1. CHEM 3310
Thermodynamics Work

2. 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

3. Heat and Work are interconvertible
Joule’s Apparatus
The amount of mechanical
Energy lost = the amount of thermal energy gained.
1 calorie = joules
ORDERS of Magnitudes of Energy
CHEM 3310

4. Change in internal energy can be a result of a chemical reaction
System: The chemical reaction
Surroundings: Everything that interacts with the system
All matter has chemical energy.
Chemical bonds are a source of energy (PE).
The movement of molecules in space is a source of energy (KE).
The vibrations and rotations of molecules is another source of chemical energy (KE).
All of these forms of chemical energy contribute in one way or another to chemical reactions.
CHEM 3310

5. Change in internal energy can be a result of a chemical reaction
System: The chemical reaction
Surroundings: Everything that interacts with the system
In a chemical reaction, when bonds are broken and new bonds are formed, the internal energy of the system changes.
Exothermic reaction – heat flows out of the system
Endothermic reaction – heat flows into the system
CHEM 3310

6. Change in internal energy can be a result of electrical work
Chemical reactions can do work on their surroundings by driving an electric current through an external load.
External
load
CHEM 3310

7. Change in internal energy can be a result of work of expansion
Chemical reactions can do work on their surroundings when the volume of the system expands during the course of the reaction.
Internal combustion engine burns a fuel with air and uses the hot gases for generating power.
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CHEM 3310

8. Work, w
Mechanical work is defined as the force exerted on an object
multiplied by the distance which the object moves resulting
from the applied force.
w = f  x
If there is no movement (x=0), then w = 0.
For infinitesimal change in position, dx,
dw = f  dx
CHEM 3310

9. Can we express the work of compression in terms of P and V?
Work, w
Define a system consisting of an ideal gas enclosed in a container with a moveable piston. Let’s evaluate the work of compression.
External pressure, Pext
Can we express the work of compression in terms of P and V? V1 V2
CHEM 3310

10. Work, w f is the force Pext is the external pressure
External pressure, Pext
f is the force
Pext is the external pressure
A is the cross-sectional area
of the piston A V1 A V2
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11. w = f  x w = Pext  A  x w = Pext V w = Pext  A  x Work, w V
External pressure, Pext
Substitutef = Pext A
w = Pext  A  x x1 A
For a change in the position of the piston, V1 x2
w = Pext  A  x V2
x =x2 - x1 V
V = V2 - V1 = A (x2 - x 1)
w = Pext V
x1 is the initial piston position x2 is the final piston position
V1 is the initial volume
V2 is the final volume
CHEM 3310

12. For an infinitesimal change
Work, w
w = Pext V
External pressure, Pext
For an infinitesimal change
In volume,
dw = Pext dV x1 A
Integrate to get
total work done. V1 x2 V2
x =x2-x1
x1 is the initial piston position x2 is the final piston position
V1 is the initial volume
V2 is the final volume
CHEM 3310

13. Work, w Sign consideration Since V1 > V2 V = V2 – V1 < 0
External pressure, Pext
For work of compression, the system gains energy. Work
should be a positive quantity. x1 A
Since V1 > V2
V = V2 – V1 < 0 V1 x2 V2
We modify the equation.
x =x2-x1
x1 is the initial piston position x2 is the final piston position
V1 is the initial volume
V2 is the final volume
Add negative sign!
CHEM 3310

14. V1 is the initial volume V2 is the final volume
Work, w
Similarly, for work of expansion, the system (gas) does work. The system loses energy. Work should be a negative quantity.
External pressure,
Pext
V > 0 V2 V1
Work is negative for work of expansion!
System loses energy
V1 is the initial volume
V2 is the final volume
CHEM 3310

15. Work, w
w = f  x
Mechanical work
PV work done by or on a gas confined in a piston and cylinder configuration.
CHEM 3310

16. Summary of work and heat
q < 0
system releases heat
q > 0
system absorbs heat
w < 0
work done by the system on the environment
w > 0
work done on the system by the environment
Surroundings
HEAT
WORK
w < 0; Work is done by the system (expansion)
q < 0; Heat flows out of the system.
System loses heat
System
E < 0
E < 0
E > 0
E > 0
w > 0; Work is done on the system (compression)
q > 0; Heat flows into the system. System gains heat
CHEM 3310

17. Work, w
Area under a PV curve!
The PV diagram is a useful visualization of a process. The area under the curve of a process is the amount of work done by or on the system during that process.
External pressure, Pext
External pressure, Pext
Compression is work done on the system
Expansion is work done by the system
CHEM 3310

18. Work, w Types of thermodynamic processes:
1. An isobaric process is a constant pressure process (dP=0).
As a gas is being heated slowly, the volume of the cylinder expands to maintain constant pressure.
As a gas is being cooled slowly, the volume of the cylinder decreases to maintain constant pressure.
w < 0
w > 0 V1 V2 V2 V1
V1 is the initial volume; V2 is the final volume
The area under the curve is the work done.
CHEM 3310

19. Work, w Types of thermodynamic processes:
1. An isobaric process is a constant pressure process.
This is the PV work equation
for an isobaric process.
CHEM 3310

20. Work, w Types of thermodynamic processes:
2. An isochoric process is a constant volume process. (dV=0)
The gas is being heated in a rigid container, such as a
chemical reaction being carried out in a bomb calorimeter. Since dV=0, the reaction does no work.
This is the PV work equation for an isochoric process.
The area under the curve is the work done.
CHEM 3310

21. Work, w Types of thermodynamic processes:
3. An isothermal process is a constant temperature process. (dT=0)
nRT is a constant for an
isothermal process
where n = # of moles of gas
R = gas constant
T = temperature of the gas
When the temperature stays constant, the pressure and volume are inversely proportional to one another.
The area under the curve is the work done.
CHEM 3310

22. Work, w Types of thermodynamic processes:
3. An isothermal process is a constant temperature process.
The piston is slowly moved so that the gas expands. As the gas expands, a temperature drop results. To maintain isothermal condition, heat will flow into the system. In fact, the amount of heat, q is equal to the magnitude of w. q
External pressure, Pext
w < 0 q
CHEM 3310

23. Work, w Types of thermodynamic processes:
4. An adiabatic process is when no heat is added or removed from the system. (q=0)
When gas expands very quickly that no heat can be transferred.
Eg – CO2 fire extinguisher
The area under the curve is the work done.
CHEM 3310

24. Work, w
Compare the area under the PV curve for the work of expansion done by the gas
by the following processes:
1. An isobaric process. dP=0
2. An isothermal process. dT=0
3. An adiabatic. q=0 w w w V1 V2 V1 V2 V1 V2
For each of the above processes, the gas starts with a volume of V1 and end with a volume of V2.
Work of expansion is process (or path) dependent.
Work is not a state function.
CHEM 3310

25. no work is done by the gas.
Work, w
What is the work done as a result of the free expansion of an ideal gas?
Free expansion is process where a gas expands into an insulated evacuated chamber.
During free expansion,
no work is done by the gas.
Pext = 0
CHEM 3310

26. Work, w
Calculate the work of expansion that accompanies the fusion of 1 mole of ice to form 1 mole of liquid water at 1 atm and 0oC. A decrease in volume of 1.49 mL is observed.
H2O (s)  H2O (l)
w = - (1 atm)(-1.49 x 10-3 L)
w = 1.49 x 10-3 Latm
Energy unit.
R = Latmmole-1K-1
w = J
R = Jmole-1K-1
w > 0 because dV < 0
Latmmole-1K-1 = J mole-1 K-1
1 Latm = J
CHEM 3310

27. Work, w
Calculate the work of expansion that accompanies the vapourization of water.
The molar volume of liquid water at 100. °C is 18.8 cc, while the molar volume of water vapour at 100 °C and 1 atm is 30.2 L.
H2O (l)  H2O (g)
w = - (1 atm)(30.2L – L)
w = Latm
Energy unit.
R = Latmmole-1K-1
w = J
R = Jmole-1K-1
w < 0;
Work of expansion for the vapourization of water is much bigger than the previous example.
Latmmole-1K-1 = Jmole-1K-1
1 Latm = J
CHEM 3310

28. Work, w External pressure, Pext = 1.00 atm Expect w < 0
Calculate the work done due to the thermal expansion of 1.00 mole of ideal gas
heated at a constant pressure of 1.00 atm from 0.00 °C to 100. °C.
External pressure, Pext = 1.00 atm
Expect
w < 0
V2 = L
V1 = L
T1 = 0.00 °C
T2 = 100.°C
n = 1.00 mole
w = -199 calories
CHEM 3310

29. Work, w External pressure, Pext = 1.00 atm Expect w < 0
Calculate the work done due to the thermal expansion of 1.00 mole of ideal gas
heated at a constant pressure of 1.00 atm from 0.oC to 100. °C.
External pressure, Pext = 1.00 atm
Expect
w < 0
T1 = 0 °C
T2 = 100. °C
n = 1 mole
CHEM 3310

30. Work, w External pressure, Pext Expect w > 0
Calculate the work compression of 2.00 moles of an ideal gas from 1.00 bar to bar at 25.0 °C. (1 bar = atm)
External pressure, Pext
Expect
w > 0
P1 = bar
P2 = 100. bar
Use this equation with pressure, and you should get the same answer, w = 23.0 kJ
T = 298 K
n = 2.00 moles
P1 = atm
P2 = 98.7 atm
w = 23.0 kJ
CHEM 3310

31. The system does 20.4 kJ of work.
Example:
The combustion of octane yields the following reaction. Calculate the work done by
the system under standard temperature and pressure condition (i.e. STP
condition is 0 °C and 1 atm)
2 C8H18 (l) + 25 O2 (g)  16 CO2 (g) + 18 H2O (g)
n = 34 – 25 = 9 moles of gas
Under STP condition, the volume that is occupied by 1 mole of gas is
22.4 Lmole-1.
V2 - V1 = 9 moles  22.4 Lmole-1
= L
w = - 1 atm  L = L atm = - (201.6 L atm)( joule L-1 atm-1) = kJ or
w = - 9 moles  J mole-1 K-1  K = kJ
The system does 20.4 kJ of work.
CHEM 3310

32. Summary:
Internal energy, E:  All the energy of the system (chemical, potential, kinetic, etc.)
Thermal energy:  The part of the internal energy that changes temperature
Gases: Possess translational, vibrational, rotational motions, all contributing to thermal energy
Ways to change internal energy of gas system:
            Heat, q
            Work, w
   Both heat and work
State: The values of P, V, T, E of the system
Change of State: By altering P, V, T, and E
CHEM 3310