- #1
nonequilibrium
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- 2
Hello,
So presume we have a system in which a chemical process A + B -> X + Y is happening. We allow it to be a non-equilibrium process (so there will be an entropy production inside the system) but for ease we presume the system is characterized by the usual variables E, V, N_A, ..., N_Y (and the homogenous T, P, \mu_i), i.e. no local densities.
In a book I found that \mathrm d S = \frac{1}{T} \left( \mathrm d Q - \sum \mu_i \mathrm d N_i \right) where they regard the first term as an entropy flux, i.e. an equilibrium process (I presume Q is simply the energy the system gets from an environment in equilibrium). Hence they explicitly draw the distinction \mathrm d_e S = \frac{\mathrm d Q}{T} which is the entropy flux from the environment, and \mathrm d_i S = - \frac{1}{T} \sum \mu_i \mathrm d N_i, which is the entropy produced internally, by the chemical process (remember: non-equilibrium).
But I was wondering: are they then neglecting energy production from the chemical reaction? Or am I overlooking something? For example, is it allowed, in a more general case, for there to be a ``\mathrm d_i Q'' which would stand for the energy produced in the chemical reaction? Hence in that case d_e S would go unchanged and we would have \mathrm d_i S = \frac{1}{T} \mathrm d_i Q - \frac{1}{T} \sum \mu_i \mathrm d N_i.
Hence if we write \mathrm d_i S = \sum_j X_j J_j (= entropy production in terms of thermodynamic forces X_j and currents J_j) we would have that the heat production would have the thermodynamic force \frac{1}{T} (which is notably different from the thermodynamic force for heat conduction, being \nabla \frac{1}{T} or sometimes written as \sim \nabla T (Fourier's law!))
The thing I'm also wondering about: I'm saying "the energy created by the chemical reaction" but of course there is no real energy created: the energy was there all along. So does it make sense to say that thermodynamically energy was created, but fundamentally there was not?
So presume we have a system in which a chemical process A + B -> X + Y is happening. We allow it to be a non-equilibrium process (so there will be an entropy production inside the system) but for ease we presume the system is characterized by the usual variables E, V, N_A, ..., N_Y (and the homogenous T, P, \mu_i), i.e. no local densities.
In a book I found that \mathrm d S = \frac{1}{T} \left( \mathrm d Q - \sum \mu_i \mathrm d N_i \right) where they regard the first term as an entropy flux, i.e. an equilibrium process (I presume Q is simply the energy the system gets from an environment in equilibrium). Hence they explicitly draw the distinction \mathrm d_e S = \frac{\mathrm d Q}{T} which is the entropy flux from the environment, and \mathrm d_i S = - \frac{1}{T} \sum \mu_i \mathrm d N_i, which is the entropy produced internally, by the chemical process (remember: non-equilibrium).
But I was wondering: are they then neglecting energy production from the chemical reaction? Or am I overlooking something? For example, is it allowed, in a more general case, for there to be a ``\mathrm d_i Q'' which would stand for the energy produced in the chemical reaction? Hence in that case d_e S would go unchanged and we would have \mathrm d_i S = \frac{1}{T} \mathrm d_i Q - \frac{1}{T} \sum \mu_i \mathrm d N_i.
Hence if we write \mathrm d_i S = \sum_j X_j J_j (= entropy production in terms of thermodynamic forces X_j and currents J_j) we would have that the heat production would have the thermodynamic force \frac{1}{T} (which is notably different from the thermodynamic force for heat conduction, being \nabla \frac{1}{T} or sometimes written as \sim \nabla T (Fourier's law!))
The thing I'm also wondering about: I'm saying "the energy created by the chemical reaction" but of course there is no real energy created: the energy was there all along. So does it make sense to say that thermodynamically energy was created, but fundamentally there was not?