state functions

8/12/2019 State Functions

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Topics in Mathematics with Applications to Chemistry

Lia Vas

State Functions

A state function is a function of the parameters of the system which only depends upon theparameters values at the endpoints of the path. Thus, the change in a state function is apath independent. For example, the change of temperature is a state function. To illustratethat it is path independent, consider that we can raise the temperature of water for 10 degreeson a few different ways: 1) heat the water by 10 degrees, b) heat the water by 100 degrees andwait till it cools down to 10 degrees above the initial temperature, c) stir the water until thetemperature gets raised by 10 degrees etc.

In thermodynamics, a state function, state quantity, or a function of state, is a propertyof a system that depends only on the current state of the system, not on the way in which

the system got to that state. For example, internal energy, enthalpy and entropy are statequantities. In contrast, mechanical work and heat are process quantities because they describequantitatively the transition between equilibrium states of thermodynamic systems.

When a system changes state continuously, it traces out a path in the state space. Thepath can be specified by noting the values of the state parameters as the system traces out thepath, perhaps as a function of time, or some other external variable. Thus, we can treat statefunctions as vector functions.

Example from Wikipedia: Let us consider the pressure P(t) and the volume V(t) asfunctions of time from time t0 to t1. We can now form all sorts of functions of time which wemay integrate over the path. For example if we wish to calculate the work done by the system

from time t0 to time t1 we calculate

W(t0, t1) =state 1

state 0 P dV =

t1t0

P(t)dV(t)

dt dt.

It is clear that in order to calculate the work Win the above integral, we will have to knowthe functions P(t) and V(t) at each time t, over the entire path. Thus, the work W is not astate function here.

Let us suppose now that we wish to calculate the work plus the integral ofV dP over thepath. We would have:

(t0, t1) = t1t0

PdVdt

dt+ t1t0

VdPdt

dt= t1t0

d(P V)dt

dt= P(t1)V(t1) P(t0)V(t0).

It can be seen that the integrand can be expressed as the exact differential of the functionP(t)V(t) and that therefore, the integral can be expressed as the difference in the value ofP(t)V(t) at the end points of the integration. The product P V is therefore a state function ofthe system.

Thus, in order to determine if a given function is a state function or not, it is key to seewhen a line integral of a vector function is independent of path.

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Plane vector functions. Consider a vector function f= (P, Q).Assume that the deriva-

tives ofP and Q are continuous. We shall say that f is conservativevector field if and onlyif it is gradient of some scalar function F(t),

f= F.

In this case, the scalar functionFis call a potential function of f .If f= (P, Q),the conditionf= Fgives us that P=FxandQ = Fy.SinceFxy =Fyx ifF is twice differentiable function,this gives us that Py = Qx. Conversely, it can be shown that if Py = Qx the vector function

is conservative assuming that the domain of fis simply-connected (intuitively, the domain is

connected and has no holes). Thus to check if f = (P, Q) is conservative, it is sufficient tocheck if

Py =Qx.

In addition, let Cbe a smooth curve r(t) = (x(t), y(t)) with endpoints r(a) = (x(a), y(a))

and r(b) = (x(b), y(b)). Assume that f is defined on a simply-connected region. Then the line

integral C

P dx+Qdy=C

f(t)dr

is independent of path if and only if f = (P, Q) is conservative (i.e. ifPy = Qx). In this case,

if f= F, C

P dx+Qdy=C

f(t)dr=CF dr= F(r(b)) F(r(a))

For a given conservative vector field (P, Q), a potential function F can be found by inte-grating Pwith respect to x and noting that the integration constant has to depend on y.Thendifferentiating the result and equating it with Q,we obtain a potential function F.

Space vector functions. Consider the vector function f= (P,Q,R).Assume that all thederivatives ofP, Qand R are continuous. Similarly as in the two dimensional case, we say thatf isconservativevector field if and only if it is gradient of some scalar function F(t), f= F,called apotential functionof f .If f= (P,Q,R),this condition gives us thatP=Fx, Q= Fyand R= Fz. Since Fxy = Fyx, Fxz = Fzx, and Fyz = Fzy for Fa twice differentiable function,this gives us that Py = Qx, Pz = Rx and Qz = Ry. Conversely, it can be shown that if thesethree conditions hold that the vector function is conservative under the assumption that thedomain of fis simply-connected. Thus to check if f= (P,Q,R) is conservative, it is sufficientto check if

Py =Qx, Pz =Rx and Qz =Ry.

Let C be a smooth curve r(t) = (x(t), y(t), z(t)) with endpoints r(a) = (x(a), y(a), z(a))andr(b) = (x(b), y(b), z(b)). Assume that f is defined on a simply-connected region. Then theline integral

CP dx+Qdy+Rdz=

C

f(t)dr

is independent of path if and only if f= (P,Q,R) is conservative. In this case, if f= F,C

P dx+Qdy+Rdz=C

f(t)dr=CF dr= F(r(b)) F(r(a))

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The Fundamental Theorem of Calculus. In one-dimensional case, the condition f =F boils down to f = F. Thus the one-dimensional analogue of the formulas for the lineintegrals above is the well known formula

ba

f(t)dt= ba

F(t)dt= F(t)

|ba=F(b)

F(a)

Practice Problems.

a) Check if given vector functions are conservative. If they are, find their potential functions.

1. f= (xey, yex)

2. f= (x3y4, x4y3 + 2y).

b) Find a potential function of given vector functions and use it to evaluate

Cf dr along

given curve C.

1. f= (x3y4, x4y3), C is x=

t, y= 1 +t3,0 t 1.2. f= (y, x+z, y), C is the line segment from (2, 1, 4) to (8, 3,1).3. f= (2xz+ sin y, x cos y, x2), C is spiral x= cos t, y= sin t z=t, for 0 t 2.

c) Show that the line integral is independent of path and evaluate it.

1.

C2x sin ydx+ (x2 cos y 3y2)dy where Cis any path from (-1,0) to (5, 1).

2.

Cydx+ (x+z)dy+ydzwhere Cis any path from (2, 1, 4) to (8, 3,1).

Solutions.

a) 1. Not conservative. 2. Conservative. F = 14

x4y4 +y2 +c.

b) 1. F = 14

x4y4 + c,

C = F(1, 2)F(0, 1 ) = 40 = 4 2. F = xy + yz+ c,C=F(8, 3,1) F(2, 1, 4) = 21 6 = 15 3. F=x2z+x sin y,

C=F(1, 0, 2)

F(1, 0, 0) = 2 0 = 2.c) 1. F = x2 sin y y3C= F(5, 1) F(1, 0) = 25sin1 1 0 = 20.04 2. See part

b) 3.C

= 15

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state functions

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