Is heat transfer and work done per unit mass path-independent since it can be expressed in terms of intensive thermodynamic properties of a system?

Question

The argument given for $du = Tds-Pdv$ being applicable to irreversible processes too (even though it was derived using dQ = TdS for a reversible process, from Clausius' inequality. For a non reversible process it would be dQ<TdS) is that the equation is expressed in terms of intensive thermodynamic properties and so the process path doesn't matter since it's all state functions.

Does this mean that work and heat transfer, per unit mass, are both independent of path? i.e if it's being stated that the Pdv term is state dependant(since it's a combination of intensive thermodynamic properties), isn't that saying that the work per unit mass, which is also Pdv, is path independent

Some additional background: The source of the confusion was the statement made at 4:50 in the lecture here. Which was that since the relation contains intensive properties only, it's not dependant on path and is always valid. So I thought that if any relation involving only intensive properties is path independant, maybe so should work per unit mass = Pdv (v = specific volume)

This has been clarified in the bottom note in chemomechanics' answer (the accepted answer).

Answer 1

Given that heat is path dependent and heat per unit mass $dq = \frac{dQ}{M}$, heat per unit mass being path independent would imply that the total mass of the system $M$ was path dependent, which is clearly untrue (for a closed system at the very least).

In general the identifications of heat and work (total or per unit mass) with $TdS$ and $-PdV$ hold only for reversible paths and it is only the combination that is path independent.

Answer 2

The argument given for $du = Tds-Pdv$ being applicable to irreversible processes too...is that the equation is expressed in terms of thermodynamic properties and so the process path doesn't matter since it's all state functions.

The argument is that since internal energy is a state property, any change in internal energy between equilibrium states is the same for any process, reversible or irreversible. That argument does not extend to the individual components of the change in internal energy, namely heat and work, which are individually process dependent.

That said, the equation

$$dU=TdS-Pdv\tag{1}$$

is equivalent to the first law for the case of a reversible process, provided that $P$ is the equilibrium pressure of the gas, in which case all the terms are system properties.

Does this mean that work and heat transfer, per unit mass, are both independent of path?

No. It means that the combination of heat and work in the first law are path independent, not that they are individually path independent. Even if heat and work are both reversible, there can still be an infinite number of reversible paths between two equilibrium states.

Hope this helps.

Answer 3

Does this mean that work and heat transfer, per unit mass, are both independent of path?

No, it means only that their sum ($u$, the internal energy per unit mass) is path independent.

isn't that saying that the work per unit mass, which is also Pdv, is path independent

(Edited to correct an error.) No; just because $T\,ds -P\,dv$ adds up to $u$ and $\frac{Q}{m}+ \frac{W}{m}$ also adds up to $u$ doesn’t mean the components are individually equal! That’s true only for reversible processes. Furthermore, $P\,dv$ is not path-independent, although $PV$ is.

Make sure not to confuse the external pressure $P_\mathrm{ext}$ in general work expressions ($P_\mathrm{ext}dV$) with the system pressure $P$. They aren’t necessarily the same, and using the same variable for each can cause a lot of confusion.

A possible useful note from our discussion: The "contains state variables" argument is really meant to apply to the definition $U=TS-PV+\sum_i\mu_iN_i$. Each term there is a product of state variables, so each term and the sum are state variables. From that definition, the Gibbs–Duhem equation, and $dN_i=0$ for closed systems, we get $dU=T\,dS-P\,dV$. This equation combined with references to "containing state variables" could lead one to conclude incorrectly that $P\,dV$, for example, is a state variable.

Answer 4

The equation dU=TdS-PdV refers to two thermodynamic equilibrium states that are differentially separated from one another. The path between these two neighboring states can be direct, or it can be very tortuous and irreversible (in terms of the P-, V-, and Q variations), as long as, in the end, the two states are differentially separated and each in thermodynamic equilibrium.

Answer 5

Th equality $dU=TdS -pdV$, or more generally, $$dU=TdS+\sum_k Y_kdX_k \tag{1}\label{1}$$ does not describe a process, instead it describes the $U=U(S,X_1, X_2.,,)$ internal energy function and its partial derivatives $T=\frac{\partial U}{\partial S}$ and $Y_k=\frac{\partial U}{\partial X_k}$ as the intensive physical quantities conjugate to the extensive charges $S, X_1,X_2,..$. The $\eqref{1}$ can be also viewed as being the difference of the internal energies between two infinitesimally close equilibrium states but again it does not describe any process between those states. Neither the differential $dU$ or the other differentials $TdS$, $Y_kdX_k$ are state functions because they depend on two (here nearby) states and not on one.

It is possible to view the infinitesimal quantities $dS, dX_k$ as being added to the system during a process that started from the sate $\{S, X_k\}$ but to conclude that the internal energy change is $dU$ per $\eqref{1}$ you need other assumptions, as well, the most important being are

(1) that as the result of such transfer the intensives $\{T,Y_k\}$ do not change or if they change the result is at least 2nd order negligible and

(2) the extensive variables are conserved, that is if $dX_k^0$ transported (exchanged with the environment) then $dX_k^0=dX_k$ is also the local infinitesimal change - very nontrivial assumption

(3) denote by $dS^0$ the transported entropy from the outside, then for a reversible process $dS^0=dS$ and for an irreversible process $dS^0 < dS$ where $dS$ is the local change measured at the equilibrium state $S+dS, \{X_k+dX_k\}$ whose energy now is $$U(S+dS,X_1+dX_1, X_2+dX_2,..)= U(S,X_1,X_2,..)+TdS+\sum_kY_kdX_k\tag{2}\label{2}.$$ (Note that $dS$ shows up in $\eqref{2}$ and not $dS^0$; the question is how do we know from $dS^0$ what should we have for $dS$ when the two are not equal?)