1. CHEM 3310
Thermodynamics

2. CHEM 3310

3. Thermodynamics
The study of energy changes accompanying physical and chemical processes.
From the laws of thermodynamics, one can:
Predict the results of chemical reactions
Ascertain whether or not a reaction is possible
Predict quantitatively the effect of changes in the experimental conditions
CHEM 3310

4. What is Energy?
1. Kinetic Energy (KE) – the energy of motion due to the motion of the system's particles (translations, rotations, vibrations)
2. Potential Energy (PE) – the stored energy in an object by virtue of its position or composition (chemical bonds, electric energy of the atoms within the molecules)
Energy cannot be created nor destroyed. The total energy of the universe is a constant.
Energy can, however, be converted from one form
to another or transferred.
CHEM 3310

5. The total energy in any isolated system is constant.
Terminology:
System is the portion of the physical world that is selected for a thermodynamic study. Examples:
A beaker of water
A polluted lake
Surroundings
System
Surroundings is the portion of the universe with which the system interacts.
Isolated System = System + Surroundings
The total energy in any isolated system is constant.
CHEM 3310

6. E = KE + PE Internal Energy, E
All substance possess internal energy, because the molecules of all substances all have kinetic energy and potential energy.
Internal energy, represented by E, is essentially the thermal energy of the particles making up the system (all the atoms or molecules of the body due to their random motion and position).
It is the sum of the kinetic and potential energies of the particles that form the system.
E = KE + PE
CHEM 3310

7. E = KE + PE Internal Energy, E T1 < T2 E = 3/2nRT
In an ideal gas the molecules are so far apart that intermolecular forces can be ignored. Therefore, there is no potential energy. So all the energy is kinetic.
E = KE + PE
The simplest system is an ideal gas.
Kinetic Molecular Theory assumes that the temperature of a gas is directly proportional to the average kinetic energy of its particles.
At T1
At T2
T1 < T2
The internal energy of an ideal monatomic gas is therefore directly proportional to the temperature of the gas.
R = gas constant
T = temperature in Kelvin
E = 3/2nRT
CHEM 3310

8. Internal Energy, E
Calculate the internal energy of a pressurized bottle of Helium gas with volume m3 at 1.01x107 Pa. You can use the gas law  PV=nRT  to express the internal energy in terms of pressure and volume:
0.368 m x m x m = m3
E = 3/2nRT = 3/2PV
= 3/2 (1.01x107 Pa)( m³)
= 7.54 x 105 Pa m³
= 7.54 x 105 J
(1 atm = kPa)
CHEM 3310

9. E = f(T) Internal Energy, E
The internal energy of other systems that are more complex than the ideal gas are harder to measure.
But the internal energy of the system is still has a unique
value for each temperature.
E = f(T)
Where f is monotonically increasing function.
As the temperature of the system increases,
the internal energy of the system also increases.
CHEM 3310

10. Temperature is a state function.
Temperature, T
Use a thermometer to measure a beaker of water
at room temperature.
T = 21.5 °C
This measurement describes the state of the system
at a particular moment in time.
It doesn’t tell us how the water got to room temperature.
Temperature is a state function.
CHEM 3310

11. Internal Energy, E
Because the internal energy of the system has a fixed value
for any temperature, internal energy is also a state function.
(E depends on the state of the system in terms of P, V and T.)
Any change in the internal energy of the system is equal to the difference between its initial and final values.
Esystem = Efinal - Einitial
CHEM 3310

12. The First Law of Thermodynamics
The total energy of the universe is constant.
Energy can be transferred from the system to its surroundings and vice versa, but it cannot be created nor destroyed.
The energy lost by the system is gained by the surroundings such that
- Esystem = Esurroundings Euniverse = Esystem + Esurroundings = 0
Surroundings
System
CHEM 3310

13. There are two ways to change the internal energy of a system:
By flow of heat, q
Heat is the transfer of thermal energy between the system and the surroundings
2. By doing work, w
Work can be converted into heat and vice versa.
q and w are process dependent, and are not state functions.
CHEM 3310

14. Conversion: 1 calorie = 4.184 joules
Units of Energy
Calorie: One calorie is defined as the amount of heat required to raise the temperature of 1 g of water by 1 degree celsius.
Joule: One joule is the work done when a force of one Newton is used to move an object one meter J = 1 Nm
Conversion: 1 calorie = joules
CHEM 3310

15. Heat, q
A form of energy
Passes from one body to another as a result of a difference in temperature
Heat flows from a higher temperature to a lower temperature
Heat cannot be measured, but the effects which it produces is measurable.
CHEM 3310

16. Measurement of the Effects of Heat
The measurements of heat effects is known as calorimetry.
The instrument used is the calorimeter.
System : the molecules inside the reaction chamber
Surroundings : the container holding the sample and the water bath
Isolated System : the calorimeter
The condition under which
the measurement is made is
constant volume. This is a
schematic of a
bomb calorimeter
CHEM 3310

17. Experiment 8 – Measure E of a combustion reaction
Parr Bomb

18. Experiment 8 – Measure E of a combustion reaction
Combustion of Cyclohexanol

19. q  nAT q = nACAT Measurement of the Effects of Heat
In the isolated system containing Substance A
q  nAT
where nA is the number of moles of substance A T = T2 – T (T2 = final temperature, T1 = initial temperature)
Introduce a proportionality constant, CA, such that
q = nACAT
CA is the molar heat capacity; it is characteristic of the substance.
CHEM 3310

20. q = mAcAT Molar Heat Capacity Specific Heat
is the quantity of heat necessary to raise the temperature of
one mole of the material by one degree Celsius.
Specific Heat
is the quantity of heat necessary to raise the temperature of
one gram of the material by one degree Celsius.
q = mAcAT
where mA is the mass of substance A T = T2 – T1
cA is the specific heat of substance A
CHEM 3310

21. Conversion: 1 calorie = 4.184 joules
Units of Heat
Calorie: One calorie is defined as the amount of heat required to raise the temperature of 1 g of water by 1 degree Celsius.
Specific heat of water = 1.00 cal g-1 °C-1
Molar heat of water = 18.0 cal mole-1 °C-1
Conversion: 1 calorie = joules
Specific heat of water = 4.18 J g-1 °C-1
CHEM 3310

22. Heat Capacity
varies as a function of temperature 100 K is not the same as 300 K
is different if the heating is done under constant pressure or under constant volume conditions
Heat capacity measured under constant pressure is Cp Heat capacity measured under constant volume is Cv
For liquids and solids, there is little difference between Cp and Cv.
For gas, Cp > Cv.
CHEM 3310

23. Sign Convention
The internal energy and temperature of a system decrease(E < 0) when the system loses heat to its surroundings.
The internal energy and temperature of a system increase (E > 0) when the system gains heat from its surroundings.
q > 0 when heat is added to the substance
q > 0 when T2 > T1 (i.e. T>0)
q < 0 when heat is lost by the substance
q < 0 when T2 < T1 (i.e. T<0)
CHEM 3310

24. Specific heat, Cp, of Pb is 0.0308 cal g-1 °C-1
Example:
100.0 g of Pb heated to 100 °C is dropped into g of water at 0oC. What is the final temperature? Assume that heat transfer is between the two substances.
Specific heat, Cp, of Pb is cal g-1 °C-1
Specific heat, Cp, of water is 1.00 cal g-1 °C-1
Answer:
Heat lost by the Pb = Heat gained by the water
- qPb= qwater
- mPbcPb (T2-T1)Pb= mwatercwater (T2-T1)water
T2 = ?
T1 = 100 °C
T2 = ?
T1 = 0 °C
T2 = 23.6 °C
CHEM 3310

25. Latent Heat Effects
The heat effects associated with changes in physical state such as fusion and vapourization are known as latent heat effects.
Heat of Fusion - the heat required to melt a substance at its normal melting temperature
- the energy required to reorganize the intermolecular structure of a substance.
Let’s heat up 1 g of water!
CHEM 3310

26. Latent Heat Effects
Heat of Vaporization – the heat required to vaporize a substance at its normal boiling temperature
Heat of fusion of ice is cal/g at 0oC and 1 atm.
Heat of vaporization of water is cal/g 100oC and 1 atm.
CHEM 3310

27. Sign Convention
The internal energy and temperature of a system decrease (E < 0) when the system loses heat.
The internal energy and temperature increase (E > 0) when
the system gains heat from its surroundings.
Surroundings
HEAT
System loses heat
System
E < 0, q < 0
E > 0, q > 0
System gains heat
CHEM 3310

28. Calculation : Example 1:
A 1.00 g sample of the rocket fuel hydrazine, N2H4, is burned in a
bomb calorimeter containing g of water. The temperature rises
From °C to °C. Taking the heat capacity of the calorimeter to be 200. cal °C-1, calculate:
q for the combustion of the 1 g sample
The molar heat of combustion of hydrazine.
Answer:
(a) calories
(b) kcal/mole

29. Calculation: Example 2:
When 5.00 g of sodium hydroxide is added to 100. g of water, the temperature rises from 25.0oC to 37.5 °C. Calculate the molar heat of reaction for the process
NaOH (s)  Na+ (aq) + OH- (aq)
taking the specific heat of water to be 1.00 cal g-1 °C-1 and that of NaOH to be 0.48 cal g-1 °C-1.
Answer:
-10.2 kcal/mole

30. Calculation: Example 3:
An experiment is designed to measure the heat of fusion of ice.
25 g of ice at 0 °C was dropped into 195 g of water at 30 °C. The water is contained in a copper calorimeter of mass 100 g. The final temperature was 18 °C.
Given the specific heat of copper is cal g-1 °C-1, find the heat of fusion of ice.
Answer:
Heat of fusion of ice = 80.1 cal/g

31. Calculation: Example 4:
Steam at 100. °C is condensed in a large calorimeter. The heat capacity of the calorimeter is expressed as “water equivalent” to 272 g.
The calorimeter contains 2.82 kg of water at 5.0 °C. The final temperature of 27.8 °C is reached after 115 g of steam has been condensed. Find the latent heat of vaporization of water as given by these data.
Answer:
Heat of vaporization of water = 540 cal/g

32. Next we will talk about work! HEAT WORK System E < 0, q < 0
Surroundings
HEAT
WORK
Work is done by the system
System loses heat
System
E < 0, q < 0
E < 0, q < 0
E > 0, q > 0
E > 0, q > 0
Work is done on the system
System gains heat
CHEM 3310