9.equation of ther

8/13/2019 9.Equation of Ther

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An example, “heating water on a stove”

Question: Calculate the change in entropy of water heated from 20oC to 100oC on a stove.

method 1: as per topic 08. A reversible path is constructed involving a series of heat baths between

initial and final temperatures T i and T f with the water brought into contact with each bath in turn. The

thermodynamic definition of entropy relates changes of entropy to reversible heat transfers, which can

be applied along this alternative path between T i and T f :

∆S =

T f

T i

d q R

T =

T f

T i

C pdT

T

The change of entropy of the water is the same as in the irreversible process that actually occurs since

entropy is a state function and the initial and final equilibrium states match.

method 2: Brute force solution applying the central equation:

∆S = f

i

dU

T + f

i

p

T dV

We know only T i and T f and pi = p f , but not U i and U f or V i and V f . Therefore we need to expand dU

and dV in terms of p,T.

∆S =

f

i

∂ U ∂ T

p

dT +

∂ U ∂ p

T

d p

T +

f

i

p

T

∂ V

∂ T

p

dT +∂ V

∂ p

T

d p

pot is open to atmosphere, so the natural choice of path is p = const = pi = p f for which d p = 0.

∆S =

T f

T i

∂ U ∂ T

p

+ p

∂ V ∂ T

p

T

dT =

T f

T i

∂ (U + pV )

∂ T

p

T

dT =

T f

T i

∂ H ∂ T

p

T

dT =

T f

T i

C p

T

dT

method 3: The water is heated at constant pressure so it is likely to be more convenient to work with

H=U+PV rather than U, which combined with the central equation gives

dH = dU + pdV + V d p = (T dS − pdV ) + pdV + V d p = T dS +V d p

⇒ dH = T dS +V d p

The above is an equivalent form of the central equation. We can integrate this choosing the path p =const

to get

p=const

dS =

p=const

dH

T

Writing H = H (T ,P)⇒ dH =

∂ H ∂ T

p

dT +

∂ H ∂ p

T

d p = C pdT +

∂ H ∂ p

T

d p. Thus along the path p =const

dH | p = C pdT

⇒ ∆S = T f

T i

C p

T dT as found previously.

In methods (2) & (3) we have not to had to think explicitly about an equivalent reversible path for the

irreversible process as done in method (1). The central equation deals with this implicitly.

Another example, “electrical work”

Consider an emf E driving a current I through a resistor R. The power delivered is E I = I 2 R. Suppose

the resistor, regarded as the system, completely fills a box with adiabatic walls. The power is dissipated

irreversibly in the resistor, which therefore heats up.

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From the 1st law the change in internal energy of the system in time interval ∆t is ∆U = (q = 0)+w = I 2 R∆t (assuming the volume is fixed so that no mechanical work is done). We can relate changes of

U to changes of temperature for a constant volume process with dU = C V dT |V const. The change in

temperature is then:

I 2 R∆t = T f

T i

C V dT = C V ∆T

where the last equality follows if C V can be approximated as constant (independent of temperature)⇒ ∆T = I 2 R∆t /C V .

What is the change of entropy?

method 2 The change of entropy can be calculated with the central equation dS = dU /T |V const ⇒ dS =

C V dT /T |V const ⇒ ∆S = T f

T i

C V

T dT with T f calculated above.

method 1 The same answer can be found by thinking of a reversible process leading to the same change

of state variables between the initial and final equilibrium states, in this case a change of temperature

∆T . This could be achieved by putting the resistor successively in thermal contact with a sequence of

heat reservoirs, starting with one at T i and ending with one at T f giving:

∆S =

T f

T i d q R/T =

T f

T i C vdT /T

ie we have considered an alternative process where heat enters the system reversibly but no work of any

kind is done.

The connection between Entropy, Counting, and Probability

a preview of subject matter to be developed in next term’s statistical mechanics course

gas everywhere

2V

gas vacuum

V V

break

partition

partition

The Second Law shows that the total entropy of an iso-

lated system must increase and is maximised at equilibrium.

In statistical physics we expect a system to change from a

less probable initial state to a more probable final state. Thissuggests there is a relationship between probability and en-

tropy. As an example, consider the Joule expansion of a

large number of atoms, N, of an ideal gas (see figure). Im-

mediately after the partition is removed the gas is initially

all on the left hand side of the container, whereas in the final

state the gas has an equal density everywhere. Denote the

left hand side of the container by A and the right hand side

by B with V A = V B = V for simplicity. A long time after the

partition has been removed and assuming the system is free

to explore all possible states available to it (respecting energy conservation), the probability that any one

molecule is in A is 1/2. The chance that all the molecules are in A is (1/2) N . This is the probability of finding the system in its initial state a long time after the wall has been removed. Denote the total number

of possible states available with energy E and all the particles in A, to be Ω A ≡Ω( N ,V A, E ). Denote the

total number of states with energy E, N particles, and volume V A +V B with the particles located anywhere

within volume V A +V B as Ω A+ B ≡Ω( N ,V A+ B, E ). Assume that all the accessible microstates are equally

likely. ThenΩ A

Ω A+ B

= Ω( N ,V , E )

Ω( N ,2V , E ) =

1

2

N

(1)

Suppose that there is a relationship between the entropy and the number, Ω, of microstates available to

the system; S = f (Ω) where f is some arbitrary function. If we consider 2 isolated systems we know

that the total entropy is the sum of the entropies of each state, S = S 1 + S 2, whereas the total number of

microstates available grows much faster, as the product Ω1Ω2. Therefore we require that,

f (Ω1Ω2) = f (Ω1) + f (Ω2) (2)

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Only a logarithm can achieve this¿ Formally: differentiate EQN 2 first with respect to Ω1 and then with

respect Ω2 (eg by changing the volume of system 1 and then the volume of system 2) we have:

f (Ω1Ω2)Ω2 = f (Ω1) (differentiating EQN 2 w.r.t. Ω1) (3)

f (Ω1Ω2)Ω1Ω2 + f (Ω1Ω2) = 0 (differentiating EQN 3 w.r.t. Ω2) (4)

=⇒ f ( x) x + f ( x) = 0 (substituting Ω1Ω2 = x in EQN 4) (5)

=⇒ d ( x f

( x))dx

= 0 (6)

=⇒ f ( x) = k B

x(k B is an arbitrary constant at this stage) (7)

=⇒ f ( x) = k B l og( x) + const (8)

Thus we see that if S and Ω are related the relationship must be of the form S = k B l og(Ω) + S 0. Setting

the constant S 0 to zero defines a natural zero for the entropy; if the state is unique (eg the system is in

a quantum ground state) the entropy is zero. As the temperature is increased more states are accessible

and the entropy increases (we will return to this when we discuss the 3rd law). Call S defined in this way

the statistical entropy:

S = k B l og(Ω) Definition of statistical entropy for fixed E (9)

Taking the logarithm of our expression for an Ideal gas, EQN 1, we have:

S A+ B −S A = k B Nl og(2) (10)

S A+ B is the statistical entropy of the gas freely occupying the whole volume, which is the final equilibrium

state of the system. S A is the statistical entropy when the gas is restricted to be in volume V A, which

is the initial state of the system. The expression is identical to the expression for the change in the

thermodynamic entropy (page 2) if we identify k B N A = R (where N A is the number of atoms in a mole).

k B is the Boltzmann constant.

General definition of statistical entropy

It is often more convenient to work with systems at constant temperature than at constant energy. The

collection of states available is known as a canonical ensemble. A micro-canonical ensemble refers to

the collection of states with a fixed total energy. It will be proved next term that a more general statistical

definition of entropy that is also applicable to the canonical ensemble is given by:

S = −k B ∑i

pi l og( pi) general statistical definition of entropy (11)

The sum is over all microstates accessible to the system. pi is the probability that the system is in

microstate i. If we specialise to the micro-canonical ensemble where the total energy is fixed, all the

microstates have a-priori equal probability. If there are Ω possible states, pi = 1/Ω is the probability of

the system being in any one of them. Substituting this into the above expression for the statistical entropy

we get the expression for the entropy of a micro-canonical ensemble given in EQN 9.

Heat and work in the statistical picture

The mean energy of a system can be written as probability of a given energy × that energy: < E >=

∑i pi ε i where ε i is the energy of each (quantum mechanical) microstate. The differential of E is,

d < E >=∑i

d pi ε i +∑i

pi d ε i

The first sum on the right hand side can be interpreted as heat supplied to the system and the second sumcan be interpreted as the work done. < E > corresponds to the internal energy, U (proof in Stat.Mech.).

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Note

THE REST OF THIS TERM’S THERMODYNAMICS COURSE IS NOT DEPENDENT ON THE

ABOVE STATISTICAL INTERPRETATIONS. THEY ARE GIVEN HERE AS A POINT OF REF-

ERENCE TO HELP DEVELOP A BETTER INTUITION WHEN WORKING WITH ENTROPY. THE

RELATIONSHIP BETWEEN STATISTICAL MECHANICS AND THERMODYNAMICS WILL BE

DEVELOPED MORE RIGOUROUSLY NEXT TERM.

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9.equation of ther

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