Tuesday, March 31, 2020

Looking At Entropy and The 2nd Law Of Thermodynamics

Two general statements applicable to heat engines-cyclical machines, provide a basis for the Second Law of Thermodynamics: 1)  The Kelvin-Planck statement, and 2) the Clausius Statement. Hence, it is worthwhile to first examine each in turn:

I) The Kelvin -Planck statement:

It is impossible to construct a heat engine, operating in a cycle, which produces no other effect than absorption of thermal energy from a hot reservoir and the performance of an equal amount of work.

Does a refrigerator not qualify? NO! Because it is simply a heat pump (cf. Fig. 1) operating in reverse. Thus in this case the engine absorbs heat Qc from the low temperature or cold reservoir and expels heat Qh to the hot reservoir. In other words, it can be depicted by Fig. 1 with the thermodynamic arrows reversed



Then, the work done is: - W = Qc - Qh or, Qh = Qc + W (e.g. heat given up to hot reservoir equals the heat absorbed from the cold reservoir).


II): The Clausius statement:

It is impossible to construct a cyclical machine that produces no other effect than to transfer heat continuously from one body to another at higher temperature.

*Entropy:

Given the Kelvin-Planck and Clausius statements regarding the 2nd law of thermodynamics, it is evident that since a perfectly 100% efficient cannot be constructed, then inefficient engines will reign and that means waste heat given off or manifesting as increased disorder. This disorder,  which refers to cohesion of states in matter is what we call entropy, often denoted by the symbol S.

All isolated systems then tend to a state of disorder and entropy is a measure of that disorder. A general result (from a field of physics known as statistical mechanics) which can be stated is:

“The entropy of the universe tends to increase in all natural processes.”

In thermodynamics at this level, however, what most concerns us is the change in entropy of a system.  A general principle to do with this can be stated:

The change in entropy DS of a system depends only on the properties of the initial and final equilibrium states.”

Example

In the case of an engine performing a Carnot cycle  (Fig. 2)running between hot and cold temperatures Th  and Tc

One finds:

DS =  Q h/Th   -  Q c / Tc  

And since we showed previously for the Carnot cycle:

Q h/Th   =   Q c / Tc  

Then:  DS =  0

One can generalize to state that for any reversible cycle:



dQ r /T  = 0

Which implies that the entropy of the universe remains constant in any reversible process.

 Quasi-static reversible process (Ideal Gas):

Of more practical application is the quasi-static reversible process, say applied to an ideal gas. In this case, we consider an ideal gas which goes from an initial state of temperature and volume (Ti, Vi)   to a final state (Tf, Vf).    

By the first law of thermodynamics:

dQ = dU + dW = dU + p dV

For an ideal gas:

dU = n C v,m dT and P = nRT/V

So that:

dQ = n C v,m dT +  nRT (dV/V)

To integrate the preceding, we need to divide through by T:

Þ  dQ/ T = n C v,m dT/T +  nR (dV/V)

And this is to be integrated between limits corresponding to the initial (i) and final (f) states. Thus the change in entropy, DS :

DS = òf i   dQ/ T = n C v,m   òf i   dT/T +  nR  òf i  (dV/V)


DS =   n C v,m    ln (Tf/Ti)  + nR ln (Vf/Vi


Change in Entropy for Reversible Process:

In the case of a real, irreversible thermodynamic process, consider:-

a)The case of heat conduction and the one way loss of heat (Q) from a hot reservoir (at temp. Th ) to a cold sink (at temp. Tc). Then at the cold sink  heat increases by Q  / Tc  while at the hot source, heat decreases by Q / Th  . The change in entropy is then:


DS =  Q  / Tc  - Q /Th    or, since Tc  < Th  :

 D S >  0

b) Free expansion:

We consider a treatment of a system equivalent to an isothermal, reversible expansion such that W = 0, Q = 0 and DU =  0, we have:

DS = òf i   dQ/ T = 1/T òf i   dQ


Here: dQ = W(i® f) = nRT  ò Vf Vi  (dV/V) = nRT ln (Vf/Vi

Note:

In the preceding case, the process must be performed very slowly to approximate an adiabatic free expansion.

Thought Challenge:

Technically speaking, the 2nd law of thermodynamics applies only to closed systems. Solar radiation injects on average 1360 watts per square meter onto the Earth's surface, or 1360 J each second per sq. meter.

Given this fact, and that plants absorb a good deal of this for the process of photosynthesis, show why the creationist argument that "evolution violates the 2nd law of thermodynamics" doesn't hold up. (Hint: NO quantitative work is needed!)

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