# Entropy change during wet compression

1. Apr 24, 2013

### kulkajinkya

I have a quick question. Consider the process S2 to S1 in the figure below

Since this represents wet (isentropic) compression, mathematically we have ΔS=0 (assuming adiabatic compression). But if we consider the process in a physical way, we are going from a region of less disorder (mixture of vapour and water) to a state of more disorder (superheated vapour). Hence physically speaking entropy should increase since we are going to a highly disordered state from a low disorder state. So here is my paradox. How can the process have mathematically 0 change in entropy, but physically it does seem to increase?

2. Apr 24, 2013

### Jano L.

If the compression is done slowly and without heat exchange, that the entropy is constant follows from the second law of thermodynamics. There is no connection to disorder in this theory.

This connection is a subject of statistical physics, where the statistical entropy is connected to disorder in a mathematical way. You have to quantify the disorder by a number which is proportional to the number of microstates of the system that are compatible with the macroscopic variables, like pressure and volume.

When you do that, you will find out that the "disorder" is proportional to the volume raised to power of N, where N is the number of molecules of the gas.

Now, in order to perform the transition to superheated vapour, we have to push the piston and thus shrink the volume V. This decreases the number of accessible microstates and hence "disorder". If the statistical entropy is to have the same value as the thermodynamic entropy, the decrease of the accessible states has to compensate exactly for the increase of them due to higher kinetic energy of the molecules.

3. Apr 24, 2013

### kulkajinkya

I didn't get this line: 'If the compression is done slowly and without heat exchange, that the entropy is constant follows from the second law of thermodynamics. There is no connection to disorder in this theory'.

4. Apr 24, 2013

### Jano L.

The above behavior of the entropy can be calculated in thermodynamics without any use of idea of "disorder", if entropy means "thermodynamic entropy". Thermodynamic entropy is a concept that does not depend on such statistical notions.

5. Apr 24, 2013

### kulkajinkya

So essentially you mean to say that on one hand we are increasing disorder by increasing the temp, but simultaneously decreasing disorder by reducing the volume? Like balancing both of them and getting a 'net' result of an 'unchanged' disorder, right?

6. Apr 24, 2013

### Staff: Mentor

Just consider isentropic compression of a gas without a phase change. This corresponds to the portion of path S2 to S1, starting from the 100% saturated vapor location. The temperature is increasing, which acts to increase the entropy, but the volume is decreasing, which acts to decrease the entropy. The two effects cancel out so that there is no change in entropy. Are you still wondering why decreasing the volume of a gas acts to decrease its entropy?

Chet

7. Apr 24, 2013

### kulkajinkya

Thank you people for the answers!