1. Thermodynamics Change in Internal Energy, E
CHEM 3310
Thermodynamics Change in Internal Energy, E

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. The first law of Thermodynamic states:
E = Efinal – Einitial E = q + w
Difficult to determine
Measureable
The change in the internal energy is equal to the sum of the heat gained or lost by the system and the work done on or by the system.
Image credit:
CHEM 3310

4. The first law of Thermodynamic states:
The internal energy change is equal to the heat
absorbed plus the work done.
E = q + w
System gains energy when
heat flows into the system or work is done on the system
System loses energy when
heat flows out of the system or work is done by the system
Path (process) independent
E a state function
w is not a state function
Path (process) dependent
q is not a state function
CHEM 3310

5. CAUTION: The first law of Thermodynamic states:
The internal energy change is equal to the heat
absorbed plus the work done.
E = q + w
CAUTION:
Some books and website use this equation, where change in internal energy is U.
This is only true when the work, w, is work done BY the system.
CHEM 3310

6. Example:
An expanding gas does kJ of work while it absorbs 800. kJ of heat.
What are:
(a) q
(b) w
(c) E for the gas.
Answer:
Gas absorbs 800. kJ of heat, q = 800. kJ q > 0 because the gas gains energy when heat is absorbed.
(b) Gas does kJ of work, w = kJ w < 0 because the gas loses energy when it does work.
(c) E= q + w = (800. kJ)+( kJ) = kJ.
The gas’s change in internal energy is a lost of 700. kJ. of energy.
CHEM 3310

7. The first law of Thermodynamic states:
For a chemical reaction,
Reactants  Products
State 1 E1
State 2 E2
 E = E2 – E1
System gains energy when
 E > 0, or E2 > E1
System loses energy when
 E < 0, or E2 < E1
CHEM 3310

8. Change in internal energy equals to its heat change.
Example:
One gram of a substance is burned in a bomb calorimeter. The total heat evolved in the process was 300. kJ.
(a) How much work was done in the process?
(b) What was E of the substance after it was burned?
Heat measured in a bomb calorimeter is measured under a constant volume condition.
Change in internal energy equals to its heat change.
Since this is under the special condition of constant volume, we include a subscript v to remind us!
E= qv
Answer:
Since dV=0, w=0
(b) E= qv = kJ
CHEM 3310

9. qv = nCvT Recall the discussion of heat capacity q = nCT
If we heat up n moles of gas while keeping its volume constant, the result of adding heat to the system is an increase of its temperature.
qv = nCvT
The subscript v reminds us that:
heat measured is under constant volume condition
Cv is the amount of heat energy which must be absorbed by one mole of a gas at constant volume to raise the temperature of the gas by one degree.
Since E = qv,
For infinitesimal changes,
Integrate, and assume that Cv is constant over T1 to T2.
CHEM 3310

10. Work for an adiabatic process
For an adiabatic process, q=0,
E = q + w = w
Since
Work in an adiabatic expansion or compression of an ideal gas is
CHEM 3310

11. E and Heat Capacity Cv = 3/2R
From the Kinetic-Molecular Theory of Gases, we can relate kinetic energy with internal energy.
Kinetic energy, KE, of one mole of monatomic ideal gas (eg – He, Ne, Ar, Kr, Xe) is directly proportional to its temperature in Kelvin.
KE = 3/2RT
(i.e. Each degree of freedom (x, y, z) in translation contributes ½ RT.)
Consider one mole of monatomic gas that is being heated from T1 to T2 under constant volume.
E = KE = 3/2R(T2-T1)
Since
E = n Cv(T2-T1) = Cv(T2-T1) when n=1
Equate the two equations,
This means that the Cv for gases such as: He, Ne, Ar, etc.
Cv = 3/2R = 3/2(1.987) =2.98 cal deg-1 mole-1
Cv(T2-T1) = 3/2R(T2-T1)
Cv = 3/2R
CHEM 3310

12. Summary: E = qv Cv = 3/2R E = nCv(T2-T1)
Change in internal energy of a system, E, can be determined by heat flow, q, and work done, w.
E = q + w
For a process that is carried out under constant volume condition, since dV=0, w=0.
E = qv
where qv is the heat flow under a constant volume
environment.
The subscript v reminds is that the heat measured is under constant volume condition.
In terms of heat capacity, assuming that Cv is constant over the temperature range from T1 to T2.
E = nCv(T2-T1)
For a monatomic gas,
Cv = 3/2R
CHEM 3310