1. 1st LAW OF THERMODYNAMICS
Rohazita Bt Bahari

2. HEAT CAPACITY
Why is important to study heat capacity?. By studying heat capacity, we will able to answer questions such as:
How much will the system temperature increase when is provided?
Is this temperature increase proportional to the mass of the system?
Is this temperature increase related to the nature of the substance?
Does it take any difference if the heat is provided at Constance pressure or volume

3. Heat Capacity
Heat Capacity is the amount of heat required to increase the temperature of an object by 1°C.
Heat Capacity Constant Volume, Cv
Constant Pressure, Cp

4. Heat Capacity At Constant Volume
Heat can be exchanged at constant volume by heating up on ideal gas enclosed in a container with fixed volume. (V=0), no work (W=0)
Q = ∆ H = n Cv ∆T (n = no of mole)
Why n? : because it gas
Heat Capacity:
Cv = Qv/n ∆T

5. Constant Volume,Cv Example : Situation:
1)Container with close lid (no changing volume + no work)
2) Put heat (q)into it, what happen? What going to change?

6. Constant volume, Cv Change: PRESSURE TEMPERATURE Remain: No of moles

7. Constant Pressure, Cp Cp = Q/n ∆T
To make sure pressure constant, put a movable frictionless piston on top of a cylinder and mass to provide the pressure with ideal gas.
Q = ∆ H = n Cp ∆T (n = no of mole)
Why n? : because it gas
Heat Capacity:
Cp = Q/n ∆T

8. Constant Pressure,Cp Example : Situation:
1)Container with close lid (no changing volume + no work)
2) Put heat (q)into it, what happen? What going to change? What remain?

9. Constant Pressure,Cp Change: VOLUME TEMPERATURE Remain: No of moles

10. RELATIONSHIP BETWEEN Cp and Cv
For SOLID and LIQUIDS, the Cp and Cv can be considered approximately the same:
Cp = Cv

11. RELATIONSHIP BETWEEN Cp and Cv
For GASES Cp > Cv If the same amount of energy is provided to the same amount of gas, the temperature will increase more if the volume is held constant, instead of the pressure

12. Examine Monoatomic Gas
∆U = Q- W
Q = ∆U + W
In constant Volume Situation,
Qv = ∆U , why? Gas that’s trapped and can’t do any work.
Then, Qv = ∆U = 3/2 nR ∆T
Cv = Qv/n∆T
= 3/2 R (heat capacity for constant volume)

13. Examine Monoatomic Gas
∆U = Q- W
Q = ∆U + W
In constant Pressure Situation,
Start with W, because work will no be zero
W = P∆T = nR∆T
But, the change energy, ∆U = Q – W ,so
Qp = ∆U + W
= 3/2nRT + nR∆T
Cp = Qp/n ∆T
= 3/2 R + R
= 5/2 R

14. RELATIONSHIP BETWEEN Cp and Cv
For IDEAL GASES Cp – Cv = R For Monoatomic Ideal Gases: Cp = 5/2 R Cv = 3/2 R (Cp > Cv)

15. MEASURES CONVERSION OF MECHANICAL ENERGY (WORK) INTO HEAT
Joule Experiment
1843- James Psescott Joule
He measured how much heat energy is going to be provided by doing certain mechanical actions. So, he did the experiment.
Joule experiment :
MEASURES CONVERSION OF MECHANICAL ENERGY (WORK) INTO HEAT

16. Joule Experiment Height falls doing work : mgh W = F gravity x hight
(mg) x (h)
Pinwheel transforms energy into heat (random motion of water)
Result :
Joule shows heat and work are different forms of energy. The total being conserved.

17. Joule Experiment
Video

18. Joule Experiment Unit SI energy: I Joule = 1kg m2/s2
1 calorie = J
(energy to heat ig water 1°C)

19. Joule Experiment To determine
 - A sample of a gas (the system) was placed in one side of the apparatus and the other side of the apparatus was evacuated.
-The initial temperature of the apparatus was measured.
-The stopcock was then opened and the gas expanded irreversibly into the vacuum. 

20. Joule Experiment
-Because the surroundings were not affected during the expansion into a vacuum, w was equal to zero.
-The gas expanded rapidly so there was little opportunity for heat to be transferred to or from the surroundings.
- If a change in temperature of the gas occurred, heat would be transferred to or from the surroundings after the expansion was complete, and the final temperature of the surroundings would differ from the initial temperature.

21. Joule Experiment To determine Joule coefficient,
Joule experiment gave and
because the changes in temperature that occurred were too small to be measured by the thermometer.

22. The Joule-Thompson Effect
The closed system of constant composition
Where the Joule-Thompson coefficient, μ (mu) is define as
This relation will prove useful for relating the heat capacities at constant volume and for a discussion of liquefaction of gases.

23. Joule-Thomson experiment
To determine
Insulated wall
Initial State
Final State

24. Observation of the Joule-Thompson Effect
The analysis of the Joule-Thompson coefficient is central to the technological problems associate with the liquefaction of gases. We need to be able to interpret its physical and measured it.
The apparatus used to for measuring The Joule-Thompson effect. The gas Expands through the porous barrier, Which acts as a throttle, and the whole apparatus is thermally insulated.
This arrangement corresponds to an Isenthalpic expansion
(expansion at constant enthalpy). Whether the expansion results in a heating
or a cooling of the gas depends on the conditions

25. Definiton of the isothermal Joule- Thompson coefficient
The isothermal Joule –Thompson coefficients is the slope of the enthalpy with respect to changing pressure the temperature being held constant.
coefficients are related by

26. A schematic diagram of the apparatus used for measuring the isothermal Joule-Thompson coefficient. The electrical heating required to offset the cooling arising from expansion is interpreted as ∆H and used to calculate , which is then converted to μ.

27. Real gases have nonzero Joule-Thompson coefficient
Real gases have nonzero Joule-Thompson coefficient. Depending on the identity of the gas, pressure, the relative magnitudes of the attractive and repulsive intermolecular forces, and the temperature, the sign of the coefficient may be either positive or negative.
+ve sign → dT is –ve when dp is –ve
(in which case the gas cool on expansion)
Heating effect (μ<0) at one temperature show at cooling effect (μ>0) when the temperature is below their upper inversion temperature,T1.
A gas typically has two inversion temperatures, one at high temperature and the other at low.

28. The principle of Linde refrigerator
The gas at high temperature is allowed to expand through a throttle, it cools
and is circulated past the incoming gas.
That gas is cooled, and its subsequent expansion cools it still further.

29. In thermodynamics, the Joule–Thomson effect  describes the temperature change of a real gas or liquid (as differentiated from an ideal gas) when it is forced through a valve or porous plug while kept insulated so that no heat is exchanged with the environment. This procedure is called a throttling process or Joule–Thomson process. 
At room temperature, all gases except hydrogen, helium and neon cool upon expansion by the Joule–Thomson process; these three gases experience the same effect but only at lower temperatures.

30. The throttling process is commonly exploited in thermal machines such as refrigerators and air conditioners.
Throttling is a fundamentally irreversible process. The throttling due to the flow resistance in supply lines, heat exchangers, regenerators, and other components of (thermal) machines is a source of losses that limits the performance.

31. Throttling Is Irreversible (explanation with real-life example)
Now, imagine there are students in a class room and as the bell rungs, they started moving out. Now if the door is opened partially then their will be clusters form by the students. Similarly in flow of fluid there is restriction to flow.
Now, as students form clusters they pushing other students and in the same way fluid particles start rubbing with other molecules and as the result friction is their and as well know friction is one of the biggest reason for any process to make it irreversible.

32. Examples Of Throttling Process
Flow through a partially opened valve as in IC engines
Flow through a very small opening
Flow through a porous plug

33. Characteristics Of Throttling
No work is done, W = 0 (no work transfer)
No Heat Transfer
Irreversible Process
(Process is adiabatic, Q = 0)
No change in enthalpy from state one to step two,
h1 = h2 (Isenthalpic Process)

34. No work transfer
Explanation:
As we know there is work transfer in turbines due to very large pressure difference. But in the case of throttling their is very low pressure differences so the work we get is very small and this work is lost in overcoming friction. So, here we generally neglect the work transfer.

35. No Heat Transfer Explanation:
Imagine you open the freeze and took out water bottle and just within the fraction of seconds you put back that bottle, now what is the temperature difference between the two states of the bottle, it is approx. Zero. It means heat transfer needs some time, now what happened in throttling is that device length is very small and the fluid is always pushed ahead due to the bulk coming and due this there is not much time for the heat transfer. As we are considering the steady flow and bulk is not accumulated in the device so heat transfer is also neglected.

36. Isenthalpic Process The word ISENTHALPIC means same Enthalpy.
An Isenthalpic process is an adiabatic reversible process. The entropy change in a isentropic process is zero.
Literally, an isotropic process is a process in which the entropy change is zero. However, in thermodynamic “isentropic” refers to a process that is both adiabatic and reversible. A process that has an entropy change of zero is not necessarily both adiabatic and reversible process.

37. Joule-Thomson experiment
Adiabatic process
Rearranging;
Isenthalpic process

38. Joule-Thomson experiment
To determine
Joule-Thomson coefficient,

39. Adiabatic means No heat transfer, Q = 0 ∆U = Q + W W = ∆U
Adiabatic Process
Adiabatic means
No heat transfer, Q = 0
∆U = Q + W
W = ∆U

40. Video
During the expansion, the speed of the particles doesn’t change, so the temperature remain same.

41. Perfect Gases Perfect gas: one that obeys both of the following:
U is not change with V at constant T
If we change the volume of an ideal gas (at constant T), we change the distance between the molecules.
Since intermolecular forces are zero, this distance change will not affect the internal energy.

42. Perfect Gases
For perfect gas, internal energy can be expressed as a function of temperature (depends only on T):
An infinesimal change of internal energy,
From

43. This shows that H depends only on T for a perfect gas
Perfect Gases
For perfect gas, enthalpy depends only on T;
Thus;
An infinesimal change of enthalpy,
This shows that H depends only on T for a perfect gas
From

44. Perfect Gases The relation of CP and CV for perfect gas; Perfect gas
PV=nRT, thus or

45. Perfect Gases
For perfect gas,
- Since U & H depend only on T.

46. Perfect gas and First Law
First law; For perfect gas;
Perfect gas, rev. process, P-V work only

47. Reversible Isothermal Process in a Perfect Gas
First law:
Since the process is isothermal, ∆T=0;
Thus, ∆U = 0
First law becomes;

48. Reversible Isothermal Process in a Perfect Gas
Since , thus:
Work done;
rev. isothermal process, perfect gas OR

49. Reversible Adiabatic Process in Perfect Gas
First law,
Since the process is adiabatic,
For perfect gas, becomes,
Since

50. Reversible Adiabatic Process in Perfect Gas
Since , thus
By integration;

51. Reversible Adiabatic Process in Perfect Gas
If is constant, becomes; OR
rev. adiabatic process, perfect gas

52. Reversible Adiabatic Process in Perfect Gas
Alternative equation can be written instead of
Consider

53. Reversible Adiabatic Process in Perfect Gas
Since
rev. adiabatic process, perfect gas, CV constant
where

54. Reversible Adiabatic Process in Perfect Gas
If is constant, becomes
rev. adiabatic process, perfect gas, CV constant

55. Calculation of First Law Quantities
1. Reversible phase change at constant P & T.
Phase change: a process in which at least one new phase appear in a system without the occurance of a chemical reaction (eg: melting of ice to liquid water, freezing of ice from an aqueous solution.)

56. Calculation of First Law Quantities
2. Constant pressure heating with no phase change

57. Calculation of First Law Quantities
3. Constant volume heating with no phase change
Recall:

58. Calculation of First Law Quantities
4. Perfect gas change of state: H & U of a perfect gas depends on T only;

59. Calculation of First Law Quantities
5. Reversible isothermal process in perfect gas
H & U of a perfect gas depends on T only;
Since , thus

60. Calculation of First Law Quantities
6. Reversible adiabatic process in perfect gas
If Cv is constant, the final state of the gas can be found from:
where
Adiabatic process