Thermodynamics: Pure Substances

Properties of Pure Substances

Chapter 2

Pure Substance

In Chemistry you defined a pure substance as an element or a compoundSomething that can not be separatedIn Thermodynamics we’ll define it as something that has a fixed chemical composition throughout

Examples

Ice in equilibrium with waterAirAir in equilibrium with liquid air is not a pure substance – Why?Is sea water in equilibrium with water vapour a pure substance?

Boiling Points of Selected Liquids

ProductBoiling Point

at Atmospheric Pressure(oC) (oF)

Acetaldehyde CH3CHO 20.8 69Ammonia -35.5 -28.1

Carbon dioxide -57 -70.6Ethanol 78.4 173

Freon refrigerant R-11 23.8 74.9Freon refrigerant R-12 -29.8 -21.6Freon refrigerant R-22 -41.2 -42.1

Helium -269 -452Nitrogen -196 -320Oxygen -183 -297Water 100 212

Water, sea 100.7 213

Why is boiling and condensation of liquids

relevant?

Phases of Pure Substances

We all have a pretty good idea of the what the three phases of matter are, but a quick review will help us understand the phase change process

Consider what happens when we heat water at

constant pressure

Piston cylinder device –maintains constant pressure

Liquid Water

T

v

1

2

5

3 4

1. Compressed liquid

2. Saturated liquid

3. Saturated mixture

4. Saturated vapor

5. Superheated vapor

Two Phase Region

Compressed Liquid

Superheated Gas

Sometime a pure substance can decrease in temperature after critical point

What happens at constant T?

Consider ideal gas law:PV=nRT

Applies only in gaseous state away from melting point (high volume region)Increasing pressure can condense a gas

… more later

P-V diagram

P

v

T2>T1

T1

Critical Point

Saturated Liquid-vapour region

Superheated vapourCompressed

liquid region

Critical Point

Above the critical point there is no sharp difference between liquid and gas!!

Property Diagrams

So far we have sketched T – v diagram P – v diagram What about the P – T diagram?

What is the difference between paths?

Combine all three

You can put all three properties P T V

On the same diagram

Contracts on Freezing Expands on Freezing

Property Tables

P - pressureT - temperaturev – specific volumeu – specific internal energyh – specific enthalpy h = u + Pvs – specific entropy -define in Chapter 6

Saturated Liquid and Saturated Vapor States

Saturation Properties

Saturation Pressure is the pressure at which the liquid and vapor phases are in equilibrium at a given temperature.

Saturation Temperature is the temperature at which the liquid and vapor phases are in equilibrium at a given pressure.

Table A-4 and A-5

A-4 - pg 830 Saturated water temperature table

A-5 - pg 832 Saturated water pressure table

Examples

Cengel & Boles 3-1 (page 125)A rigid tank contains 50kg of saturated liquid water at 90°C. Determine the pressure in the tank and the volume of the tank.

1. How do we start?

SolutionWater is saturated.… so we know thepressure – Psat

Use the(A-4)So Psat = 70.14kPa, & v=0.001036m3/kgTank Volume = specific volume x mass

=50x0.001036m3=0.0518m3

Water is saturated.… so we know thepressure – Psat

Use the Temperature table (A-4)

Example 2: C&B Ex 3.2

A piston-cylinder device contains 0.06m3 of saturated water vapour at 350kPa. Determine the temperature and the mass of the vapour inside the cylinder.

All water vapour BUT saturated …

Solution

Vapour is saturated.… so we know thetemperature – Tsat

Use the Pressure table (A-5)

V.… so we know thetemperature – TUse the Pressure table (A-5)So Tsat = 138.86K, & v=0.52422m3/kgMass of vapour = volume/specific volume

=0.06/0.52422kg=0.114kg

PresskPa300325350375400

Vapour Not liquid!

State Variables

Once the Pressure, temperature and volume are known, other state variables are determined Internal energy Enthalpy Entropy For each phase of the pure substance

u u uh h hs s s

fg g f

fg g f

fg g f

g stands for gasf stands for fluidfg stands for the difference between gas and fluid

Internal Energy

Enthalpy

Entropy

Liquid-Vapour Mixtures:Quality

xmass

massm

m msaturated vapor

total

g

f g

Fraction of the material that is gas

x = 0 the material is all saturated liquid

x = 1 the material is all saturated gas

x is not meaningful when you are out of the saturation region

X = 0 X = 1

Example (C&B – 3-4)

A rigid tank contains 10.0kg of water at 90°C. If 8.00kg of the water is in the liquid phase, and the rest is in the vapour phase, determine:(a) the pressure in the tank(b) the volume of the tank

What is the fixed point – where we start?

SolutionContents of tank are in themixed phase regionLiquid and vapour at 90°C, and P=PsatP=70.183kPa (from table A-4)In this case we need specific volume of liquid AND gas at these conditionsvf=0.001036m3/kg; vg=2.3593m3/kgSo V=8kgx0.001036m3/kg+2kgx2.3593m3/kg

=4.73m3

90°C

Superheated Properties

Table A-6

Page 834

In the superheated region, there is only vapourTable looksa little different

Linear Interpolation

A B

100 5

200 10

130 X

5105

100200100130

x

Linear InterpolationEg: What is the saturation temperature for superheated steam at P=0.15MPa?

P1=0.1MPa, T1=99.61°CP2=0.2MPa`, T2=120.21°C… from tables

C

TTPPPPTT

TTTT

PPPP

91.109)61.9921.120(61.99

)(

1.02.01.015.0

1212

11

12

1

12

1

Example – putting it all together

Direct Solar Steam Generation

How do we solve?Consider just superheated steam region

0.1MPa0.5kg/s

P along 6m collectorowing to fluid dynamics!

Pinlet > 0.11MPa

Tsat>101.67°C

Texit~179.88°C Pinlet=1MPa, Tsat= 179.88°C

Compress

Tsat from interpolation: Tsat, 0.11MPa=99.61+(120.21-99.61)*(0.11-0.1)/(0.2-0.1)

Power needed?U179.88-U101.67For 0.5kg EACHEACH SECOND

Power needed?Power = Energy / timeSo need first INTERNAL ENERGY change per kg steam from inlet to outlet temp. at 0.1MPa (ignoring friction losses)

kgkJUkgkJ

UUUU

TBOTHforTTTT

UUUU

/8.2508/9.2627

)(150200

15088.179

67.101

15020015088.179

12

1

12

1

Power = Energy / timeSo need first INTERNAL ENERGY change per kg steam from inlet to outlet temp. at 0.1MPa (ignoring friction losses)U=118.9kJ/kg0.5kg/sPower=59.9kW

Equations of State

Equations vs Tables

The behavior of many gases (like steam) is not easy to predict with an equationThat’s why we have tables like A-4, A-5 and A-6Other gases (like air) follow the ideal gas law – we can calculate their properties

Ideal Gas Law

PV=nRT Used in your Chemistry class From now on we will refer to the gas

constant , R, as the universal gas constant, Ru , and redefine R=Ru/MW

PV=mRT R is different for every gas Tabulated in the back of the book

Ideal Gas Law

Pv = RT This is the form we will use the most

When does the ideal gas law apply?

The ideal gas equation of state can be derived from basic principles if one assumes:

1. Intermolecular forces are small 2. Volume occupied by the particles is small

These assumptions are true when the molecules are far apart – ie when the gas is not dense

Criteria

The ideal gas law applies when the pressure is low, and the temperature is high - compared to the critical valuesThe critical values are tabulated in the Appendix

Virial Equation of State

...5432 vTd

vTc

vTb

vTa

vRTP

EndFireworks are beautiful Are they Exothermic or endothermic reactions.?

Tutorial Problems

1.7c, 1.8e, 1.9, 1.12, 1.16c, 1.17c, 1.23c, 1.62, 1.65, 1.66, 2.4c, 2.88, 2.89