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By Kent W. Mayhew

Logarithmic Functionality by Kent W. Mayhew

If entropy is not necessary to understand thermodynamic then why does its logarithmic functionality seemingly explain so much?

The answer is best understood by examining something like a gass ability to work. Consider that a closed system of gas is to do work, and part or all of this work is done onto the atmosphere as defined by W=PdV, which is lost work.Also see expanding piston-cylinder

Consider that the gas is an ideal monatomic gas with a kinetic energy defined by 3PV/2. Furthermore let us assume that the gas expands isothermally as it does work onto the atmosphere. In other words the expansion is slow enough (quasi-static) that a sufficient amount of thermal energy (AKA heat) can pass through the systems walls, hence the gas can remain isothermal during its expansion.

If the ideal gass kinetic energy is 3PV/2 =3NkT/2 and the gas remains isothermal during expansion then that gass kinetic energy does not change during expansion. One may ask how can this be because the  expanding  gas is doing work onto the surrounding atmosphere as defined by PdV.

Let us reexamine the expansion of a piston-cylinder as was discussed in our blog on expanding & compressing piston-cylinder. As the gas expands its pressure decreases while its volume increases. Furthermore, for a given pressure change (dP) the amount of work that the ideal gas can do onto the surrounding atmosphere decreases, as the gass volume increases. And in terms of the expandimng system's parameters we can (wrongly) write: (see parameters)

W = (NkT)In(Vf/Vi)    (1)

Although (1) defines the work done by such an isothermal  expansion in terms of the expanding systems parameters, the correct way  irreversible work (lost work) done onto the atmosphere remains: (see parameters)

W=(PdV)atm  = PatmdVsys     (2)

Eqn (2) defines the work done onto the atmosphere (with all the parameters being the atmospheres).  However eqn (1) defines the work in terms of expanding system's parameter, which is not exactly correct because the work is done onto the surrounding atmosphere! (see parameters)

Rate of work

What happens if one was to consider the rate of work being done onto the atmosphere. Understandably the greater the pressure difference between the expanding system and surrounding atmopshere is, then the faster the rate of work being done would be. So for the rate of work (dW/dt), one would obtain a logarithmic function along these lines:

dW/dt = -AIn(Pf/Patm)           (3)

where A is some constant. The negative sign tells us that the rate of work done onto the surrounding atmosphere by the expanding system is a decreasing function of time. Reason being as the pressure within the system approaches that of the surroundings, then the amount of time required for each unit pressure change (dP) increases.

Again we have not defined how the work is done. However eqn (3) applies to an expanding gas whose pressure is significantly greater than that of the surrounding atmosphere, and it is this gas's pressure difference that drives the expansion.  In such a situation, since the initial pressure represents the greatest pressure difference then the rate of work would be greatest during the initial expansion.

Consider that the piston-cylinder is fully insulated. Since it is fully insulated then its temperature will no longer remain constant. Specifically as the gas does work onto the surrounding atmosphere its temperature will decrease. Since temperature and pressure are now both decreasing, then its ability to do work is now decreasing at a rate that decreases significantly faster than that described by eqn (3). Again this would be a logarithmic function

Next imagine that the gas inside a piston-cylinder is being heated, and as such the pressure inside of the expanding piston-cylinder is only infinitesimally greater than that of the atmosphere e.g. a frictionless piston-cylinder. In this case the rate of work would be approximately constant.

We realize that logarithmic functionality is part of empirically measured thermodynamics, however to attribute this functionality to entropy walked us down a dangerous illogical path.

Rate of energy

Imagine that you are considering a set amount of energy that disperses. An example would be an exothermic reaction whose energy disperses into its surroundings. The rate of energy transfer into the surroundings would be a logarithmic function of the temperature difference between the surroundings and that associated with the location of the exothermic reaction. Accordingly we might expect the following equation to define the rate of heat transfer:

dE/dt = -BIn(Tf/Tsurrounding)           (3)

where B is some constant. Again the negative sign tells us that the rate of heat transfer is a decreasing function of time.

Discussion

In previous traditional thermodynamics the above logarithmic decreases would have been wrongly attributed in some fashion to entropy. It does explain why some traditional authors think one understanding for entropy is that it describes the quality of a systems energy i.e. Atkins. Even so, one begins to understand why entropy lacks clarity and why it is contemplated as having real importance when it may be nothing more than a poorly construed mathematical contrivance.

Blog:Logarithmic       This website is copyright of Kent W. Mayhew who in 2018 resides in Ottawa Ontario Canada

This website is full of new ideas, which are the property of Kent W. Mayhew.

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