# Q reversible in thermodynamics

Could someone please help me to understand why Q reversible is a state function while Q is a path function(where Q is the heat energy absorbed or released by the system) in relation to thermodynamics. This would also help me to understand the basic definition of entropy. Much thanks in advance!

• Where did you read $Q_(rev)$ is a state function?
– Paul
Commented Feb 9, 2017 at 11:48
• I read it in a book of mine...it said that entropy too is a state function since it is defined as Q(rev)/T..and since Q(rev) is another state function.. Commented Feb 9, 2017 at 12:36
• You may have misunderstood ,because Q_rev is not a state function but Q_rev/T is
– Paul
Commented Feb 9, 2017 at 12:56

Q reversible is not a state function. Imagine a great big Carnot cycle and a tiny little Carnot cycle (on a P-V diagram), both starting at the same state. $\Delta S$ is zero for both cycles, but the big cycle does much more reversible work and has much larger Q reversible than the tiny little Carnot cycle.

• The answer is not straightforward. How do you reach a conclusion based on a cycle, in which both Q and T change (within the cycle) (and $\Delta S=0$), with the heat dQ=TdS? (or, if you want to use S=Q/T, then T has to be constant)
– user126422
Commented Feb 9, 2017 at 17:05
• You know that more heat is transferred and more work is done in the large cycle than the small cycle. Suppose that for the small cycle the size of the cycle is zero, so that no heat is transferred and no work is done at all. Certainly, any Carnot cycle of finite size will have net heat transfer and work done. Yet the change in entropy for the two cases is zero. So the only conclusion you can reach from this comparison is that Q reversible is not a state function. Commented Feb 9, 2017 at 17:17
• @AlbertAspect: Are the initial and final states of the two reversible process paths the same (namely, the initial state is the same as the final state of the system for these cycles)? Are the amounts of heat transferred in these two reversible process paths the same? If the answer to the latter question is "no," then a different amount of reversible heat is transferred in one process path than the other between the same two initial and final thermodynamic equilibrium states of the system. This means that Q rev is not a state function. Commented Feb 9, 2017 at 17:28

The general definition is $dS=\frac{dQ}{T}$. This means that $S=\frac{Q}{T}$ is only valid along an isotherm. If you move only along the same isotherm, then $Q$ would be indistinguishable from a function of state. As Chester showed on his answer, this is not valid in the general case. More explicitly,
Q being a function of state requires that $Q_f=Q_i+∫dQ$ for any path, and this is impossible because for a cycle $Q_i=Q_f$, thus $∫dQ=0$, which is not the case, for instance, in a carnot cycle.

• It doesn't seem very complicated to me. All I had to provide was a single example that shows that, for two different reversible process paths between the same pair of initial and final thermodynamic equilibrium states of a system, the amounts of reversible heat are different. That would certainly prove that the reversible Q is not a state function. Commented Feb 9, 2017 at 18:17
• Yes, our arguments are the same. I do not know why I still find it convoluted (but not incorrect). I did not get the right idea on my first reading. I moved my comment to my answer.
– user126422
Commented Feb 9, 2017 at 18:31