Why does dS = dQ / T in Entropy?

[Laix]

While I think I have somewhat an understanding of Entropy. (A measurement of a given state by its tendency proceed to it's most "probable" state), I have somewhat of a problem reconciling it with temperature.

Specifically why:
dS = dQ / T

rearranging the equation a bit...

1/T = dS/dQ, which appears to me that to indicate that at smaller temperature, it is "easier" to increase the entropy of a system. Easier in this case means requires less energy. Am I correct in this assumption?

Now, anticipating a bit, suppose this was qualitatively true, why the directly inverse relation 1/T, why not 1/T^2, or e^(-T). Also, supposing this were to be true, is there any quantitative usefulness to this? That is what sort of phenomenas can we predict defining Entropy this way. Is Entropy conserved? What sort of useful properties does it have?

 

[olgranpappy]

While I think I have somewhat an understanding of Entropy. (A measurement of a given state by its tendency proceed to it's most "probable" state), I have somewhat of a problem reconciling it with temperature.

Specifically why:
dS = dQ / T

That equation does not always hold. In the situations where it does hold you may take the equation $$\frac{dS}{dE}=\frac{1}{T}$$ as defining the temperature.

 

[FredGarvin]

The rule of thumb I remember from my grad thermo was:

1 kJ of heat from a 1000 K reservoir is more organized than 1 kJ of heat from a 300 K reservoir.

The definition was somewhat explained to me by using a carnot cycle and the efficiency. Here's the Reader's Digest very condensed version:

$$\eta = \left[\frac{W_{net}}{Q_{in}}\right]$$

$$= \left[\frac{Q_{in}-Q_{out}}{Q_{in}}\right]$$

$$= \left[1 - \frac{Q_{out}}{Q_{in}}\right] = \left[1 - \frac{T_{l}}{T_{h}}\right]$$

Therefore

$$\left[\frac{Q_{out}}{Q_{in}}\right] = \left[\frac{T_l}{T_h}\right]$$

$$\left[\frac{Q_{out}}{T_l}\right] = \left[\frac{Q_{in}}{T_h}\right]$$

 

[olgranpappy]

Fred, I like your quote about engineering... of course there is also a plus side to being an engineer as opposed to being a student: they pay you instead of you paying them

 

[FredGarvin]

Fred, I like your quote about engineering... of course there is also a plus side to being an engineer as opposed to being a student: they pay you instead of you paying them

Thanks. I wish I could remember where it was from. I grant you that the getting paid part does soften the blow.

 

[pkleinod]

Reply to question by Laix:

The essential content of the second law of thermodynamics is
encapsulated by the statement that all actual processes are
irreversible in the sense that it is impossible to restore **all**
systems to their original states. The entropy, S, is a measure of the
degree of irreversibility of any particular process. Furthermore, it
is formulated in such a way as to be a function of state, (e.g.
S(E,V)); i.e the change in entropy in any process can be computed
from the values of the state variables before and after the process
without paying attention to the particulars of the process. This is
useful.
Since total entropy increases in all spontaneous processes, one can
calculate the entropy change for some feasible process to see whether
it would actually occur. This is also useful. There are axillary
functions of state that can be derived from the entropy which can
tell us whether a given process will occur or not: the Gibbs free
energy could tell whether a particular chemical reaction will occur
spontaneously under conditions of constant temperature and pressure
and it tells us also how much useful work can (in principle) be
extracted from the process. This is useful. Would you try to put out
a fire of burning charcoal by pouring petrol on it? Obviously not.
But would you be tempted to pour water on it to put it out? Be
careful! If the temperature is high enough, there is a
spontaneous process leading to the production of carbon monoxide and
hydrogen, so you would be adding fuel to the flames! Wouldn't it be
useful to know what this temperature is before trying to extinguish a
fire in a nuclear reactor? Useful!

To get a quantitative measure of the degree of irreversibility of a
spontaneous process, consider restoring the system to its
original state by putting it in contact with a standard system made
up of weights and pulleys and reservoirs such that the restoration
process is carried out *reversibly*. The measure of irreversibility
(in the original process) is then taken to be the heat absorbed by
the standard system divided by some single-valued function B(T) of the
temperature (B(T) could itself be used as a measure of temperature).
Division by B(T) is necessary if we are to be able to distinguish
between the degrees of irreversibility for the same amount of heat
absorbed by reservoirs of differing temperatures.

Now, why take B(T) to be T instead of, say, exp(T)? It must first
be emphasised that B(T) is a universal function: it is defined with
respect to the heat absorbed reversibly by the standard system used to
restore the original system to its prior state. It does not depend on
the original system under investigation. This means we can take *any*
system to find out what B(T) is. In particular, one can take the
perfect gas for which the energy depends only on the temperature T and
not on the volume V.
Since PV=nRT, the differential for the heat absorbed in the
reversible expansion is dQ = Cv dT + P dV. (This differential is not
exact). The definition of entropy is dS = dQ/B(T). To ensure that dS
is an exact differential, it is necessary that d(Cv/B)dV = d(P/B)/dT.
This in turn implies that (nR/V)[d(T/B)/dT] = 0. This can only occur
if B is proportional to T; hence, one can take B=T. Thus, the
thermodynamic termperature B can be taken to be the ideal gas
temperature.

I hope this is a useful way of looking at it. There are many other
ways.