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Re: [Phys-l] T dS versus dQ



John was asked to render a calculation of the changes to be expected from a rapid compression of the cylinder. I have looked at the submissions and I see some disagreements, but I will use conventional thermodynamic definitions of terms like "adiabatic" (adj.) and "heat" (v.). I envision a horizontal cylinder with a piston that moves toward the right. Be kind to me; I'm going to work this out on the screen without writing it down first. The equations may be verrückt.

I will take "rapid" to mean that the speed of the piston is comparable to, but less than the speed of sound u at the initial gas temperature To and for the time being that the total compression is accomplished in less time than Lo/u, where Lo is the initial length of the cylinder of gas of cross-sectional area A. I assume the process is carried out adiabatically and, for this computational purpose, by moving the piston at constant velocity v to a new position Lf < Lo from the end of the cylinder, at which point the piston stops moving. I will comment on the end conditions of this process eventually.

I envision, in mid process at time t after the piston starts moving, that the gas in the cylinder consists of two parts, the boundary between which is a shock propagating to the right at speed u. On the left, moving with the speed of the piston, v, is a volume of gas of length v*t having a pressure P'. On the right, in front of the shock, is a stationary cylinder of gas of length

L(t) = Lo - u*t

at the initial pressure Po and temperature To.

One might at this point be tempted to calculate P' using the adiabatic gas law equation, but that would neglect the possibly considerable macroscopic kinetic energy imparted to the left-hand gas cylinder by the rapidly moving piston. (We will, however, neglect heating across the boundary.) Instead, we will assume that the gas in each of the two cylinders is in thermodynamic equilibrium, the gas on the left having (constant) temperature T'. The only part of the system in which thermodynamic changes are occurring is at the boundary.

The mass density of the gas on the right is given by the ideal gas law: Po*Lo*A = N*R*To, where N is the number of moles of gas and R is the universal gas constant. Let the molecular weight of the gas be M, and we find the mass density is

D = N*M / Lo*A = Po*M / R*To.

The rate of mass increase of the moving cylinder is

dm'/dt = u*A*D.

(It must be the same as the rate of mass decrease of the stationary cylinder.) The rate of doing work on the system is the same as the rate of increase in macroscopic kinetic energy of the system plus the rate of increase of internal energy of the system:

P'*A*v = [(1/2)*v^2 + c*(T' - To)]*(dm'/dt)

where c is the specific heat capacity (per unit mass) at constant volume of the gas. Combining we obtain

P'*v = [(1/2)*v^2 + c*(T' - To)]*u*[Po*M / R*To]

Thus we have a sum of two terms, the first of which is a kinetic energy term that would vanish for a quasistatic process. If that term is neglected the equation can be integrated over time to yield the usual adiabatic gas law. (Of course the constants must be adjusted to the more usual ones so it comes out neatly in gammas.) The more general equation is still integrable, and the energy expended in the compression will have a kinetic component. It takes more energy to compress the gas rapidly, and its final state (at time t = L'/v) is not in thermodynamic equilibrium. After the piston stops the gas in the cylinder will equilibrate. We will assume the cylinder containing the gas is firmly anchored to a large mass, so the kinetic energy will all appear as internal (thermal) energy after equilibration.

That's my solution. Finishing it should be easy. If you want it taken forward ... "You do the math."

Leigh

Disclaimer: I have constructed this solution over three days while otherwise occupied with Olympics & family & friends. I have changed the symbols a couple of times and have used some nonstandard ones (e.g. c instead of c_v - c was originally the speed of sound). Any errors you find are solely the responsibility of other members of my family (my physicist son isn't here); they couldn't possibly be mine!