1. Thermodynamics: The First Law
자연과학대학 화학과
박영동 교수
Classical thermodynamics

2. Thermodynamics: the first law
2.1 The conservation of energy
2.1.1 Systems and surroundings
2.1.2 Work and heat
2.1.3 The measurement of work
2.1.4 The measurement of heat
2.1.5 Heat influx during expansion
2.2 Internal energy and enthalpy
2.2.6 The internal energy
2.2.7 Internal energy as a state function
2.2.8 The enthalpy
2.2.9 The temperature variation of the enthalpy

3. Thermodynamic Systems, States and Processes
Objectives are to:
define thermodynamics systems and states of systems
explain how processes affect such systems
apply the above thermodynamic terms and ideas to the laws of thermodynamics

4. Thermodynamic universe
The Universe
The universe consists of the system and surroundings.
surroundings
the system is the region of interest; its region is defined by the boundary.
the rest of the world is its surroundings.
The surroundings are where observations are made on the system.
system

5. Various Systems state, state functions, path, process,
open
system
matter
energy
open(열린 계)
allowed
closed(닫힌 계)
forbidden
isolated(고립 계)
isolated
closed
Some other thermodynamics terms
state,
state functions,
path, process,
extensive properties,
intensive properties.

6. A State Function and Paths
The altitude is a state property, because it depends only on the current state of the system. The change in the value of a state property is independent of the path between the two states.
The distance between the initial and final states depends on which path (as depicted by the blue and red lines) is used to travel between them. So it is not a state function.
pressure, temperature, volume, ...
heat, work

7. The First Law of Thermodynamics
-- Heat and work are forms of energy transfer and energy is conserved.
U = Q + W
change in
total internal energy
heat added
to system
work done
on the system
State Function Process Functions

8. Calculating the change in internal energy
Suppose someone does 622 kJ of work on an exercise bicycle and loses 82 kJ of energy as heat. What is the change in internal energy of the person? Disregard any matter loss by perspiration.
Solution
w = −622 kJ (622 kJ is lost by doing work on the bicycle),
q = −82 kJ (82 kJ is lost by heating the surroundings).
Then the first law of thermodynamics gives us
U = q + w = (-82 kJ )+ (-622 kJ) = -704 kJ
We see that the person’s internal energy falls by 704 kJ. Later, that energy will be restored by eating.

9. Work, and the expansion(p-V) work
Work = force • distance
pex=F/A
force acting on the piston = pex × Area
distance when expand = dy F F +y
Increase in volume, dV
(p-V work) dW

10. Total Work Done We will consider the following cases
To evaluate the integral, we must know how the pressure depends (functionally) on the volume.
We will consider the following cases
0. Constant volume work
Free expansion
Expansion against constant pressure
Reversible isothermal expansion

11. 0. Work for the constant volume process
For the constant volume process, there is no p-V work

12. Heat and Internal Energy
For the constant volume process, there is no p-V work, so
U = q + w = qV
If we add heat q to the system, the temperature of the system increases by T , and the internal energy increases by U.
Constant volume heat capacity

13. Internal Energy of a Gas
A pressurized gas bottle (V = 0.05 m3), contains helium gas (an ideal monatomic gas) at a pressure p = 1×107 Pa and temperature T = 300 K. What is the internal thermal energy of this gas?

14. molar constant volume heat capacity of monatomic gases (at 1 atm, 25 °C)
CV, m (J/(mol·K))
CV, m/R He
12.5
1.50 Ne Ar Kr Xe

15. Temperature and Energy distribution
The temperature is a parameter that indicates the extent to which the exponentially decaying Boltzmann distribution reaches up into the higher energy levels of a system. (a) When the temperature is low, only the lower energy states are occupied (as indicated by the green rectangles). (b) At higher temperatures, more higher states are occupied. In each case, the populations decay exponentially with increasing temperature, with the total population of all levels a constant.

16. A constant-volume bomb calorimeter.
The constant-volume heat capacity is the slope of a curve showing how the internal energy varies with temperature. The slope, and therefore the heat capacity, may be different at different temperatures.

17. 1. Work for free expansion case
For the free expansion process, there is no p-V work
U = q + w = q

18. 2. Work for expansion against constant pressure
U = q + w = q - p V
qp = U + p V = (U + p V )
If we define Enthalpy as, H ≡ U + pV
qp = H
Enthalpy change is the heat given to the system at constant pressure.

19. Cp, Cv for an ideal gas
For an ideal gas, U and H do not depend on volume or pressure.
For an example, for an ideal monatomic gases, U = (3/2) nRT

20. Heat capacity of CO2 and N2
The heat capacity with temperature as expressed by the empirical formula Cp,m/(J K-1 mol-1) = a+bT+c/T2. The circles show the measured values at 298 K.

21. molar heat capacities, Cp,m/(J K-1 mol-1) = a+bT+c/T2
b/(l0-3K-1)
c/(105K2)
@273K*
@298K*
@350K*
Monatomic gases
20.78
2.50
Other gases
Br2
37.32
0.5
-1.26
4.30
4.34
4.39
Cl2
37.03
0.67
-2.85
4.02
4.09
4.20
CO2
44.22
8.79
-8.62
4.22
4.47
4.84 F2
34.56
2.51
-3.51
3.67
3.77
3.92 H2
27.28
3.26
3.47 I2
37.4
0.59
-0.71
4.40
4.42
4.45 N2
28.58
-0.5
3.48
3.50
3.55
NH3
29.75
25.1
-1.55
4.15
4.27
4.48 O2
29.96
4.18
-1.67
3.53
3.62
H2O(l)
75.29
9.06
C(s, graphite)
16.86
4.77
-8.54
0.81
1.04
1.39
*calculated results are given in unit of R(8.314 J K-1 mol-1)

22. An exothermic process Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)
When hydrochloric acid reacts with zinc, the hydrogen gas produced must push back the surrounding atmosphere (represented by the weight resting on the piston), and hence must do work on its surroundings. This is an example of energy leaving a system as work.
exothermic and endothermic processes

23. Work and Heat
Work is transfer of energy that causes or utilizes uniform motion of atoms in the surroundings. For example, when a weight is raised, all the atoms of the weight (shown magnified) move in unison in the same direction.
Heat is the transfer of energy that causes or utilizes chaotic motion in the surroundings. When energy leaves the system (the green region), it generates chaotic motion in the surroundings (shown magnified).

24. The expansion(p-V ) Work
When a piston of area A moves out through a distance h, it sweeps out a volume ΔV = Ah. The external pressure pex opposes the expansion with a force pexA.

25. 3. Work for reversible isothermal expansion for an ideal gas

26. Reversible Isothermal Expansion Work for a perfect gas
The work of reversible, isothermal expansion of a perfect gas. Note that for a given change of volume and fixed amount of gas, the work is greater the higher the temperature.

27. Molecular Basis for Heat Capacity
The heat capacity depends on the availability of levels. (a) When the levels are close together, a given amount of energy arriving as heat can be accommodated with little adjustment of the populations and hence the temperature that occurs in the Boltzmann distribution. This system has a high heat capacity. (b) When the levels are widely separated, the same incoming energy has to be accommodated by making use of higher energy levels, with a consequent greater change in the ‘reach’ of the Boltzmann distribution, and therefore a greater change in temperature. This system therefore has a low heat capacity. In each case the green line is the distribution at low temperature and the red line that at higher temperature.

28. Temperature dependence of enthalpy(H) and internal energy(U)
The enthalpy of a system increases as its temperature is raised. Note that the enthalpy is always greater than the internal energy of the system, and that the difference increases with temperature.

29. Temperature dependence of enthalpy(H) and internal energy(U)
Note that the heat capacityies depend on temperature, and that Cp is greater than CV.

30. a DSC (differential scanning calorimeter)
sample
reference
A thermogram for the protein ubiquitin. The protein retains its native structure up to about 45°C and then undergoes an endothermic conformational change. (Adapted from B. Chowdhry and S. LeHarne, J. Chem. Educ. 74, 236 (1997).)

31. Cp , CV and internal energy
𝜅 𝑇 =− 1 𝑉 𝜕𝑉 𝜕𝑝 𝑇 : isothermal compressibility
𝜕𝑆 𝜕𝑉 𝑇 = 𝜕𝑝 𝜕𝑇 𝑉
𝜋 𝑇 =−𝑝+ 𝑇 𝜕𝑝 𝜕𝑇 𝑉

32. Cp , CV and internal energy

33. The Joule-Thomson effect
Work done by the gas on the first piston: W1 = p1V1 and the work by the gas on the second piston: W2 = p2V2
The change in the internal energy is: ΔU= - (p2V2 - p1V1)
So: U2 – p2V2 = U1 – p1V2
And therefore: H2 = H1

34. For an isenthalpic process:
This can be rearranged to give:

39. The Joule-Thomson coefficient is zero for an ideal gas.
For a non-ideal gas:
dH = TdS + Vdp
and at constant temperature:

40. The Joule-Thomson coefficient of the Van der Waals gas
which can be differentiated after neglect of the smallest magnitude term to give:
And using the approximation:

41. leads to:
So the Van der Waals form for the JT coefficient is:
with an inversion temperature of:
If the neglected term is included, the exact solution is:

43. 𝑝 𝑉 𝑚 𝑅𝑇 =1+ 𝐵 2 (𝑇) 𝑉 𝑚 + 𝐵 3 (𝑇) 𝑉 𝑚 2 + …
The most general equation is the virial equation :
𝑝 𝑉 𝑚 𝑅𝑇 =1+ 𝐵 2 (𝑇) 𝑉 𝑚 𝐵 3 (𝑇) 𝑉 𝑚 …
Which yields as an expression for the JT coefficient:
𝜇= 𝑇 𝜕 𝐵 2 /𝜕𝑇 𝑝 − 𝐵 2 𝐶 𝑝,𝑚

44. 𝐵 2 𝑇 = 𝑁 𝑜 0 ∞ 1− 𝑒 −𝑈(𝑟)/𝑘𝑇 2𝜋𝑟2𝑑𝑟
The virial coefficient, B2, can be found from statistical mechanics:
𝐵 2 𝑇 = 𝑁 𝑜 0 ∞ 1− 𝑒 −𝑈(𝑟)/𝑘𝑇 2𝜋𝑟2𝑑𝑟
Differentiation with respect to temperature gives a theoretical expression for the JT coefficient that makes it now a function of the potential energy of interaction of the molecules, U(r). A common representation of this potential is the Lennard-Jones potential:
𝑈 𝑟 =4𝜀 𝜎 𝑟 12 − 𝜎 𝑟 6
Thus, the JT coefficient can be related directly to a theoretical model.

45. The Maxwell relations
𝑈 𝑆, 𝑉 ;𝑑𝑈=𝑇𝑑𝑆 −𝑝𝑑𝑉;𝑇= 𝜕𝑈 𝜕𝑆 𝑉 ;𝑝=− 𝜕𝑈 𝜕𝑉 𝑆 ; 𝜕𝑇 𝜕𝑉 𝑆 =− 𝜕𝑝 𝜕𝑆 𝑉
𝐻 𝑉, 𝑇 ;𝑑𝐻=𝑇𝑑𝑆+𝑉𝑑𝑝; 𝑇= 𝜕𝐻 𝜕𝑆 𝑝 ; 𝑉= 𝜕𝐻 𝜕𝑝 𝑆 ; 𝜕𝑇 𝜕𝑝 𝑆 = 𝜕𝑉 𝜕𝑆 𝑝
𝐺 𝑉, 𝑇 ;𝑑𝐺=𝑉𝑑𝑝 −𝑆𝑑𝑇; 𝑉= 𝜕𝐺 𝜕𝑝 𝑇 ; 𝑆=− 𝜕𝐺 𝜕𝑇 𝑝 ; 𝜕𝑉 𝜕𝑇 𝑝 = − 𝜕𝑆 𝜕𝑝 𝑇
𝐴 𝑉, 𝑇 ;𝑑𝐴=−𝑝𝑑𝑉 −𝑆𝑑𝑇;𝑝=− 𝜕𝐴 𝜕𝑉 𝑇 ;𝑆=− 𝜕𝐴 𝜕𝑇 𝑉 ; 𝜕𝑝 𝜕𝑇 𝑉 = 𝜕𝑆 𝜕𝑉 𝑇

46. 4. Work for adiabatic expansion
qad = 0
U = q + w = w
If the heat capacity is independent of temperature,
U = CV T = w
CV dT = -pdV

47. 4. Work for adiabatic expansion
For an adiabatic process for an ideal gas,

48. For an ideal monatomic gas of n moles, calculate q, w, ΔU for each process.
w12 = -pb(Vb-Va)
ΔU12 = CV(T2-T1) = CV(pbVb/nR - pbVa/nR)
= CVpb(Vb - Va) /nR
q12 = ΔU12 - w12
w23= 0
ΔU23 = CV(T3-T2) = CV(pbVa/nR - pbVb/nR)
= CVpb(Va - Vb) /nR = - ΔU12
q23 = ΔU23
w31= -nRT3 ln(Va/Vb) = -pbVa ln(Va/Vb)
ΔU31 = 0
q31 = - w31
T2 = pbVb/nR
T1 = T3 = pbVa/nR = paVb/nR
pbVa = paVb