# 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? – Lapmid Feb 9 '17 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.. – physics123 Feb 9 '17 at 12:36
• You may have misunderstood ,because Q_rev is not a state function but Q_rev/T is – Lapmid Feb 9 '17 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 Feb 9 '17 at 17:05
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.