Even if the system is isolated and there is no heat exchange with surroundings, shouldn't the decrease/increase of pressure result in increase/decrease of entropy?
Does this property of an isolated system means that increase of pressure means equivalent decrease in volume, so after the process the $pdV$ i.e. the work is still the same?
Can someone explain this in more detail?
Quoted text above was original question.
Please read on further, I clarified the question.
To be more specific, situation is this.
We have an adiabaticaly isolated system and (for example) a piston inside it. If there is piston, that means that volume work or $PdV$ can be done.
So if we do work on the system, by some force from surroundings which pushes piston(compresses the gas) we would get increase of the pressure inside the system… right?
If we increase the pressure in the system, the entropy will decrease as well, altough we haven’t bring in the heat .. right?
So if we increased the pressure and mutually decreased the volume, the molecules of the gas are packed more tightly and they have lesser space to move in…This is followed by system having more internal energy now, then it had before process…Since the activity of molecules inside is higher, they are moving at higher speeds(cause of smaller volume)…
That means that system should have more internal energy after the process, right?
Taking that into consideration, and if we follow the rule that increase in pressure is also increase in temperature, that is molecules have higher activity on higher pressure and that kind of activity is measured with temperature…that means that system also has more amount of heat?
Correct me if I am wrong.
And even if it is adiabatically isolated (dQ=0) shouldn't the internal energy increase and thus amount of heat in the system increase as well when we do the work on the system (move the piston to compress the gas )?
I am talking that higher activity leads to more heat inside the system even if it is prevented (isolated) to exchange heat with its surroundings.
I hope you understand me now.
Best Answer
By definition, $dS =\frac{dQ}{T}$, so an adiabatic process doesn't change entropy. But you can find more details at http://en.wikipedia.org/wiki/Adiabatic