1
$\begingroup$

Throttling is an adiabatic process so $\delta Q=0$.

Enthalpy is defined as $$H=U+PV\\ dH=dU+d(PV)=\delta Q+VdP$$ But in throttling (or Joule-Kelvin expansion), $dS \neq 0$ since the process is irreversible, $$dS> \frac{\delta Q}{T} \to TdS >\delta Q.$$ But my lecture notes seem to suggest that $$dH=TdS+VdP,$$ which I interprete as $\delta Q=TdS=0$ but $dS, T\neq 0$. What's going on here?

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ dH=TdS+VdP applies only to the mutual changes between dH, dS, and dP between to closely neighboring (differentially separated) thermodynamic equilibrium states. But, in an irreversible process, dQ is not equal to TdS, so your interpretation is incorrect. $\endgroup$
    – Chet Miller
    Commented Dec 3, 2021 at 15:45
  • $\begingroup$ @Chet Miller So is $dS=TdS+VdP$ the more 'intrinsic' definition of enthalpy, and thus $dH=\delta Q +VdP$ only if $\delta Q=TdS$ (i.e. the process is reversible)? $\endgroup$
    – Chern-Simons
    Commented Dec 3, 2021 at 15:54
  • $\begingroup$ I don't know what you mean by intrinsic, but if you mean that H is only strictly defined for equilibrium states, then yes. And for non-reversible processes, I only allow myself to consider finite changes: $\Delta H=\Delta U+\Delta (PV)=Q-W+\Delta (PV)$ $\endgroup$
    – Chet Miller
    Commented Dec 3, 2021 at 16:02

1 Answer 1

Reset to default
2
$\begingroup$

  1. Relation between $dQ$ and $dS$ is $$ dS \ge \frac{dQ}{T} $$ where equality holds only for reversible process.

  2. The relation $$ dH = T dS + V dP $$ is a statement about how much one function of state (enthalpy) changes when there are changes in other functions of state (here, entropy and pressure). It does not matter how the system moves between $(S,P)$ and $(S + dS, P + dP)$; the above relation gives what the change in enthalpy will be. It is a bit like walking around on a hill. If you start at one place and finish at another, then it does not matter what path you took, or even if you hopped and jumped and thus left the surface of the hill altogether at some points: no matter what you do, when you arrive at the second place the net change in your height will be the same. In this analogy $H$ is the height of the hill and $S$ and $P$ are coordinates in the horizontal directions.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.