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.
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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