Dear sir

Q is the process variable due to it is depended upon the pass. But the specification such as P,T,.... is independent of path.

Regards

M.A.Ehyaei

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)

Many process functions are piecewise functions, so is Carnot cycle 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 ) .

Process variable because it is not defined at a point.

Dear Decker,

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.

Hi Hijaz Ahmad,

Yes, Q(heat) is process variable.

Q = f(T, V, P) is a process quantity which varies with path, it has innumerable forms between the same original and terminal states, and has a unique form for fixed reversible process path.

The process or cycle we discussed is for a given path.

that is, between the same original and terminal states:

to path1, Q1=f1(T, V, P);

to path2, Q2=f2(T, V, P),

......,

to path i, Qi=fi(T, V, P);

When the given path is fixed, Q = f(T, V, P) is the system state variable.

P, V and T are all variables (two variables of T, V and P are generally independent) for any reversible process.

Heat is normally a process variable , because for example there is the heat equation.

Hi Danilo Rastovic,

Yes, Q(heat) is process variable, but When the given path is fixed, Q = f(T, V, P) is the system state variable.

So, Q is both a process variable and a state variable.

So is W(work).

Q = f(T, V, P) is a process quantity which varies with path, it has innumerable forms between the same original and terminal states, and has a unique form for fixed reversible process path.

The process or cycle we discussed is for a given path.

that is, between the same original and terminal states:

to path1, Q1=f1(T, V, P);

to path2, Q2=f2(T, V, P),

......,

to path i, Qi=fi(T, V, P);

When the given path is fixed, Q = f(T, V, P) is the system state variable.

P, V and T are all variables (two variables of T, V and P are generally independent) for any reversible process.

Hi Shufeng Zhang, thank you very much. Unfortunately, I am only mathematician, not physicist.

Hi Danilo Rastovic,

I think, relative to the scholars of other fields, mathematicians more easily become physicists.

Welcome to have a discussion.

If Qx=x holds true, it means that Q=I,and it is equal to the assertion that Q is identity operator.

Normally, it is assumed that Qx=x for every x.

The all desired equations can follow from the operator functional calculus of several variables.

Calculus is not "take for granted", ΔQ/T can not be turned into dQ/T. That is, the so-called "entropy " doesn't exist at all.

It is well known that calculus has a definition.

Any theory should follow the same principle of calculus; thermodynamics, of course, is no exception, for there's no other calculus at all, this is common sense.

Based on the definition of calculus, we know:

to the definite integral ∫T f(T)dQ, only when Q=F(T), ∫T f(T)dQ=∫T f(T)dF(T) is meaningful.

As long as Q is not a single-valued function of T, namely, Q=F( T, X, …), then,

∫T f(T)dQ=∫T f(T)dF(T, X, …) is meaningless.

1) Now, on the one hand, we all know that Q is not a single-valued function of T, this alone is enough to determine that the definite integral ∫T f(T)dQ=∫T 1/TdQ is meaningless.

2) On the other hand, In fact, Q=f(P, V, T), then

∫T 1/TdQ = ∫T 1/Tdf(T, V, P)= ∫T dF(T, V, P) is certainly meaningless. ( in ∫T , T is subscript ).

We know that dQ/T is used for the definite integral ∫T 1/TdQ, while the definite integral ∫T 1/TdQ is meaningless, so, ΔQ/T can not be turned into dQ/T at all.

that is, the so-called "entropy " doesn't exist at all.