The reasons are that:
1) As we know, Variational Calculus is about Functional, however, here, we have known Q=f(T, V, P), Q=f(T, V, P) is a normal function, if you think δQ as the variation, then, what is the Functional ?
The Differential of Q can only be dQ, dQ=df(T, V, P) is a perfect differential.
△Q becoming δQ=δf(T, V, P) is taking for granted, for there is NO functional problem at all.
2) On the other hand, the problem is not whether dQ is meaningful or not here, it is (1/T))dQ is meaningless !
In △Q/T, Q and T could be any relationship( general form is X△Y) , but in dQ/T, they(Q and T) have to be functional relationships, however, dQ/T=(1/T)df(T, V, P) is meamingless.
that is △Q/T can not become dQ/T at all.
3) How can you turn △Q/T into δQ/T ?
In fact, we have known Q=f(T, V, P) , Q has nothing to do with Functional or Variational Calculus, it is just a normal function, △Q/T can not turn into δQ/T=δf(T, V, P)/T at all, δf(T, V, P)/T just doesn't make sense, for, here, δf(T, V, P)/T is actually df(T, V, P)/T .
The problem is that "Q" itself without the delta is meaningless. There is no "heat" function in thermodynamics. So, Q=f(T,V,p) is not defined. It only makes sense in a process between two points, as the amount of heat supplied to the system between two states. So, you always have δQ or △Q, and the first notation has nothing to do with variations, it is just used to refer to a process between two infinitesimally close states. Even in this case, δQ is not a true differential. That's why the authors refrain from using dQ. It depends on the process of going from A to A' close to A. But it turns out, the integral of δQ/T between two states A and B, is the same for all reversible processes imaginable between the two (the Clausius theorem). Thus, δQ/T has to relate to a total differential of some state function (because its change is path independent). This function is called the entropy, with △S = S_B - S_A = int (δQ/T) and dS = δQ/T. But still, you only have access to the entropy difference between states, and absolute entropy is only calculable if you fix the entropy of one given state. This is done through the Nernst–Simon theorem (the third law of thermodynamics).
Q is meaningful.
Please read my paper, and the comments and replies of my paper.
Q is meaningful.
The reason is that It must be (T, V, P) and only need (T, V, P) to define a state of a system.
There is a one-to-one correspondence between Q ( W or E ) and the system state (T, V, P) of a given path.
so, there must be Q= fq(T, V, P) , W= fw(T, V, P) and E= fe(T, V, P) , and
P-V chart should be P-V-T chart.
The difference among Q ,W and E is that Q = fq(T, V, P) and W= fw(T, V, P) are process quantitys which vary with path;
but E= fe(T, V, P) is the system state variable, the functional forms of E change with the paths, the result of E= fe(T, V, P) does not vary with paths between the same original and terminal states.
Because it(T, V, P) is constrained by system state equation, to a given system, T, V and P are not independent each other, in fact there are T= ft(V, P) , V= fv(T, P) and P= fp(T, V),
for example: ideal gas state equation PV=(vR)T or Van der Waals equation.
there are：
W= fw(T, V, P) is equivalent to W= fwp(T, V) or W= fwt(V, P)
or W= fwv(T, P)
Q= fq(T, V, P) is equivalent to Q= fqp(T, V) or Q= fqt(V, P)
or Q= fqv(T, P)
E= fe(T, V, P) is equivalent to E= fep(T, V) or E= fet(V, P)
or E= fev(T, P)
Q=f(T, V, P) , does it result in ΔQ=0 in cycle process ?
NO !
Many process functions are piecewise functions, so is cycle process function.
e.g. Carnot cycle process function:
Carnot cycle process function include four processes, that is two adiabatic processes and two isothermal processes :
1) In adiabatic process: ΔQ=0;
2) The two isothermal processes correspond to two functions
Q1=f1(T, V, P) and Q2=f2(T, V, P).
and
Q1=f1(T, V, P)=f1( (T—T1), ( V—V1), (P—P1) )
Q2=f2(T, V, P)=f2( (T—T3), ( V—V3), (P—P3) )
here T1, V1, P1 and T3, V3, P3 are constant .
then
ΔQ1=f1(T2, V2, P2)—f1(T1, V1, P1)
=f1( (T2—T1), ( V2—V1), (P2—P1) ) — f1( (T1—T1), ( V1—V1), (P1—P1) )
=f1( (T2—T1), ( V2—V1), (P2—P1) ) — 0
=f1( (T2—T1), ( V2—V1), (P2—P1) )
ΔQ2=f2(T4, V4, P4)—f2(T3, V3, P3)
=f2( (T4—T3), ( V4—V3), (P4—P3) ) — f2( (T3—T3), ( V3—V3), (P3—P3) )
=f2( (T4—T3), ( V4—V3), (P4—P3) ) — 0
=f2( (T4—T3), ( V4—V3), (P4—P3) )
After a cycle, net heat ΔQ=ΔQ1+ΔQ2（= net work ΔW ) .
So is W=f(T, V, P).
And, please notice that , in thermodynamics, when we're discussing "process" or "cycle", we're discussing a process or a cycle of a specific path(trajectory).
The arbitrary cycle that a set of element Carnot cycles " replace" ( in fact, it cann't be replaced) is just a cycle of a specific path(trajectory).
We never discuss "process" or "cycle" in general, for that is meaningless.
Since Entropy change can be analysed by the Calculus of Variations and the Pontryagin's Maximisation Principle at least in a closed system, we use the Qdot=f(Vdot,rdot,pdot), the notation is mine!!!!, not seriously, to analyse the variations with time. Isn't it true?
Soumitra K. Mallick
- Badges
- Science topic
- Similar topics
- Philosophy
- Philosophy Of Science
- Reasoning