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We see that the total amount of work that can be extracted in an isothermal process is limited by the free energy decrease, and that increasing the free energy in a reversible process requires work to be done on the system. If no work is extracted from the system then
[itex]\Delta A \leq 0[/itex]
and thus for a system kept at constant temperature and volume and not capable of performing electrical or other non-PV work, the total free energy during a spontaneous change can only decrease.
This result seems to contradict the equation [itex]dA = -S dT - P dV[/itex] as keeping [itex]T[/itex] and [itex]V[/itex] constant seems to imply [itex]dA = 0[/itex] and hence [itex]A = [/itex] constant..
is that your assumption?stevendaryl said:The discussion in Wikipedia is misleading, because of course, a system can't do any work if the volume is not allowed to change. I think that the issue is resolved by assuming that there is more than one system involved. The volume of the composite system doesn't change, but the volumes of the smaller subsystems can change, and Helmholtz free energy is a measure of the capacity of the subsystem to do work.
You are correct that no P-V work is done. So, if the only focus is P-V work, and the P-V work is zero, then the inequality is satisfied. But there are other kinds of work that can be done on the system (or the system can do on the surroundings), and, if these are operative, then the inequality applies. For a more detailed derivation and description of all this, see Denbigh, Principles of Chemical Equilibrium.tze liu said:Sorry,i want to ask a question here
the note said the volume is "fixed" here.
if the volume if fixed,how comes the work done(because no change of volume) here
i totally get lost here
thank
""But there are other kinds of work that can be done on the system""Chestermiller said:You are correct that no P-V work is done. So, if the only focus is P-V work, and the P-V work is zero, then the inequality is satisfied. But there are other kinds of work that can be done on the system (or the system can do on the surroundings), and, if these are operative, then the inequality applies. For a more detailed derivation and description of all this, see Denbigh, Principles of Chemical Equilibrium.
Yestze liu said:""But there are other kinds of work that can be done on the system""
thank
do u means those work done doesn't change the volume of the system?
when I study thermodynamic,Chestermiller said:Yes
Electrochemical work, stirring worktze liu said:when I study thermodynamic,
i never heard that the work done doesn't change the volume of the system
So what is such "work done" looks like and how does this "work done" act on the system?
do u have any example?thank
why those work doesn't change the volume?Chestermiller said:Electrochemical work, stirring work
When you stir a liquid in a rigid closed container, does the volume of the container or the liquid change?tze liu said:why those work doesn't change the volume?
oh i seeChestermiller said:When you stir a liquid in a rigid closed container, does the volume of the container or the liquid change?
Don't forget also the possibility of electrochemical work like a battery.tze liu said:oh i see
the Work here is the free mechanical work.
but not contributed to the change of volume.
thank
Chestermiller said:Don't forget also the possibility of electrochemical work like a battery.
is this sentense incorrect?Chestermiller said:Don't forget also the possibility of electrochemical work like a battery.
You can have a current-carrying wire coming into the container and another wire exiting the container, with the current driven by a battery situated inside the container (elecrochemical). In this case, the internal energy of the system in the container changes (i.e., the battery is running down). The current can drive a motor outside the container. So the system is doing work.tze liu said:is this sentense incorrect?
even that the system is thermal isolated and the volume is fixed
there are STILL SOME electrochemical work or other type of mechanical work here
so the delta U is not zero?
tze liu said:Sorry,i want to ask a question here
the note said the volume is "fixed" here.
if the volume if fixed,how comes the work done(because no change of volume) here
i totally get lost here
thank
Chestermiller said:You can have a current-carrying wire coming into the container and another wire exiting the container, with the current driven by a battery situated inside the container (elecrochemical). In this case, the internal energy of the system in the container changes (i.e., the battery is running down). The current can drive a motor outside the container. So the system is doing work.
No. If you had checked out Denbigh like I suggested, you would have found that T is the temperature of the reservoir and also the temperature of the system at the interface with the reservoir. It is also the initial and final temperature of the system in the initial and final equilibrium states. But, during the process, it is not the temperature of the system (which is non-uniform) except at the interface with the reservoir.tze liu said:By the way,Is that true the reverior and the system both are at the same constant temperature in my case?
however, there is dF=d(U-TS) in my equation.Chestermiller said:No. If you had checked out Denbigh like I suggested, you would have found that T is the temperature of the reservoir and also the temperature of the system at the interface with the reservoir. It is also the initial and final temperature of the system in the initial and final equilibrium states. But, during the process, it is not the temperature of the system (which is non-uniform) except at the interface with the reservoir.
Chet
If the temperature is not uniform within your system during the process, then F (which is a function of equilibrium state) is undefined. If you want to define F within your system when it is not at equilibrium, then you have to integrate the local free energy per unit mass over the mass of your system.tze liu said:however, there is dF=d(U-TS) in my equation.
how can they use the temperature in the reservoir but not using the temperature inside the system?
there are two T in this case
the T in the reservoir and the other T in my system which can be described by F free energy.
why the equation dF=d(U-TS) ignore the temperature change inside the system.
it seemsChestermiller said:No. If you had checked out Denbigh like I suggested, you would have found that T is the temperature of the reservoir and also the temperature of the system at the interface with the reservoir. It is also the initial and final temperature of the system in the initial and final equilibrium states. But, during the process, it is not the temperature of the system (which is non-uniform) except at the interface with the reservoir.
Chet
if the temperature is constant in my process, i assume this temperature is T2Chestermiller said:If the temperature is not uniform within your system during the process, then F (which is a function of equilibrium state) is undefined. If you want to define F within your system when it is not at equilibrium, then you have to integrate the local free energy per unit mass over the mass of your system.
As I said, see Denbigh.tze liu said:it seems
if the temperature is constant in my process, i assume this temperature is T2
and the environment temperature is T1 ( T1 cannot be T2,if so, then there is no heat transfer)
why it should be dF=d(U-T1S) but not dF=d(U-T2S) ?i don't know why they use the environment temperature in the equation but not using the system's temperature( assume the system has constant temperature)
It is only at the boundary that they match. Normal to the boundary, the gas temperature is varying. And the average gas temperature is different from its temperature (and the reservoir temperature) at the boundary.tze liu said:Oh I see.than you very much .There are still problems here.if the temperature at the beginning are equal to the reverior,how comes the heat transfer between them at the beginning?(because it violates the zero law)
Chestermiller said:Here is the direct quote from Denbigh: "Consider the special case (a) that the only heat transferred to the system is from a heat reservoir which remains at the constant temperature T; (b) that the initial and final temperatures, ##T_1## and ##T_2##, of the system are equal, and are equal to the temperature T of the reservoir."
Note that this description fails to mention the important fact that the temperature of the system is not uniform during the process, or that it is equal to T only at the boundary with the reservoir (except in the initial and final states).
that means the heat is transfer from the boundary to the gas,and cause the heat transfer from the heat bath to the system boundary?Chestermiller said:It is only at the boundary that they match. Normal to the boundary, the gas temperature is varying. And the average gas temperature is different from its temperature (and the reservoir temperature) at the boundary.
tze liu said:Look very strange.are you sure the temperature are equ
that means the heat is transfer from the boundary to the gas,and cause the heat transfer from the heat bath to the system boundary?
thank youjim mcnamara said:@tze liu - there is a free to download pdf version of what @Chestermiller referred to.
Hope you can get and read the book.
https://archive.org/details/ThePrinciplesOfChemicalEquilibrium
Chet - FWIW some folks in places outside the US have trouble finding or getting to sources via links - countrywide IP blacklistings, limits on search engines, etc.
On the bath side of the boundary, the thermal conductivity of the heat bath liquid is taken to be very large, so that the magnitude of the temperature gradient in the bath liquid is very small, even for a finite heat flux. This is an inherent characteristic of an ideal heat bath. So there can be heat flow in the bath to- and from the boundary even though the temperature gradient is tiny.tze liu said:that means the heat is transfer from the boundary to the gas,and cause the heat transfer from the heat bath to the system boundary?
oh i seeChestermiller said:On the bath side of the boundary, the thermal conductivity of the heat bath liquid is taken to be very large, so that the magnitude of the temperature gradient in the bath liquid is very small, even for a finite heat flux. This is an inherent characteristic of an ideal heat bath. So there can be heat flow in the bath to- and from the boundary even though the temperature gradient is tiny.
If the gas is compressed, its temperature rises above that of the bath fluid, or if is expanded the opposite happens. Or if you electrically heat the gas, the temperature will rise above that of the bath fluid. Another example is a chemical reaction in the gas that is exotheic or endothermic. This all results in heat flow. So the heat flow is the effect, not the cause.tze liu said:oh i see
what causes the transfer of heat?
the different temperature between the gas and the system boundary created a heat flow?
thank youChestermiller said:If the gas is compressed, its temperature rises above that of the bath fluid, or if is expanded the opposite happens. Or if you electrically heat the gas, the temperature will rise above that of the bath fluid. Another example is a chemical reaction in the gas that is exotheic or endothermic. This all results in heat flow. So the heat flow is the effect, not the cause.
The equation should really read $$C_v=\left(\frac{\partial U}{\partial T}\right)_v=T\left(\frac{\partial S}{\partial T}\right)_v$$ This a physical property relationship for the material that applies to any process. The problem with the example is the Q in the equation, which is path-dependent.tze liu said:thank you
i get confused in this part also
did they do something wrong?
for example Cv=(dQ/dT)v=T(dS/dT)v
however this is true only for reversible process
but the question doesn't state it is reversible
do u think the Q in the textbook here is wrong also?Chestermiller said:The equation should really read $$C_v=\left(\frac{\partial U}{\partial T}\right)_v=T\left(\frac{\partial S}{\partial T}\right)_v$$ This a physical property relationship for the material that applies to any process. The problem with the example is the Q in the equation, which is path-dependent.
but the definition of heat capacity is not dU/dt but dQ/dt~~Chestermiller said:The equation should really read $$C_v=\left(\frac{\partial U}{\partial T}\right)_v=T\left(\frac{\partial S}{\partial T}\right)_v$$ This a physical property relationship for the material that applies to any process. The problem with the example is the Q in the equation, which is path-dependent.