/sci/ - Science & Math

[Anonymous No.3694239]

hey /sci/

So I've been studying thermodynamics, and there's something that I can't get my head around. I don't understand why maximum work is achieved via reversible expansion of a gas (where external pressure is maintained infinitesimally smaller than internal pressure).

What does reversible even mean in this context? Why isn't irreversible expansion reversible?

Any help would be much appreciated

[Anonymous No.3694273]

anyone? :<

[Anonymous No.3694304]

forever alone

[Anonymous No.3694308]

reported for discussing science on /sci/

[Anonymous No.3694312]

>What does reversible even mean in this context?

Equillibria, brah.

where delta G = delta H - T (delta S)

in room temperature,

if delta H greater than +50, reaction is insignificant
if delta H lower than -50, reaction goes to completion

if delta H between -50 and +50, then reaction reaches chemical equilibria

[Anonymous No.3694318]

>Why isn't irreversible expansion reversible?

its all about the delta G's brah

when you want to find out about the spontanenity of a reaction, calculate the gibbs free energy ( delta G)

if delta G is positive, then reaction is not feasible

if delta G is negative, then reaction is spontaneous

[Anonymous No.3694328]

>>3694308

yeah, I've sort of realised that only a small minority of this board actually try to learn science lol :(


So if reversible expansion does a certain amount more work than the equivalent irreversible expansion (against fixed, constant pressure), is there a difference between their internal energies? As: dU = dq + dw , and obviously dw is different for the two, is there a difference in dq?

[Anonymous No.3694330]

>I don't understand why maximum work is achieved via reversible expansion of a gas

Equillibria brah.

expansion makes delta S (system) positive, but is also likely to cause delta H to be positive, in room temperature, delta S is no more than 100 J per K per mol, so - T (delta S) would be no more than - 300 x 100 or delta S will have no more influence than 30 kj per mol. In this case delta H would be the decisive factor where any enthalpy beyond say +50 kj per mol will be unfeasible or below -50 kj per mol to be spontaneous.

you want to maximize the expansion (delta S system), while keeping delta H (enthalpy change) low, so as to maximize work done with the least amount of temperature imput.

its easy brah.

[Anonymous No.3694335]

>>3694330

or in the case of a combustion engine, which is probably the example you have been given,

with the least amount of pressure exerted (not so much temperature input)

[Anonymous No.3694349]

im going now OP,

keep chugging brah

[Anonymous No.3694458]

thermodynamics isnt a tough concept.

smells like pre-university in here.

[Anonymous No.3694836]

^^

[Anonymous No.3694889]

>>3694458
lol'd hard.

[Anonymous No.3694982]

can anyone shed any light on this:

So if reversible expansion does a certain amount more work than the equivalent irreversible expansion (against fixed, constant pressure), is there a difference between their internal energies? As: dU = dq + dw , and obviously dw is different for the two, is there a difference in dq?

[Anonymous No.3695080]

>>3694982
>is there a difference between their internal energies
no, because of conservation of energy.

an example for the reversible process is
>the whole system is
>a) you and you have some energy (chemical energy, coal) to spent.
>a given a container of gas with one movable wall. beyond the wall, there is a spring
now what you do:
>heating up the walls of a container of gas with your energy from outside
>temperature T of gas goes up, the gas presses against the wall and the spring gets clinched
this is reversible, because if you cool the container from outside (take the heat away again), then the spring will be stronger than the gas pressure and everythink will go back to the starting porsition. you have your energy outside and the box is small again. no entropy change btw. and energy of the system is always the same.

an irreversible process is the following:
>the wall of the box is totally loose (no spring this time) and it will blow up the box until pressure is equal form inside and from outside.
>no ∆Q in this case, no work ∆W is done here and the entropy inside the box goes up because there is more spece for the same amount of particles

the total U is at all times the same in both cases btw.

[Anonymous No.3695092]

>>3695080
-in the second case I mean no ∆Q from outside, the box blows up because of the gas pressure until equilibrium.
-and no work is done, because the wall in the second case behaves like a sheet of paper and it's free to move it around (it just seperates the gas in the box from the outside)

[Anonymous No.3695098]

>>3695080

ahhh I think I'm finally starting to get it.

So for reversible: dU = 0, dw = -dq
And for irreversible: dU = dw = dq = 0

thanks :)

[Anonymous No.3695134]

>>3695080

>no entropy change btw

lol'd hard

>look at exhaust from a car
>think that the atoms in the fumes behave more orderly than the petrol used to produce it

[Anonymous No.3695141]

>no ∆Q in this case, no work ∆W is done here and the entropy inside the box goes up

For fucks sake man. Do you even theromdynamics?

How the fuck can you get a change in entropy without a change in either ∆Q or ∆W??????

What fucking multi-stasis compound is this that can exist in seperate phases all in the same conditions????

Do you even know how ∆S is calculated? It is the gradient for the relation between Temperature in Kelvins plotted against ∆G. OMG.

[Anonymous No.3695288]

>>3695134
The process I described is reversible, so the entropy necessarily doesn't change. Wouldn't you agree?
Your example doesn't correspond to my first one.

>>3695141
>How the fuck can you get a change in entropy without a change in either ∆Q or ∆W??????
The equation of state f(V,P,T)=0 (for example PV=NkT in our case) describes the system in equilibrium. For quasistatic process where you go from equilibrium to equilibrium, the change of entropy is dS=dQ/T, so from dQ=0 you get dS=0. However if you go from one equilibirum state to another non-quasistatically, then not even temperature is well defined in classical thermodynamics. In such a process you can get from low to high entropy without heat flow.
If you pop a balloon with helium in a room full of air, then the gas will spread and then the entropy will grow and grow. consider this process from 3 seconds after the ballon-explosion to 5 seconds after the balloon explosion. clearly the internal energy is constant, no work is done so also ∆Q=0. but the entropy of course gets bigger - because this is not a process from equilibrium to equilibrium.
You'll find other examples, exspecially for processes where dT=0. I think entropy of mixing is one.
http://en.wikipedia.org/wiki/Entropy_of_mixing

>>3695098
yes, except for the fact that there is another possibility. If you work, such that ∆W-->∆Q and nothing else happens (such that you can't get the energy back, like if you stir the pudding) then that's irreversible too.

[Anonymous No.3695307]

As an addon to
>>3695098
(OP I think?)
maybe it's best for you to internalize the Kelvin formulation of the second law of thermodynamics
http://en.wikipedia.org/wiki/Second_law_of_thermodynamics
(look for Kelvin statement)
The statement "whose sole result is" is quite important here, otherwise you'll confuse yourself on how you can get your energy back in the reversible expansion, once you heated the box.

[Anonymous No.3695509]

^^

[Anonymous No.3695523]

>I don't understand why maximum work is achieved via reversible expansion of a gas
Reversible means you can run the reaction backwards without any extra energy input.

I know the Feynman lectures has one thought-experiment proof of why a no machine can be more efficient than a reversible one. Basically, if there's any non-reversible machine that beats a reversible equivalent, you can get free energy forever by setting up a net-energy-positive closed loop. But that violates conservation of energy.