# I Why there is a " > " sign in the entropy of an isolated system

1. Jul 19, 2017

### tze liu

it seems it is not reasonable
Sorry

it said that delta S>=0
in an isolated system. i know why they put " = " sign, but i don't know why it is still possible that dS>0 in an isolated system

because in an isolated system,it is said that dQ=0 (because no heat transfer between the system and the enviroment?)
by the same logic,dQrev is also 0 in this case
So there should not have " > " sign in an isolated system?
any entropy change in any isolated system is 0 because dQrev is 0
so delat S=0 in an isolated system but not deltaS>=0?
what is the difference between dQrev and dQirrev ?

#### Attached Files:

• ###### 13769422_10154400692839515_1628218858161132827_n.jpg
File size:
70 KB
Views:
103
2. Jul 19, 2017

### DrDu

The classical example is that of stirring a viscous fluid with a paddle. Even if the system is thermally isolated, the work done on the system (stirring of the paddle) will be converted into internal energy. Basically, the friction will heat up the system (also no heat is transferred to the system by the boundaries!) . So its entropy will increase. As friction is dependent on velocity, this mechanism is absent when you only allow for reversible changes of the system. So if you want to bring the system from the same initial to the same final state, you have to provide the system with external heat, i.e. this process cannot be done both reversibly and adiabatically.

3. Jul 19, 2017

### tze liu

but why the dQ (not reversible one) is 0 in such case?

4. Jul 19, 2017

### DrDu

If you do it, say, in a Dewar, where should the heat come from?

5. Jul 19, 2017

### tze liu

i don't understand why the system is thermally isolated in this case.

can u define what is the "system" here

if the friction heats up the system that means even dQ is not zero?
because dQ is defined as the heat transfer in an irreversible path .

6. Jul 19, 2017

### DrDu

Think of the fluid only as forming the system. The paddle may be as light as you want, so that it's heat capacity can be neglected. Furthermore, it can be idealized to not to conduct heat. Hence the system remains perfectly thermally isolated in the process. No change of the temperature of the surrounding will change the outcome of the experiment. This is how we define adiabatically isolated. In an isolated system, by definition, there is no heat flow into or out of the system.
In fact, this is how we define internal energy, namely the change in internal energy $\Delta U$ is defined as the work done on the system in an adiabatic process: $\Delta U:=W_\mathrm{adia}$. Now the first law states that U is a state function. As energy is conserved, and U certainly is a form of energy, we define heat as $Q:=\Delta U -W$ for a general (non-adiabatic) process.

Be carefull because "heating up" does not necessarily mean that heat is transferred to the system, rather, it only means that its temperature rises.

7. Jul 19, 2017

### tze liu

thank.
however the defination of entropy change is int dQ(rev)/T

that mean dQrev is not zero in this case?

8. Jul 19, 2017

### DrDu

No, it isn't. Say the viscous fluid is water. Then you can use the paddle to increase its temperature by, say 10 deg. You could do this reversibly, e.g. doing no work at al, simply slowly heating the water. Assuming that the heat capacity C is approximately constant, $\Delta S= \Delta U \int_{T_1}^{T_2} 1/T dT=\Delta U \ln (T_2/T_1)=W_\mathrm{adia}\ln (T_2/T_1)$, where $W_\mathrm{adia}$ is the amount of work done in the adiabatic process to achieve the same change of temperature form $T_1$ to $T_2$.

9. Jul 19, 2017

### tze liu

do u mean dQrev is the heat transfer between the systems but dQ is the heat transfer from outside to system.
the defination of entropy change is S2-S1= int dQrev/T>= dQ/T
if it is an isolated system, that means there is no change in dQ between the enviroment and the system.
however there is heat transfer between the system dQrev
do u mean any type of heat transfer between the system is reversible(which represents by d Qrev),so dQrev is not zero
so S2-S1=int (dQrev)/T is not 0 (which greater than 0)?
if Qrev is zero here ,then S2-S1 is definitely equal to zero.

So the formula should be S2-S1=0 no matter what S2 and S1 is
it is hard to understand what is the difference between dQrev and dQ
that means dQrev works in the heat transfer inside the system only but dQ is only responable for the heat transfer between the environment and the system?

10. Jul 19, 2017

### DrDu

In an isolated system, dQrev is 0. The only way to calculate the entropy change using dQrev is to consider a reversible process instead, which can't be isolated.
I choose one where W=0, i.e. all the change of U is due to Qrev, which must equal Wadia in the irreversible,adiabatic case.

11. Jul 19, 2017

### tze liu

do u agree that the definition of entropy change in general is Sb-Sa=intregrel(a to b) dQv(rev)/T (where it is >= intregrel (a to b) dQ(irrev)/T)

b and a represents any two points at any path of states.

#### Attached Files:

• ###### 13769422_10154400692839515_1628218858161132827_n.jpg
File size:
40.7 KB
Views:
94
12. Jul 19, 2017

### tze liu

do u mean the definition "Sb-Sa=intregrel(a to b) dQv(rev)/T"" is only for non-isolated system?
this definition cannot apply in the isolated system?
but there is another definition of change of entropy such that it doesnt involve dQrev here?

so even in such isolated system, even dQrev is zero but the change of entropy still greater than zero?

13. Jul 19, 2017

### tze liu

still very confusing~~sorry >__<

14. Jul 19, 2017

### tze liu

consider a cycle in a system W including of one reversible path and a non-reversible path where the reversible path call p1 and a non-reversible path call p2.
where they are connected together with point a and point b .
Now, we know the entropy change is ONLY DEFINED in a reversible path p1 but not in p2.
we assume the W is an isolated system and no heat can transfer between the system and the environment,
(1) the definition of the change of entropy is S2-S1= integral(a-> b) dQrev/T where we know it is >= integral(a->b) dQ/T
(2) the entropy change of an isolated system between point a and b has some probability greater than zero.
where dQ is the heat transfer in the irreversible path.
however, we know it is an isolated system, from the definition of an isolated system, there is no any heat transfer or work done, so dQ=0

so int (a->b)dQrev/T>=0 but dQrev is also zero because there is no heat transfer in an isolated system.
so according to the =(1) the change of entropy S is definitely zero because if S2-S1= integral(a-> b) dQrev/T=S2-S1= integral(a-> b) (0)/T=0,there is 0 possibility that it is greater than zero.

however from the result of the (2), the entropy change of an isolated system is S>=0 and have no probability greater than 0, so the definition of entropy is contradicted t o each other?

15. Jul 19, 2017

### DrDu

Entropy is a state variable. That means that its value only depends on the actual equilibrium values of p, V, T ... but not on how you get there. However, you can only calculate the entropy change when going from state A to B, using your formula for a reversible path, not for an irreversible one.

16. Jul 19, 2017

### tze liu

consider a cycle in a system W including of one reversible path and a non-reversible path where the reversible path call p1 and a non-reversible path call p2.
where they are connected together with point a and point b .
Now, we know the entropy change is ONLY DEFINED in a reversible path p1 but not in p2.
we assume the W is an isolated system and no heat can transfer between the system and the environment,
(1) the definition of the change of entropy is S2-S1= integral(a-> b) dQrev/T where we know it is >= integral(a->b) dQ/T
(2) the entropy change of an isolated system between point a and b has some probability greater than zero.
where dQ is the heat transfer in the irreversible path.
however, we know it is an isolated system, from the definition of an isolated system, there is no any heat transfer or work done, so dQ=0

so int (a->b)dQrev/T>=0 but dQrev is also zero because there is no heat transfer in an isolated system.
so according to the =(1) the change of entropy S is definitely zero because if S2-S1= integral(a-> b) dQrev/T=S2-S1= integral(a-> b) (0)/T=0,there is 0 possibility that it is greater than zero.

however from the result of the (2), the entropy change of an isolated system is S>=0 and have no probability greater than 0, so the definition of entropy is contradicted t o each other? so S=0 or S>= 0 is the actual formula of an isolated system?

17. Jul 19, 2017

### DrDu

If you consider only adiabatic paths, in general there is no reversible path joining A and B, so you can't determine Qrev.

18. Jul 19, 2017

### tze liu

if not?

19. Jul 19, 2017

### DrDu

You need a non-adiabatic processes to determine S. You simply can't determine the change of S as long as you only consider adiabatic processes.

20. Jul 19, 2017

### Lord Jestocost

Depending on the initial state immediately after isolation, even in isolated systems "internal things" can happen between the various parts of the system, e. g., reactions between components, redistribution of components by means of diffusion, phase transition etc., until the overall system has attained its absolute thermodynamic equilibrium state. That’s the general meaning of ΔSsystem >= 0 for isolated systems.

21. Jul 19, 2017

### DrDu

Another example is free expansion of a gas: you have two cylinders (thermally isolated), one filled with gas, the other vacuum. You suddenly puncture the membrane between the two cylinders. If the gas is ideal, p*V doesnt change and so doesnt T. Neither does internal energy. Nevertheless entropy increases.
To calculate the entropy change you can again consider a reversible process, this time without thermal insulation, where the gas is slowly brought from V1 to V2 at constant T. As U is a state variable and didn't change in the adiabatic process, Qrev must equal -W in the new process.

22. Jul 19, 2017

### tze liu

even in isolated systems "internal things" can happen between the various parts of the system, e. g., reactions between components, redistribution of components by means of diffusion, phase transition etc.,
------------
if this is true ,why dQ is zero here? there is dQ because there are "internal things" happen

Last edited: Jul 19, 2017
23. Jul 19, 2017

### Lord Jestocost

In an isolated system dQ = 0 because no energy is transfered between the system and the enviroment. Whatever happens inside an isolated system results in an increase of the system's entropy.

24. Jul 19, 2017

### Staff: Mentor

Do you know how to determine the change in entropy of an isolated system that has experienced an irreversible change?

Are you aware that, in determining the change in entropy for an isolated system that has experienced an irreversible process, the reversible path you choose (to evaluate the integral of $dS=dQ_{rev}/T$) does not have to bear any resemblance whatsoever to the actual path followed by the irreversible process (as long as it passes through the same two end states), and this includes paths for which the system is not isolated. In fact, for any reversible path between these same two end states, $dQ_{rev}$ is not zero. Here is a link that gives a recipe for determining the entropy change for an irreversible process, and includes examples applied to two isolated systems: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

Last edited: Jul 20, 2017
25. Jul 21, 2017

### tze liu

why!??? if dQrev doesnt actually exist in the process, why we can still use this "dQrev" to calculate the entropy.
then where is this dQrev coming from

dQrev is defined as the heat transfer between the system and the environment in the reversible process