- #1

zenterix

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- Homework Statement
- It can be shown (with a simple exactness test) that the differential ##\frac{\delta q_{rev}}{T}=dU+\frac{nRT}{V}{dV}## is exact.

This differential is ##dS##, the differential of entropy.

- Relevant Equations
- Since this differential is exact, it comes from a state function (ie, a path independent function), ##S## and ##\oint dS=0##.

That is, for any cylic reversible process the entropy doesn't change.

If a process is irreversible, on the other hand, then

$$\oint \frac{\delta q}{T}\leq 0=\oint dS\tag{1}$$

Apparently, from this equation we can conclude that

$$dS \geq \frac{\delta q}{T}\tag{2}$$

How do we mathematically justify the step from (1) to (2)?

Next, consider an isolated system.

Consider the expression ##\Delta S>0##.

$$\int dS=\Delta S> \int \frac{\delta q}{T}=0$$

But isn't ##\delta q_{rev}## also zero in an isolated system?

$$\oint \frac{\delta q}{T}\leq 0=\oint dS\tag{1}$$

Apparently, from this equation we can conclude that

$$dS \geq \frac{\delta q}{T}\tag{2}$$

How do we mathematically justify the step from (1) to (2)?

Next, consider an isolated system.

We can restate the second law by saying that the entropy increases in a spontaneous process in an isolated system because for an isolated system, ##\delta q=0## and therefore ##\Delta S>0## for a spontaneous process. The entropy of an isolated system (in which the internal energy is constant) can continue to increase as long as spontaneous processes occur.

Consider the expression ##\Delta S>0##.

$$\int dS=\Delta S> \int \frac{\delta q}{T}=0$$

But isn't ##\delta q_{rev}## also zero in an isolated system?