet Miller May 21 '18 at 11:11
  • $\begingroup$ Thank you sir, for both of the answer and the comment !!! $\endgroup$ – The Monk May 21 '18 at 11:28

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