Physical Chemistry Week 1

8/9/2019 Physical Chemistry Week 1

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Thermodynamic Definitions

Basic Physical Chemistry - Thermodynamics 1


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■ A thermodynamic system is defined as any part

of the Universe under consideration.

Thermodynamic System

2Basic Physical Chemistry - Thermodynamics

■ It may be something as simple as a beaker ofwater or as complicated as an entire galaxy!


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■ Thermodynamic surroundings are defined as

everything other than the thermodynamic

system. In other words, the entire rest of the

Universe.

Thermodynamic Surroundings



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■ The Universe is therefore the system plus its

surroundings.

The Universe



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The observable Universe is a sphere with a radius of ca. 4.66 × 1010 light years.

A light year is 9.46 × 1015 m. The observable Universe is thus ca. 880 Ym or

880,000,000,000,000,000,000,000,000 m across.

The Vastness of the Universe


Image: NASA


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Fe2O3 + 2Al 2Fe + Al2O3

Image: Nikthestunned


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Image: NASA / ESA

super nova


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■ Surroundings are assumed infinite and remain at

constant temperature and pressure.

■ The vast size of the Universe validates thisassumption.


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■ The boundary may be actual or notional.

Boundary Conditions


■ It controls transfer of work, heat and matter fromthe system to the surroundings and vice-versa.

■ The boundary may or may not impose

restrictions on such transfers.


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■ An open system may exchange both energy and

matter with its surroundings.

Open, Closed and Isolated


Open


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■ A closed system may exchange energy but not

matter with its surroundings.



Closed


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■ An isolated system may exchange neither

energy nor matter with its surroundings.



Isolated


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■ A diathermic system allows heat flow into or out

of the system.

Diathermic and Adiabatic


■ An adiabatic system prevents heat flow into orout of the system.

Hot Cold

Diathermic Adiabatic

Hot Cold


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Diathermic and Adiabatic Walls


Both containers are attempts

to create adiabatic walls. Theflask is a good approximation

to an adiabatic calorimeter –

the cardboard cup isn’t…

Which will keep coffee

warmer for longer?


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■ Isothermal implies constant temperature, T .

■ Isobaric implies constant pressure, p.

■ Isochoric implies constant volume, V .

Isothermal, Isobaric and Isochoric



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■ A state function describes the state of a system.

State functions


Pressure, p

Volume, V

Temperature, T

Mass, m

Quantity, n

Internal Energy, U

Enthalpy, H

Entropy, S

Gibbs Energy, G

■ The following are all state functions:


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■ A state function describes the current state of a

system.

Path functions


■ How the system came to be in that particularstate is of no consequence.

■ Functions governing transition between states

are called path functions.

Heat, q Work, w


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■ The state of a system is changed by the supply

or removal of energy in the form of heat or work.

Heat and Work


Work energy transfer is

uniform molecular motion

Heat energy transfer is

random molecular motion


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■ Path functions are positive when energy enters

the system.

Heat and Work Conventions


■ Path functions are negative when energy leavesthe system.

■ They are often defined as:

heat supplied to the system, qin mechanical work done on the system, w on


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0th Law of ThermodynamicsTemperature



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■ Developed around 1850 by Rudolf Clausius.


The First Law of Thermodynamics

■ Adaptation of the Law of Conservation of Energy

for thermodynamic systems.

■ Defined thermodynamic energy or internal

energy for a thermodynamic system.

■ Concerns energy changes and lead to thedefinition of a new state function called enthalpy.


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■ Also developed around 1850 by Rudolf Clausius

following initial work by Sadi Carnot in 1824.

■ Describes the direction in which all processesspontaneously occur, e.g. a hot object loses heat

to its surroundings.

■ Lead to the definition of a new state function

called entropy which fundamentally measuresdisorder.

The Second Law of Thermodynamics



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■ Developed in 1912 by Walter Nernst.

■ Considers the entropy of perfect crystals.

■ Lead to the definition of zero on the entropyscale thus enabling determination of absolute

entropy.

3rd Law of Thermodynamics



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■ Temperature, which lay at the heart of this new

science of thermodynamics, was routinely

measured by thermometer.

■ Think how a thermometer works…

A Crisis of Temperature



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■ “When two objects are separately in

thermodynamic equilibrium with a third object,

they are in thermodynamic equilibrium with each

other.”

■ Whenever two objects are in contact with one

another energy will flow between them until they

reach a state of thermodynamic equilibrium.Upon reaching this state we say that the two

objects are at the same temperature.

The Zeroth Law of Thermodynamics



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■ “When two objects are separately at the same

temperature as a third object, they are at the

same temperature as each other.”

■ This sounds obvious but it has vital ramifications

for the measurement of temperature by

thermometer.

■ The obvious nature of this statement explainswhy it did not become a law of thermodynamics

until well after the other three laws.

The Zeroth Law of Thermodynamics



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1st Law of ThermodynamicsEnthalpy



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■ Stated by Rudolph Clausius in 1850.



■ “Energy can neither be created nor destroyed,

merely changed from one form into another.”

■ Every system possesses an internal energy, U .

DU = qin + w on [1]


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Evolution of Gas from a Reaction

Zn (s) + 2HCl (aq) ZnCl2 (aq) + H2 (g)


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The Work Done by an Expanding Gas


p = F / A

w = Fd

\ F = pA

\ w = pAd

w = pDV

ExternalPressure, p

Cross-section, A

ExpandingGas

Frictionless piston moves through a distance d

w on = – pDV [2]


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DU = qin + w on [1]

w on = – pDV [2]

DU = qin – pDV

■ Expansion work clearly depends on p and DV .

Image: NASA


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Free Expansion

DU = qin – pDV

But in space p = 0

DU = qin

However, relatively veryfew experiments arecarried out in space!

Image: NYNAS


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Reactions at Constant Volume

DU = qin – pDV

But at constant volume DV = 0

D

U = qv [3]

Constant volume reactors must withstand massivepressure changes – heavy chemical industry, £ € $.

Image: Lilly M


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Reactions at Constant Pressure

■ Now we define a new state function, enthalpy:

DU = qin – pDV

\ DU + pDV = qin at contant pressure

\ DH = DU + D( pV )

H = U + pV

DH = DU + pDV + V D p

But, at constant pressure, D p = 0 hence V D p = 0\ DH = DU + pDV

DH = qp [4]

Product rule:d uv

d x = u

dvd x

+ v dudx


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Reversible IsothermalExpansion of an Ideal Gas



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The Ideal Gas Equation

pV = nRT

p is pressure

V is volume

n is quantity

R is the ideal gas constant, 8.314 J K –1 mol –1

T is absolute temperature


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The Ideal Gas Equation

p = nRT / V

Consider 1 mol of an ideal gas at 298 K

p = 1 mol × 8.314 J K –1 mol –1 × 298 K / V

p = 2477.572 J / V

Using the above equation, calculate the pressure

of the gas for volumes of 1, 3, 5, 7, 9 and 11 m3.

Answer: 2478, 826, 496, 354, 275 and 225 Pa

respectively.


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Isothermal Expansion

External

Pressure, p

One mole of

Gas

External pressure, p = 2478 Pa volume = 1m3

Now imagine the external pressure suddenly drops

to 826 Pa. What happens to the gas?

Ambient temperature = 298 K


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0

500

1000

1500

2000

2500

1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 11.00


p /

P a

V / m3

w = pDV826 Pa × 2 m3

w = 1652 J

w = 450 Jw = 550 Jw = 708 J

w = pDV

496 Pa × 2m3

w = 982 J

w total = 1652 + 982 + 708 + 550 + 450 J = 4342 J



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0

500

1000

1500

2000

2500

1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 11.00


p /

P a

V / m3



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0

500

1000

1500

2000

2500

1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 11.00


p /

P a

V / m3

p = nRT / V



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0

500

1000

1500

2000

2500

1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 11.00

p /

P a

V / m3

w = 1652 J

w = 450 Jw = 550 Jw = 708 J

w = 982 J

w on = –4342 J



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0

500

1000

1500

2000

2500

1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 11.00

p /

P a

V / m3

1239 J826 J

619 J496 J

413 J 354 J 310 J 275 J 248 J 225 J

w on = –5004 J



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0

500

1000

1500

2000

2500

1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 11.00

p /

P a

V / m3

w on = –5428 J



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0

500

1000

1500

2000

2500

1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 11.00

p /

P a

V / m3

w on = –5672 J



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0

500

1000

1500

2000

2500

1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 11.00

p /

P a

V / m3

w on = –5803 J



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Reversible Isothermal Expansion

0

500

1000

1500

2000

2500

1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 11.00

p /

P a

V / m3

w on = –5941 Jw on = –5941 J


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14

w on for Reversible Isothermal Expansion

w on = – pdV

But, pV = nRT

So, p = nRT / V

w on = –(nRT / V) dV

Now consider an expansion from V i to V f

dV V

w on = –nRT

V f

V i


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15


dV V

w on = –nRT

V f

V i

w on = –nRT V f

V i

lnV

w on = –nRT (lnV f – lnV i)

w on = –nRT ln V

V f

i

[5]

d x x

= ln x + c

ln A

−lnB

=ln

A

B


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w on = –nRT ln V

V f

i

w on = –5941 Jw on = –1 mol × 8.314 J K

–1

mol –1

× 298 K × ln(11)


■ Reversible change is a theoretical construct.

■ Determines maximum possible expansion work.


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Heat Capacity



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Heating Objects

Heat Capacity, C , is the heat energy required toraise the temperature of an object by 1 °C or 1 K.

Low

Heat

Capacity

High

Heat

Capacity

C = q / DT [6]


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Definitions of Heat Capacity

Heat capacity is an extensive property.

Specific Heat Capacity, C s = C / m units J K –1 kg –1

Molar Heat Capacity, C m = C / n units J K –1 mol –1

Specific Heat capacity and Molar Heat Capacity

are intensive properties.


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130

140 390

440

710

710

13402000

2010


520

Specific heat capacities of substances / J K –1 kg –1

880

1670


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Isochoric and Isobaric Heat Capacities

Isochoric heat capacity is defined as:

C V = qV / DT = DU / DT [7]

Isobaric heat capacity is defined as:

C p = qp / DT = DH / DT [8]


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H m = U m + pV m

H = U + pV

For an ideal gas: pV = nRT

pV m = RT


RT = 8.314 J K –1 mol –1 × 298 K ~ 2.5 kJ mol –1

H m = U m + RT

This is not negligible for a gas.

C p,m & C V,m Relationship for an Ideal Gas


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For a change of temperature DT

H m = U m + RT

DH m = DU m + R DT

DH m

DT =

DU m

DT +R


C p,m = C V,m + R [9]

C p,m & C V,m Relationship for an Ideal Gas

where R = 8.314 J K –1 mol –1

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Physical Chemistry Week 1