Where is the work done coming from in Helmholtz free energy

[Chestermiller]

but the definition of heat capacity is not dU/dt but dQ/dt~~

That's the freshman physics definition of heat capacity. When we got to thermodynamics, we learned that, if work is being done, the equation doesn't apply as well. The subscript v on Cv means that we can measure this physical property of the material by doing an experiment at constant volume and determining the amount of heat added. Then the measured Cv can be used with the other equations to do calculations even if the volume isn't constant.

 

[tze liu]

That's the freshman physics definition of heat capacity. When we got to thermodynamics, we learned that, if work is being done, the equation doesn't apply as well. The subscript v on Cv means that we can measure this physical property of the material by doing an experiment at constant volume and determining the amount of heat added. Then the measured Cv can be used with the other equations to do calculations even if the volume isn't constant.

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.

the example in my textbook implies that dQ is reversible but forget to use dQrev?

 

[Chestermiller]

the example in my textbook implies that dQ is reversible but forget to use dQrev?

Yes. But, now that you are using thermodynamics, it is no longer valid to use the definitions of the heat capacities in terms of Q. You should use them only in terms of U, H, and S. otherwise, you are going to get confused.

 

[tze liu]

8

did they make mistake again?

 

[tze liu]

Yes. But, now that you are using thermodynamics, it is no longer valid to use the definitions of the heat capacities in terms of Q. You should use them only in terms of U, H, and S. otherwise, you are going to get confused.

seems in this charter, they use the same logic where ds T=dQ again here

 

[Chestermiller]

8View attachment 208098did they make mistake again?

This is correct. Note the constant p.

 

[tze liu]

This is correct. Note the constant p.[/QUOT

constant pressure doesn't imply the heat transfer is reversible?

 

[tze liu]

This is correct. Note the constant p.

constant pressure doesn't imply the heat transfer is reversible? that means Cp=(dQ/dT)p less or equal to T(dS /dT)p

 

[Chestermiller]

seems in this charter, they use the same logic where ds T=dQ again here

dQ=TdS only for a reversible path. A reversible path using dQrev is the only way we know of for calculating $\Delta S$ between two thermodynamic equilibrium states of a system.

 

[Chestermiller]

constant pressure doesn't imply the heat transfer is reversible? that means Cp=(dQ/dT)p less or equal to T(dS /dT)p

I don't know about that. I NEVER use differentials for an irreversible path.

 

[tze liu]

This is correct. Note the constant p.

how are u know it is correct.

because it seems they assume it is reversible here?
Cp=(dQrev/dT)p =T(dS /dT)p

however, if the dQ is not dQrev (for example: Cp=(dQ/dT)p)
this relationship seems to be wrong?

 

[Lord Jestocost]

The heat capacity at constant volume, C_V, is the ratio δq/dT for a process in a closed constant-volume system with no nonexpansion work - that is, no work at all. The first law shows that under these conditions the internal energy change equals the heat: dU = δq. (Eq. 5.3.9). We can replace δq by dU and write C_V as a partial derivative:

C_V = (dU/dT )_V

From: THERMODYNAMICS AND CHEMISTRY, SECOND EDITION, by Howard DeVoe, Associate Professor Emeritus, University of Maryland (http://www2.chem.umd.edu/thermobook/)

 

[Chestermiller]

how are u know it is correct.

because it seems they assume it is reversible here?
Cp=(dQrev/dT)p =T(dS /dT)p

however, if the dQ is not dQrev (for example: Cp=(dQ/dT)p)
this relationship seems to be wrong?

Please forget about using Q or Qrev. You need to switch to the following definitions to successfully use heat capacities in thermodynamics: $$C_v=\left(\frac{\partial U}{\partial T}\right)_v$$
$$C_p=\left(\frac{\partial H}{\partial T}\right)_p$$With these definitions, you can NEVER go wrong.
If you stick with Q, you are going to get confused when you consider processes that are neither at constant volume or constant pressure.

 

[tze liu]

The heat capacity at constant volume, C_V, is the ratio δq/dT for a process in a closed constant-volume system with no nonexpansion work - that is, no work at all. The first law shows that under these conditions the internal energy change equals the heat: dU = δq. (Eq. 5.3.9). We can replace δq by dU and write C_V as a partial derivative:

C_V = (dU/dT )_V

From: THERMODYNAMICS AND CHEMISTRY, SECOND EDITION, by Howard DeVoe, Associate Professor Emeritus, University of Maryland (http://www2.chem.umd.edu/thermobook/)

Closed constant volume has no pv work but seems there may have other type of work done?

 

[Chestermiller]

how are u know it is correct.

because it seems they assume it is reversible here?
Cp=(dQrev/dT)p =T(dS /dT)p

however, if the dQ is not dQrev (for example: Cp=(dQ/dT)p)
this relationship seems to be wrong?

The total heat that flows is not required to be reversible. If the equation were written correctly, it would read:
$$C_p=\frac{Q}{(T_2-T_1)}=\frac{(S_2-S_1)}{\ln{(T_2/T_1)}}$$
where the subscripts 1 and 2 refer to the initial and final thermodynamic equilibrium states of the system, and where the pressure in state 1 is the same as the pressure in state 2, and the externally applied pressure to the system over the entire process path is the same as the in equilibrium states 1 and 2. Note that the heat does not have to be applied reversibly for this equation to be correct. During the process, the system (which is initially in thermodynamic equilibrium state 1), can be placed in contact with a reservoir at temperature $T_2$ and held in contact with this reservoir until the system reaches equilibrium in thermodynamic equilibrium state 2. The equation I wrote applies to this irreversible process equally well as to a process that took the system from state 1 to state 2 reversibly.

 

[Chestermiller]

Closed constant volume has no pv work but seems there may have other type of work done?

This equation is correct, by definition, irrespective of whether work is being done or heat is being added (or removed). It is a physical property relationship, independent of any process.