The fundamental thermodynamic relation

[Andrew Davies]

Why is it that the The fundamental thermodynamic relation dU = Tds - PdV works in general even though Tds > dQ for irreversible processes. Likiewise PdV >/= dW for irreversible processes. Do the two irreversible effects cancel out. How does this happen physically.

 

[netheril96]

Why is it that the The fundamental thermodynamic relation dU = Tds - PdV works in general even though Tds > dQ for irreversible processes. Likiewise PdV >/= dW for irreversible processes. Do the two irreversible effects cancel out. How does this happen physically.

Who tells you dU = Tds - PdV works in general?It definitely works only for reversible process!

 

[DrDu]

No, it really works in general (it has to be generalized if there are chemical reactions or electric fields, however), at least when you use it to calculate the change in energy between the starting and end point of the process.
But let me consider a simple example: suppose you are stirring a glass of some viscose liquid. The work done is not equal to -pdV obviously, ideally the volume does not change at all. Nevertheless all the work done on the liquid will eventually end up increasing the temperature of the liquid. So this is a completely irreversible process where all the work done is converted into entropy.
You may consider a reversible process (at least as judged from the liquid sub-system considered) where I heat up the liquid by bringing the liquid into contact with a heat bath whose temperature slowly increases (which can be due to a reversible or irreversible process). If the heat capacity of the heat bath is very small compared with the capacity of the liquid, I could imagine that the increase of temperature of the heat bath is also realized by stirring some viscous liquid (or rubbing the outside of the container). So the energy change dU is really the same in both cases, only that in one case the irreversible step takes place inside the container (then W neq -pdV=0 and TdS=W neq Q=0) and in the other case outside (then W =-pdV=0 and TdS=Q). In both cases dU=TdS-pdV and dU=Q+W are true.

 

[netheril96]

No, it really works in general (it has to be generalized if there are chemical reactions or electric fields, however), at least when you use it to calculate the change in energy between the starting and end point of the process.
But let me consider a simple example: suppose you are stirring a glass of some viscose liquid. The work done is not equal to -pdV obviously, ideally the volume does not change at all. Nevertheless all the work done on the liquid will eventually end up increasing the temperature of the liquid. So this is a completely irreversible process where all the work done is converted into entropy.
You may consider a reversible process (at least as judged from the liquid sub-system considered) where I heat up the liquid by bringing the liquid into contact with a heat bath whose temperature slowly increases (which can be due to a reversible or irreversible process). If the heat capacity of the heat bath is very small compared with the capacity of the liquid, I could imagine that the increase of temperature of the heat bath is also realized by stirring some viscous liquid (or rubbing the outside of the container). So the energy change dU is really the same in both cases, only that in one case the irreversible step takes place inside the container (then W neq -pdV=0 and TdS=W neq Q=0) and in the other case outside (then W =-pdV=0 and TdS=Q). In both cases dU=TdS-pdV and dU=Q+W are true.

Your example feels weird to me
Moreover,in most irreversible process,the matter involved is not homogeneous,such as the free expansion,which means they cannot be described with a single P,T or something else,so the formula dU=Tds-pdV definitely fails since there is no T and P.

"During free expansion, no work is done by the gas. The gas goes through states of no thermodynamic equilibrium before reaching its final state, which implies that one cannot define thermodynamic parameters as values of the gas as a whole. For example, the pressure changes locally from point to point, and the volume occupied by the gas (which is formed of particles) is not a well defined quantity."

----- http://en.wikipedia.org/wiki/Free_expansion

 

[DrDu]

You are right. But, as I wrote, in thermodynamics the initial and final states are necessarily equilibrium states, so you can use it to calculate the change of U and S between the initial and final state.
The example of stirring appears e.g. in the book on thermodynamics by Max Planck (who did his thesis on thermodynamics) and thus is quite a classic.
In the case of free expansion, you know that Delta U=0 as neither work nor heat is exchanged. On the other hand, you can calculate int (-p dV) for a reversible path connecting the initial and final state and thus get Delta S.

 

[netheril96]

You are right. But, as I wrote, in thermodynamics the initial and final states are necessarily equilibrium states, so you can use it to calculate the change of U and S between the initial and final state.
The example of stirring appears e.g. in the book on thermodynamics by Max Planck (who did his thesis on thermodynamics) and thus is quite a classic.
In the case of free expansion, you know that Delta U=0 as neither work nor heat is exchanged. On the other hand, you can calculate int (-p dV) for a reversible path connecting the initial and final state and thus get Delta S.

You are just constructing a reversible process connecting the innitial and terminal state,but it doesn't mean that the formula applies to the corresponding irreversible process

 

[sawan.patnaik]

how can we explain for perfect gas (dU/dV)=0, using maxwell's thermo relation?

please help me, i can't solve this.

 

[netheril96]

how can we explain for perfect gas (dU/dV)=0, using maxwell's thermo relation?

please help me, i can't solve this.

$${\left( {\frac{{\partial U}}{{\partial V}}} \right)_T} = {\left( {\frac{{\partial U}}{{\partial S}}} \right)_V}{\left( {\frac{{\partial S}}{{\partial V}}} \right)_T} + {\left( {\frac{{\partial U}}{{\partial V}}} \right)_S} = T{\left( {\frac{{\partial p}}{{\partial T}}} \right)_V} - p$$

Substitute the $$pV = nRT$$ in and you will get zero

 

[nonequilibrium]

You are just constructing a reversible process connecting the innitial and terminal state,but it doesn't mean that the formula applies to the corresponding irreversible process

The argument one uses to say it also does apply for the irreversible process, is that the quantities you deal with (U,S,V) are state variables, and seeing as the reversible and the irreversible process in question have the same begin and end-point, both predict the same changes in the state variables.

The fact that U and S are well-defined state variables is another thing, but I think we just presume they are?