Chemistry 2402 - Thermodynamics

Lecture 8 : Converting Heat to Work

Lecture 9 : Free Energies

Lecture 10: Phase Diagrams

The Intensive Parameters (again)

We will also want to know how entropy changes when we change energy, particle number and volume – to this end we need:

NVS

UT

,

TU

S

NV

1

,

T

P

V

S

NU

, TN

S

VU

,

The intensive parameters temperature, pressure and chemical potential were previously defined in terms of changes in the energy U by

NSV

UP

,

VSN

U

,

How can entropy change?

Now we can write the most general expression for a change in entropy dS in terms of changes is U, V and N as

or, making use of the intensive parameters,

Now we can relate how systems tend towards equilibrium (i.e. to maxima in S) in terms of T, P and and the kinds of changes (i.e. dU, dV and dN) that we chose to allow.

dNN

SdV

V

SdU

U

SdS

VUNUNV ,,,

dNT

dVT

PdU

TdS

1

Does behave like a real temperature?

Here’s the basic problem in thermodynamics.

UI VI NI UII VII NII

Before

? ?

After

In an isolated system, we remove an internal constraint. What happens?

Let’s consider the case when we allow the internal wall to conduct heat between subsystem I and subsystem II.

NVS

UT

,

Does behave like a real temperature?NVS

UT

,

The total entropy is S = SI + SII and dS = dSI +dSII

As the only quantity allowed to change in either region is the energy, we have

dS = dUI/TI + dUII /TII

The system is isolated, i.e.the total energy is conserved, so dUI = -dUII - i.e. what ever flows into I must have flown out of II.

So now we have dS = dUI (1/TI - 1/TII)

Equilibrium corresponds to the state at which S is a maximum, i.e. dS = 0. According to the equation, equilibrium corresponds to the state at which TI = TII.

In any spontaneous process dS > 0. If TI > TII then, according to our equation relating dS to dU, dUI<0, i.e. heat must flow from the high temperature region to the low temperature one.

Thermodynamics in Open Systems

U V N ? ? ?

An isolated system is characterised by the energy, volume and particle number which are fixed

What can one say about on open system? What properties are fixed?

Constraints and Reservoirs

When we allow exchange of a system quantity with the surroundings we shall assume that the surroundings are so much larger than our system, that the state of the surroundings remains unchanged – i.e. the surroundings represents infinite reservoirs.

If a system can exchange heat with the surroundings then energy is no longer fixed but, at equilibrium, the temperature of the system must equal that of the surroundings and so we have fixed the temperature.

If our system can expand or contract, volume is no longer fixed but at equilibrium the system pressure is equal to that of the surroundings.

Similarly, if we allow particle exchange, we no longer have a fixed number of particles, but we are effectively setting the system chemical potential to be equal to that of the surroundings.

Equilibrium at constant temperature

At equilibrium, dS = dSsys+dSbath =0

dSbath = -qin/T, where T is the temperature of the bath.

And dUsys = qin because V is fixed so no work can be done

So our equilibrium condition is dSsys - dUsys/T = 0 where T is the bath temperature. Note that we are no longer maximising Ssys.

.

Surroundings or bath

System with fixed V and N qin

T

So what ARE we maximising?

dSsys - dUsys/T = 0 → d(S-U/T) = 0 at constant T

So we could invent a new entropy = S-U/T to be maximised.Perfectly sound, but not the way it was done.

Instead, the equilibrium condition was rewritten to bedUsys - TdSsys = 0

and we defined A = U-TS such that equilibrium at constant temperature was defined as the state that minimised A at constant T, i.e.

dA = dU-TdS = 0

This A is a modification of the energy U and is called a free energy. This one is called the Helmholtz or canonical free energy.

Instead of U=U(S,V,N) we have A = A(T,V,N)

What about equilibrium at constant T and P?

Equilibrium at constant T and P, the most useful for most chemical processes, minimizes the Gibbs free energy G = G(T,P,N) with

G = U -TS +PV (Note enthalpy H = U+PV) = H - TS = A + PV = N

And

There are 7 possible free energies, each defined by a different combination of constraints. We will work with the Gibbs free energy from now.

ST

G

NP

,

VP

G

NT

,

Solid, Liquid and Vapour

Image source: Mila Zinkova & CillanXChttp://en.wikipedia.org/wiki/Image:Iceberg_with_hole_edit.jpg

Equilibrium between phases

The equilibrium state at a fixed temperature and pressure is that which minimises G. Let’s look at a sketch of G versus temperature for the solid, liquid and gas phase of a substance at a single pressure.

G

0

T

solid

liquid

gas

Tf Tb

ST

G

NP

,

Ssolid < Sliquid < Sgas

Pressure Effects on Phase Equilibrium

When pressure is decreased, G decreases too, but it decreases a different amount for each phase. Phases with bigger the molar volume (gases) are decreased more.

VP

G

NT

,

G

0

T

solid liquid

gas

Ts

In this case the liquid is now no longer the most stable phase at any temperature. Ts is the sublimation point where gas and solid are in coexistence.

The Critical Point:where liquid and gas merge

Increasing the pressure sees the liquid and gas states become increasingly similar.

At a particular value of P and T, the critical point, the liquid and gas phase become indistinguishable.

G

0

T

solid

liquid

gas

Tf

Tb

Phase Diagrams

Liquids eke out an existence in the niche between solids and gases.

Coexistence lines represent values of P and T where two phases can coexist.

P

T

solid

gas

liqui

d

critical point

triple point

supercritical fluid

Flash Quiz!

Why can’t we use easily use hydrogen or methane as a fuel?

P

T

solid

gas

liqui

d

critical point

triple point

supercritical fluid

substance Ttp (°C) Tc (°C)

hydrogen –259.4 –240.2

methane –182.5 –82.7

propane (LPG) –187.9 96.7

hexane (petrol) –95.1 234.5data from http://www.lakeshore.com/pdf_files/Appendices/LSTC_appendixI_l.pdf

Answer

P

T

solid

gas

liqui

d

critical point

triple point

supercritical fluid

data from http://www.lakeshore.com/pdf_files/Appendices/LSTC_appendixI_l.pdf

substance Ttp (°C) Tc (°C)

hydrogen –259.4 –240.2

methane –182.5 –82.7

propane (LPG) –187.9 96.7

hexane (petrol) –95.1 234.5

Why can’t we use easily use hydrogen or methane as a fuel?

Unless we cool them heaps, they are always above their critical point, which means that we cannot exploit a phase transition to a small-volume liquid. If we want small volumes, we would need extreme pressures. If we want manageable pressures, we would need to store a large volume to contain enough to provide the chemical energy we need.

The slope of coexistence lines.

The variation in G is dG = -SdT +VdP

If we want to move along the coexistence line between phases I and II, thendGI = dGII since GI = GII defines coexistence.

so that-SIdT + VIdP = -SIIdT + VIIdP

The temperature and the pressure must, of course, be the same in any phases at coexistence. Rearranging this equation, we have

III

III

VV

SS

dT

dP

The Clapeyron Equation

But G = H-TS=0 and, therefore, S = H/T

Substituting this, we arrive at the Clapeyron equation

III

III

VV

SS

dT

dP

VT

H

dT

dP

1

212 ln

T

T

V

HPP

If H and V vary little with P and T then we can integrate the Clapeyron equation to give

Pressure melting or pressure freezing?

From the Clapeyron equation Now ΔH < 0 for freezing.

Usually, ΔV < 0 also since solids are denser than liquids. That means the dP/dT is positive as in Diagram I. When ΔV > 0 as in ice or silica, then we get Diagram II

P

T

solid

gas

liqui

d

critical point

triple point

P

T

solid

gas liq

uid

critical point

VT

H

dT

dP

I II

Sample exam questions from previous years

• Provide a brief explanation of the following terms:

– Gibbs free energy

– Triple point

• Water ice melts at 0 °C at 1 atm, with an enthalpy change of 6.0095 kJmol ‑1. The volume decrease on melting is 1.65×10‑6 m3mol‑1. What change in the bulk melting temperature will be observed if an ice skater applies a pressure of 13 atm? Is this effect enough to explain the melting of ice while skating?

• Under standard laboratory conditions, the melting point of iron is 1811 K. The inner core of the earth is primarily composed of iron. It is solid, despite being at a temperature of around 5600 K. Give a brief explanation of how this is possible. Calculate an estimate of the pressure at the surface of the inner core, where solidification first occurs. You can assume that the volume increase upon melting, 0.055 cm3mol−1, and the enthalpy of fusion, 13.8 kJmol-1, remain constant.

• Napthalene melts at 353 K at 1 atm with an enthalpy change of 19 kJmol -1. The volume increase on melting is 19 × 10-6 m3mol-1. What change in the melting temperature will be observed if the pressure is raised by 100 atm?

Summary

You should now• Be able to calculate the entropy change of an equilibrium

system as a function of changes in U, V, and N.• Be able to explain the concept of free energies, and in

particular what the Gibbs free energy represents• Be able to draw and read a P-T phase diagram, understand its

meaning, and understand why the different phases occur under the conditions they do

• Be able to explain the meaning of coexistence, a triple point, and a critical point

• Understand and apply the Clayperon equation to coexistence lines

Next Lecture• Phase diagrams and solubilities