0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ In an isolated system, neither energy (including heat and work) nor matter can be exchanged with the surrounding environment. Do you mean a closed system instead? $\endgroup$ – user3814483 Jul 17 '14 at 3:26
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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 suppose I understand your question. I think it is true that the internal energy of adiabatically isolated system can increase due to work done on it by the environment. However, it is inconvenient to talk about heat (which is not a quantity of state) for such a system. Rather, what happens is perhaps the increase of entropy, not heat (meaningless). Let's think this way: before the piston is pushed, assume the system is in thermodynamic equilibrium; now somehow push the piston, the system gets stirred and undergoes a path of non-equilibrium states (heat may be generated and circulated between parts inside the system), meanwhile work has been done so the internal energy is increased; after some time, stop the piston, and now the system has to relax to a new equilibrium state, which has a larger internal energy and no-lesser entropy (entropy can not decrease for an adiabaticallly isolated system, Clausius's law). If the path is reversible (only true for infinitely slow process), then no entropy change; otherwise, entropy must increase.

$\endgroup$
0
$\begingroup$

I'm not sure what our question is exactly, so I'll try to answer what I guess your question is.

In general, the entropy of a fluid is a function of both $V$ and $T$. During an isentropic compression, the decrease in entropy from the reduction of volume is compensated by an increase due to the temperature rise. The net effect is zero.

For an ideal gas we have $$\Delta S=C_VR\ln (T/T_0) + R\ln (V/V_0)$$

With this we can calculate the temperature rise from a given volume reduction.

$\endgroup$
0
$\begingroup$

$\Delta s=\delta q/t$. Here, as the process is adiabatic it implies that there is no exchange of heat energy. So change in heat that is $\delta q$ would be zero. So accordingly in the equation if $\delta q$ is zero. So ultimately $ds$ is also zero. This implies a change in entropy.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.