[GeLe5000]

Hello.

In his book "Atkins' Physical Chemistry" (Peter Atkins and Julio de Paula, Oxford University Press, ninth edition, 2010), Dr Atkins presents the second law of thermodynamics in chapter 3, and in pages 114-116, explains the following :

When a system undergoes a transformation at constant temperature and volume, there's a function A (Helmoltz energy) whose value is always decreasing when the process is spontaneous, as is increasing the entropy of the universe.

Spontaneity ---> ΔA < 0, ΔSuniv > 0

Helmoltz energy A is defined as A = U –TS, where U is the internal energy of the system, S its entropy and T the thermodynamic temperature (in Kelvin).

The tendency to lower A in a spontaneous process is solely a tendency towards a state of greater overall entropy (entropy of the universe).

The entropy of the universe is the sum of the system's entropy and the surrounding's entropy : Suniv = Ssystem + Ssurr.

The change in the surrounding's entropy corresponds to heat received or released by the system from or into the surroundings, following the relation ΔSsurr = - ΔU / T.

ΔA is also called the "maximum work" that a system can give in a process at constant temperature and volume. wmax = ΔA = ΔU – tΔS. (remember that wmax is negative when the system produces work)

When there's an increase in the entropy of the system, TΔS > 0 and the maximum work that can be obtained from the system is greater than ΔU. Indeed, there's a flow of heat from the surroundings, which can be used as work. And the entropy of the surroundings decreases because of this loss of heat towards the system.

Dr Atkins gives an example :

The (exothermic) combustion of 1 mole of glucose at 25°C (298 K) liberates 2808 kJ mol-1 as heat (ΔU = - 2808 kJ mol-1) and the change in entropy of the system, ΔS = + 259.1 J K-1 mol-1.

C6H12O6 + 6 O2 ---> 6 CO2 + 6 H2O
ΔU = - 2808 kJ mol-1
ΔS = + 259.1 J K-1 mol-1

The energy that can be extracted as work is thus ΔA = - 2808 – (298 x 0.2591) = - 2885 kJ mol-1.
                                             
On account of the positive entropy of reaction, the system can draw in energy from the surroundings, so reducing their entropy, and make it available for doing work.

However, to calculate the change in entropy of the surroundings, we can use the relation ΔSsurr = - ΔU / T, and we get ΔSsurr= - (-2808 / 298) = + 9.4 kJ K-1 mol-1. A POSITIVE VALUE. An INCREASE in the surrounding's entropy.

And for the change in entropy of the universe, we have  ΔSuniv = - ΔA / T = - (- 2885 / 298) = + 9.68 kJ K-1 mol-1. A positive value is correct since the combustion process is spontaneous.

And finally using the relation ΔSuniv = ΔSsystem + ΔSsurr, we see that the value ΔSuniv = + 9.68 will require a positive value of ΔSsurr : 9.68 = 0.2591 + 9.4.

After all, how could we get an increase in the entropy of the universe when the surroundings are supposed to see their entropy decreasing and the increase in the entropy of the system is so small (0.2591 J K-1)?

So, there must be a problem with Dr Atkin's explanation of the supplementary decrease in Helmoltz energy when the entropy of the system increases. It can't stem from a decrease in the entropy of the surroundings, since it's actually increasing.

Or am I wrong ?

[Yggdrasil]

After all, how could we get an increase in the entropy of the universe when the surroundings are supposed to see their entropy decreasing and the increase in the entropy of the system is so small (0.2591 J K-1)?

I don't understand your question here.  The entropy of the surrounds is increasing, not decreasing.  The entropy of the surroundings increases because heat gets transferred from the systems to the surroundings.  You calculated the increase in entropy of the surroundings to be + 9.4 kJ K-1 mol-1.

Are you asking about the statement "When there's an increase in the entropy of the system, TΔS > 0 and the maximum work that can be obtained from the system is greater than ΔU"?  In this case, I think the writing is just sloppy.  Clearly if ΔS > 0, then wmax = ΔA < ΔU.  He's probably just referring to the magnitude of wmax and ΔU: |wmax| > |ΔU|, which is true.

[GeLe5000]

Thank you for your comment.

If ΔS > 0, then TΔS > 0, then (- TΔS) < 0 and ΔA is more negative. Consquently, ΔA = ΔU - TΔS = - 2885 instead of - 2808, then more work is available. It's correct I think.

But the real problem is what appears on fig. 3.17 of page 116. It is written "ΔSsur < 0" and it is not an "erreur de frappe" (in French), it's really part of Dr Atkin's demonstration (on fig. 3.16, we have ΔSsur > 0, in the case of ΔSsyst < 0). He insists that when there's an increase in entropy of the system, there's a flow of heat from the surroundings into the system because "there's room for it", which leads to a decrease of the entropy of the surroudings.

So I think that Dr Atkins can't explain the change in entropy of the universe. It's a bit disturbing and uncomfortable.

It would be nice if you could have a look into this book.

Thank you again.

[GeLe5000]

All right, I think that I understand (I hope so).
When he writes "ΔSsur < 0", it doesn't mean that it's the final state of the surroundings. It means that a flow of heat into the system will lower the actual increase in the surrouding's entropy.
All is well in the Universe, thank you very much.
But he should have thought that writing "ΔSsur < 0" on the figure could be misleading when we refer to the equation ΔSuniv = ΔSsyst + ΔSsur.
Thank you for your help.