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user796099
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I think the answer is no, there is no such thing as the Gibbs energy of the universe, and thus no way to say that it only decreases. The reason is that thermodynamics does not take account of how things got to be the way they are. If a raised weight happens to be conveniently waiting there attached to a pulley, which is attached to a mechanism that allows it to drive the pistons of a Van't Hoff equilibrium box, in which a chemical reaction with $\Delta G > 0$ takes place, thermodynamics does not ask how all that stuff got there. Unless I am mistaken, the potential energy, $mgh$, of the weight does not figure into its Gibbs energy because it is not internal energy (though I would be excited to be corrected on this if I am mistaken). Thus it is entirely possible for the weight to fall very slowly and reversibly drive the increase of free energy of the chemical system without any compensating decrease of free energy elsewhere. In the reversible limit, the falling weight does not generate any compensating entropy, either, which is frequently claimed to be necessary to avoid violating the second law. The position of the weight has no entropy associated with it.

So does this net $\Delta G > 0$ process mean that the total entropy decreases, violating the second law? No, it doesn't. The formula $\Delta G = -T\Delta S_{total}$ only applies if there is no work being done on the system other than expansion or compression work. That formula is derived by assuming that the change in enthalpy is equal to the heat absorbed from the surroundings, or $\Delta H = Q$, which is the usual situation for a reaction in a lab. In that case $\Delta S_{total} = \Delta S + \Delta S_{surroundings} = \Delta S - Q/T = \Delta S - \Delta H/T = -\Delta G/T$$$\Delta S_{total} = \Delta S + \Delta S_{surroundings} = \Delta S - Q/T = \Delta S - \Delta H/T = -\Delta G/T$$.

But if work is also done on the system, then the enthalpy change is not just the heat absorbed. You also have to add the work done.

$$ \Delta H = \Delta(U+PV) = Q + W + P\Delta V = Q + W - W_{PV} = Q + W_{non-PV}$$.

Now when you calculate the total entropy change, you get

$$ \Delta S_{total} = \Delta S + \Delta S_{surroundings} = \Delta S - Q/T = \Delta S - (\Delta H-W_{non-PV})/T = (W_{non-PV}-\Delta G)/T \geq 0$$.

The last inequality comes from $W_{non-PV} \geq \Delta G$, which follows from Clausius' inequality, $\Delta S \geq Q/T$. So the second law checks out, even though it seems like some free energy was created out of nowhere (but actually was converted from the potential energy of the raised weight).

I think the answer is no, there is no such thing as the Gibbs energy of the universe, and thus no way to say that it only decreases. The reason is that thermodynamics does not take account of how things got to be the way they are. If a raised weight happens to be conveniently waiting there attached to a pulley, which is attached to a mechanism that allows it to drive the pistons of a Van't Hoff equilibrium box, in which a chemical reaction with $\Delta G > 0$ takes place, thermodynamics does not ask how all that stuff got there. Unless I am mistaken, the potential energy, $mgh$, of the weight does not figure into its Gibbs energy because it is not internal energy (though I would be excited to be corrected on this if I am mistaken). Thus it is entirely possible for the weight to fall very slowly and reversibly drive the increase of free energy of the chemical system without any compensating decrease of free energy elsewhere. In the reversible limit, the falling weight does not generate any compensating entropy, either, which is frequently claimed to be necessary to avoid violating the second law. The position of the weight has no entropy associated with it.

So does this net $\Delta G > 0$ process mean that the total entropy decreases, violating the second law? No, it doesn't. The formula $\Delta G = -T\Delta S_{total}$ only applies if there is no work being done on the system other than expansion or compression work. That formula is derived by assuming that the change in enthalpy is equal to the heat absorbed from the surroundings, or $\Delta H = Q$, which is the usual situation for a reaction in a lab. In that case $\Delta S_{total} = \Delta S + \Delta S_{surroundings} = \Delta S - Q/T = \Delta S - \Delta H/T = -\Delta G/T$.

But if work is also done on the system, then the enthalpy change is not just the heat absorbed. You also have to add the work done.

$$ \Delta H = \Delta(U+PV) = Q + W + P\Delta V = Q + W - W_{PV} = Q + W_{non-PV}$$.

Now when you calculate the total entropy change, you get

$$ \Delta S_{total} = \Delta S + \Delta S_{surroundings} = \Delta S - Q/T = \Delta S - (\Delta H-W_{non-PV})/T = (W_{non-PV}-\Delta G)/T \geq 0$$.

The last inequality comes from $W_{non-PV} \geq \Delta G$, which follows from Clausius' inequality, $\Delta S \geq Q/T$. So the second law checks out, even though it seems like some free energy was created out of nowhere (but actually was converted from the potential energy of the raised weight).

I think the answer is no, there is no such thing as the Gibbs energy of the universe, and thus no way to say that it only decreases. The reason is that thermodynamics does not take account of how things got to be the way they are. If a raised weight happens to be conveniently waiting there attached to a pulley, which is attached to a mechanism that allows it to drive the pistons of a Van't Hoff equilibrium box, in which a chemical reaction with $\Delta G > 0$ takes place, thermodynamics does not ask how all that stuff got there. Unless I am mistaken, the potential energy, $mgh$, of the weight does not figure into its Gibbs energy because it is not internal energy (though I would be excited to be corrected on this if I am mistaken). Thus it is entirely possible for the weight to fall very slowly and reversibly drive the increase of free energy of the chemical system without any compensating decrease of free energy elsewhere. In the reversible limit, the falling weight does not generate any compensating entropy, either, which is frequently claimed to be necessary to avoid violating the second law. The position of the weight has no entropy associated with it.

So does this net $\Delta G > 0$ process mean that the total entropy decreases, violating the second law? No, it doesn't. The formula $\Delta G = -T\Delta S_{total}$ only applies if there is no work being done on the system other than expansion or compression work. That formula is derived by assuming that the change in enthalpy is equal to the heat absorbed from the surroundings, or $\Delta H = Q$, which is the usual situation for a reaction in a lab. In that case $$\Delta S_{total} = \Delta S + \Delta S_{surroundings} = \Delta S - Q/T = \Delta S - \Delta H/T = -\Delta G/T$$.

But if work is also done on the system, then the enthalpy change is not just the heat absorbed. You also have to add the work done.

$$ \Delta H = \Delta(U+PV) = Q + W + P\Delta V = Q + W - W_{PV} = Q + W_{non-PV}$$.

Now when you calculate the total entropy change, you get

$$ \Delta S_{total} = \Delta S + \Delta S_{surroundings} = \Delta S - Q/T = \Delta S - (\Delta H-W_{non-PV})/T = (W_{non-PV}-\Delta G)/T \geq 0$$.

The last inequality comes from $W_{non-PV} \geq \Delta G$, which follows from Clausius' inequality, $\Delta S \geq Q/T$. So the second law checks out, even though it seems like some free energy was created out of nowhere (but actually was converted from the potential energy of the raised weight).

Source Link
user796099
user796099
  • 139
  • 3

I think the answer is no, there is no such thing as the Gibbs energy of the universe, and thus no way to say that it only decreases. The reason is that thermodynamics does not take account of how things got to be the way they are. If a raised weight happens to be conveniently waiting there attached to a pulley, which is attached to a mechanism that allows it to drive the pistons of a Van't Hoff equilibrium box, in which a chemical reaction with $\Delta G > 0$ takes place, thermodynamics does not ask how all that stuff got there. Unless I am mistaken, the potential energy, $mgh$, of the weight does not figure into its Gibbs energy because it is not internal energy (though I would be excited to be corrected on this if I am mistaken). Thus it is entirely possible for the weight to fall very slowly and reversibly drive the increase of free energy of the chemical system without any compensating decrease of free energy elsewhere. In the reversible limit, the falling weight does not generate any compensating entropy, either, which is frequently claimed to be necessary to avoid violating the second law. The position of the weight has no entropy associated with it.

So does this net $\Delta G > 0$ process mean that the total entropy decreases, violating the second law? No, it doesn't. The formula $\Delta G = -T\Delta S_{total}$ only applies if there is no work being done on the system other than expansion or compression work. That formula is derived by assuming that the change in enthalpy is equal to the heat absorbed from the surroundings, or $\Delta H = Q$, which is the usual situation for a reaction in a lab. In that case $\Delta S_{total} = \Delta S + \Delta S_{surroundings} = \Delta S - Q/T = \Delta S - \Delta H/T = -\Delta G/T$.

But if work is also done on the system, then the enthalpy change is not just the heat absorbed. You also have to add the work done.

$$ \Delta H = \Delta(U+PV) = Q + W + P\Delta V = Q + W - W_{PV} = Q + W_{non-PV}$$.

Now when you calculate the total entropy change, you get

$$ \Delta S_{total} = \Delta S + \Delta S_{surroundings} = \Delta S - Q/T = \Delta S - (\Delta H-W_{non-PV})/T = (W_{non-PV}-\Delta G)/T \geq 0$$.

The last inequality comes from $W_{non-PV} \geq \Delta G$, which follows from Clausius' inequality, $\Delta S \geq Q/T$. So the second law checks out, even though it seems like some free energy was created out of nowhere (but actually was converted from the potential energy of the raised weight).