1. May 17, 2009

### Trenthan

1. The problem statement, all variables and given/known data

Ok, i have a cycle that goes from "a to b", via isobaric expansion, than from "b to c" vis isochoric decrease in pressure, and "c to a" via a adiabatic compression. Each occur under quasistatic conditions

I just want to double check ive got the, changes in work, heat, internal energy, enthalpy and entropy right

3. The attempt at a solution

Process "a to b", isobaric expansion
a) Work is negative due to expansion
b) Q (heat) is added to the system hence Q is positive
c) U, internal energy increases due to us adding heat
d) Enthalpy inceases, since Q is positive
e) entropy increases since volume increases due to the number of microstates increasing in the system

Process "b to c", isochoric decrease in pressure
a) W = 0, no change in volume
b) Q(heat) is negative since heat is taken from the system to decrease pressure
c) U internal energy is negative since temperature decreases as heat is removed
d) Enthalpy is negative as heat is removed
e) entropy is negative since the nubmer of possible microstates decreases since there is less energy in the system ***Can it be negative my understanding is it can only be positive or zero***??????

Process "c to a", adiabatic compression
a)W is positive since we compress the gas
b) Q = 0 since it happen so quick no heat flow into or out of the system
c) U internal energy is positive since temp increases
d) enthalpy is zero, i no this from the formula but why,, internal energy increases, and positive work is done on the gas, hence shouldnt it be positive????
e) Entropy Ive read its zero, i cannot work out how mathematically it was determined, and am a little stuck with understanding as to why??? i would have thought negative since we decrease the volume hence fewer microstates in the system

Im mainly concerned with the red parts, they seem pritty iffy, and my textbook is just mass confusion is those areas with teh explanation lolz

Thanks heeps

2. May 17, 2009

### Trenthan

from what ive been reading, entropy cant be negative, it can only increase and when it does its irreversable. Its only reversable when it is equal to zero??

Ive been reading it from the net so im not sure how reliable it is if any1 can confirm and explain why an isochoric decrease in pressure, wouldnt change entropy??

Cheers Trent

3. May 17, 2009

### Mapes

Total entropy tends to increase, and entropy tends to increase in isolated systems. In a non-isolated system, entropy decreases are unremarkable; the entropy of my coffee is decreasing right now as the cup cools to room temperature.

Reversible work doesn't increase entropy; reversible work is the $-p\,dV$ part of energy (in contrast to reversible heat transfer, which is the $T\,dS$ part).

4. May 17, 2009

### Trenthan

so in an isolated environment, leaving the system on its own heat is transfered hence, entropy increases and if we than take the item out it doesnt go back to its intial temperature i.e not reversable

On the other hand with a piston we can push it in, than pull it back out hence enthalpy can be + or negative. + when we increase volume, - when we decrease volume in my case.

That the basic jist???

Thanks by the way

5. May 17, 2009

### Trenthan

so for an "adiabatic compression", since Q = 0 no energy, heat is transfered out of the system so entropy wouldnt change...

Mathematically out that just be shown as
delta(S) = Q/T, hence Q = 0, hence delta(s) = 0, basically??

Cheers Trent

6. May 17, 2009

### Mapes

Generally, energy can be transferred by work or heat transfer. In both cases, the system energy can increase or decrease.

Heat transfer occurs when objects are at different temperatures. Along with energy, entropy is transferred. Also, entropy is created if the process is irreversible; that is, if the temperatures aren't equal.

Work occurs when there is an imbalance in pressure (or voltage, or magnetic field, or stress, or surface tension...). Energy, but no entropy, is transferred if the process is reversible; that is, if there's no difference in pressure. As before, entropy is created if the process is irreversible.

It should be obvious that there's a contradiction here: if the temperatures/pressures must be equal for reversibility, how does energy transfer occur in the first place? The resolution is that no real process is reversible. Reversibility is an ideal we sometimes assume to simplify calculation; it can be approached, but never achieved in real life.

An adiabatic reversible process is isentropic because it fulfills the requirements that Q = 0 and $S_\mathrm{created} = 0$

7. May 17, 2009

### Trenthan

So entropy for an adiabatic compression is zero, due to the process being reversable i.e adiabatic expansion, would have the complete opposite effect.

And for an isochoric process, the change in entropy is negative, hence "irreversible".

"As before, entropy is created if the process is irreversible"
, would this include entropy loss,

"entropy is given away if the process is irreversible"???

Thanks again Trent

8. May 17, 2009

### Mapes

No, the entropy change may be zero or positive depending on whether the process is reversible or irreversible.

No, an isochoric process can be reversible and the system's entropy will still decrease if the temperature decreases. The difference is that if the process is reversible (i.e., the temperature difference between the system and the surroundings is negligible), the entropy loss in the system will be exactly made up by the entropy gain in the surroundings. If the process is irreversible (i.e., a temperature difference exists), additional entropy will be created at the interface.

No, entropy can be created but not destroyed.

Hopefully these comments won't be discouraging; getting the wording precise is part of learning thermo.