Entropy in isolated composite system for irreversible process

  • #1
zenterix
629
82
Homework Statement
There is a section in the book "Physical Chemistry" by Silbey, Alberty, and Bawendi that I would like to understand.
Relevant Equations
When an ideal gas is heated reversibly, we can write

$$dU=C_VdT=\delta q_{rev}+\delta w=\delta q_{rev}-PdV=\delta q_{rev}-\frac{nRT}{V}dV$$
I am using the symbol ##\delta## in ##\delta q_{rev}## and ##\delta w## to denote an inexact differential.

$$\delta q_{rev}=C_VdT+\frac{nRT}{V}dV$$

We can turn this inexact differential into an exact differential by multiplying by the integrating factor ##\frac{1}{T}##.

$$\frac{\delta q_{rev}}{T}=\frac{C_V}{T}dT+\frac{nR}{V}dV$$

The state function associated with this exact differential is called entropy, ##S##. Thus
$$dS=\frac{\delta q_{rev}}{T}$$

The Clausius theorem says that

$$\oint \frac{\delta q}{T}\leq 0\tag{3.14}$$

This is a statement of the 2nd law of thermodynamics.

The inequality above is to be understood as follows

(a) If any part of the cyclic process is irreversible (spontaneous), we have strict inequality.

(b) If the cyclic process is reversible, we have equality.

(c) It is impossible for the cyclic integral to be greater than zero.

Starting at this point, there are a series of statements that I don't quite understand.

Equation 3.14 leads to the following inequality for a noncyclic process

$$dS\geq \frac{\delta q}{T}$$

Question 1: How do we obtain the inequality above?

If the process is reversible, then ##dS=\frac{\delta q_{rev}}{T}##, and if the process is irreversible, then ##dS>\frac{\delta q_{irrev}}{T}##.

Question 2: How do we obtain ##dS>\frac{\delta q_{irrev}}{T}##?

My last question is about the following reasoning.

Suppose we have an isolated system. Then, ##\delta q=0## and therefore ##\Delta S=\int\frac{\delta q_{rev}}{T}>0## for a spontaneous process.

Thus, as long as spontaneous processes occur, entropy keeps increasing until equilibrium, in which entropy is at a maximum.

This reasoning can be applied to a system that is not isolated by treating the system plus its surroundings as an isolated system.

Suppose we have a composite system made of a main system and surroundings, and this composite is isolated.

$$dS_{tot}=dS_{sys}+dS_{surr}$$

If we have a reversible process in which the main system gains heat from the surroundings then

$$0=dS_{sys}-\frac{\delta q_{rev}}{T_{surr}}$$

$$dS_{sys}=\frac{\delta q_{rev}}{T_{surr}}=\frac{\delta q_{rev}}{T}$$

since ##T_{surr}=T## for a reversible process.

Next consider an irreversible process in which the system gains heat ##dq_{irrev}## from its surroundings.

$$dS_{sys}=dS_{tot}-dS_{surr}$$

If the transfer of heat occurs reversibly in the surroundings, the entropy change in the surroundings is ##-dq_{irrev}/T_{surr}## and so we have

$$dS_{sys}>\frac{\delta q_{irrev}}{T_{surr}}$$

since ##dS_{tot}>0$ for a spontaneous change in the total (isolated) system.

Question 3: What does it mean for the transfer of heat to occur reversibly in the surroundings (especially given that we are considering an irreversible transfer of heat to the main system)?

We can now combine equations

$$dS_{sys}\geq \frac{\delta q}{T_{surr}}$$

where the inequality applies when the process is irreversible and the equality applies when the process is reversible.

For a finite process

$$\Delta S_{sys}\geq \int\frac{\delta q}{T_{surr}}$$

I really spent a lot of time trying to understand the snippet above, especially the part about an irreversible process. But to no avail.
 
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  • #2
So you know that the cyclic integral for a reversible integral is 0. As S is defined for equilibrium states, an irreversible process has to start and end at an equilibrium state. You can make a cyclic process out of it, closing the loop with a reversible process.
The prototype of such a process is a viscuous fluid in a container with a stirrer. You can heat up the fluid either reversibly by bringing it into contact with heat reservoirs of only slightly higher temperatures or you can increase its temperature by doing work on the stirrer (which irreversibly dissipates work into the internal energy of the fluid). Clearly the second process can be done in an adiabatic fashion. If only a part of the cyclic process going from nearby equilibrium states A to B so that T is approximately constant is irreversible, you have
## \oint dq_\mathrm{irr}/T \le \oint dq_\mathrm{rev}/T =\oint dS =0##
##\int_A^B dq_\mathrm{irr}/T +\int_B^A dq_\mathrm{rev}/T \le \int_A^B dS +\int_B^A dq_\mathrm{rev}/T## or ## dq_\mathrm{irr}/T \le dS##.
 
Last edited:
  • #3
What is meant in 3. is that the surrounding remains (approximately) in an equilibrium state, i.e. there is no appreciable temperature gradient building up in the surrounding.
 

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