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More rigorous Gibbs free energy / spontaneity relationship

More formal understanding of why a negative change in Gibbs Free Energy implies a spontaneous, irreversible reaction. Created by Sal Khan.

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• Wouldn't the irreversible process need more heat than the reversible process, as the heat taken up in the irreversible process will be used not just to move the piston, but also to move the piston against the surface, creating the friction, therefore using more heat?
• I think the mass that is left on top of the second piston is less than the one left on top of the first one, that's the only way they can end up at the same height, otherwise, with friction, the right piston would stop before reaching the same height, unless the right reservoir had a higher temperature than in the left one, and then it WOULD need more heat, like you said. But it's not the case since they have the same temperature.

rigorous proof : the equilibrium is reached when the upper part of the piston is not moving, so, when the sum of the forces it undergoes are zero (i.e. when it is pushed down as much as it's pushed up)
Let's consider the right piston, with friction : the forces it undergoes are : pressure from atmosphere, weight of the mass, friction and pressure of the gas inside.
we have mg + Pext/A + F = P1/A (where m is the mass on top on the piston, Pext is the external pressure, A is the top surface of the piston, F is the force caused by friction, if we consider it as a dry friction, P1 is the pressure inside the piston) we then have m = P1/gA - F/g - Pext/gA <=> m = nRT/VgA - Fg - Pext/A (because P1 = nRT/V)
now, n is constant, because the gaz is closed inside the piston, T is constant because it's isothermic process, V is fixed because we want to reach a certain given height. So, if F is null i.e. there's no friction, there is a certain mass m1. And it F is non null i.e. there's friction, m2 is smaller than m1. I.e : in the left piston, the mass left on top of the piston is smaller !
If I am not mistaken :p
• At Sal says that for a spontaneous process, deltaS > 0. But what about when supercooled water freezes?
• yes, that's it. spontaneity depends on dS and dH. at temperatures below the freezing point of water, then dH factor contributes more entropy to the surroundings than the entropy lost due to solidifying the liquid-thus, the net entropy change in the universe is positive and process is spontaneous
• i keep on feeling that the work done by the irreversible process should be less but when i think about how in the PV diagram the start and end point of both graphs for reversible and irreversible processes are the same they make me wonder that the piston must have moved up to the same height in the irreversible process as well so would that mean both the reversible and irreversible processes did the same amount of work as they displaced the piston by the same change in volume? but just thinking that they would do the same amount of work gives me an uneasy feeling
can some one help me?
• here's how i see it : they did the same amount of work, but in the right piston experiment, some of the work was used to counter the friction. The remaining work was used to move the mass to the top. This mass being lower than the one on the piston on the left, as I explained in my answer to Jamie Martin.
• Sal, at you state that delta SiR is = to QR/T1 = delta SR because they move to the same spot on the PV plot, and that QiR/T1 is less than QR/T1, but, how can this be? If QiR/T1 is less than QR/T1 and Qfriction is generated, this means that the engine would perform less work. They would not be at the spot on the PV diagram.
• It would mean that the engine would have to draw more Q from the heat source to perform the same amount of work due to the inefficiency.
• why is it assumed that every irreversible process is spontaneous?
• This is not the case, as per I understand it.
On the other hand, we know that only irreversible processes are spontaneous.
(1 vote)
• In one of Sal's earlier videos, we've been told that Reversible = quasi-static AND frictionless. OK. This in turn means that Irreversible = not Quasi-static OR with friction

But then in the "More rigorous Gibbs free energy..." at , Sal explains that BECAUSE the process is irreversible, there is friction, and that this friction generates heat. Does this mean that irreversible = with friction ?
If not, let us suppose it is frictionless, but still irreversible BECAUSE not quasi-static.
How can we show that Q will still be less that in the reversible case ?
• Notice that the external pressure is "modelled" by weights on the roof of the canister. When you remove pebbles (in the reversible case) the system lifts the roof just a little, and has to perform work against almost all the remaining pebbles, thereby supplying them with potential energy.

Since there is equilibrium throughout, the external matches the internal pressure, and that is why you can plot the work (that what you do against the external pressure) in the state diagram of the system (plotting the internal (!) pressure).

In constrast, if you remove the heavy weight, the system has to "lift up" a less heavy reamining weight until a pressure balance is reached. Think of the removed weights in either case as being removed horizontally ("through the wall") onto a shelf of corresponding height, next to the canister. You see that you had to transfer more internal energy into potential energy alongside this process in the revesible case, since the pebbles' height on these shelfs surmounts in total the one of the two big weights. That is why you have to draw less heat from the reservoir to compensate for that, in the irreversible case. That is true even without friction. The videos are unfortunately not terribly clear about that connection.
• if enthalpy change is positive and entopy change is negative then what is the condition for the spontanity of a process?
(1 vote)
• Then the process can't be spontaneous under any conditions.
Because, delG=delH -T*delS
delH>0, delS<0. And T can't be less than 0 (Kelvin Scale)
Hence both the terms are positive. Thus, delG>0.
This implies process non spontaneous. :)
• at how can we we lose temperature in absence of reservoir? i did not get it.!
(1 vote)
• because the piston has moved up and kinetic energy is transformed to mechanical work, so the average kinetic energy inside the cylinder becomes less..
• hey please can you tell why do we need del G for finding out the spontaneity of the process in place of del S??
after all for finding out del G we do have to know del S...
• The change in entropy is only part of the story. If entropy increases, a reaction will be spontaneous at all temperatures if it is exothermic. However, if the reaction is endothermic, then an increase in entropy on its own is not sufficient for the reaction to be spontaneous - a high temperature is needed as well. DeltaG takes both enthalpy and entropy into account.
(1 vote)
• The irreversible engine has to draw more heat from the reservoir to come to the same point(say,from A to B) as the reversible engine,on the plot.Then how is Q r > Q ir ?
(1 vote)
• No, the irreversible engine does not have to draw more heat because part of the heat required to get to B is generated by friction in the engine itself.
Thus, it withdraws less heat as compared to the reversible engine. :)