Chapter 6 Thermodynamics:
The First Law
6-1 Systems, States, and Energy
What is Thermodynamics?
Thermodynamics
(熱力學):
The study of the
transformation of energy
during chemical reaction
and physical process
Heat of Reaction
Steam Engine
Thermodynamics Laws
The 0th Law: Thermal Equilibrium
The 1st Law: Conservation of Energy
The 2nd Law: Entropy
The 3rd law: Absolute Zero Temperature
初識 The First Laws
Conservation of Energy
動能+位能=常數
Energy can be changed from one form to another, but it
cannot be created or destroyed.
熱力學研究中,如何討論能量保守?
熱和能量之間有什麼關係?
在化學反應實驗中,如何討論能量保守?
System & Surroundings
Varieties of Systems
Definition of Internal Energy
Internal Energy U:The total energy of a
system includes kinetics energy, intermolecular
potential energy, bond energy, ionization
energy ….et al.
Conservation of Energy
Isolated systems
ΔUsys = 0
Open systems
ΔUsys + ΔEsurr = 0
ΔUsys = -ΔEsurr
ΔUsysΔEsurr
ΔUsys �� ΔEsurr
The First Law of
Thermodyanmics
ΔUsys = q + w
q 及 w 皆以系統為中心
增加系統內能者為正值
減少系統能量者負值Heat q Work w
ΔUsys
Example
An automobile engine does 520 kJ of work and
loses 220 kJ of energy as heat. What is the
change in the internal energy of the engine?
A chemical reaction known to release 1.78 kJ
of heat take place in a constant-volume
container. What are ΔUsys , q and w?
What is work(功)?
Expansion Work
The work of expansion
against an external force
Nonexpansion work
The work does not
involve a change in
volume
Expansion Work
VP
A
VAP
lFw
Work
ex
ex
∆−=
∆−=
∆×=
×=
distanceforce
Example
Suppose a gas expands by 0.5L against a pressure of
1.20 atm. How much work is done in the expansion?
Energy Conversion Table
1 Pa*m3 = 1 kg/m/s2 * 1 m3 = 1 J
1 L*atm = 10-3 m3* 101325 Pa = 101.325 Pa*m3 =101.325 J
Reversible & Irreversible Expansion
Example
To calculate the work of
reversible, isothermal
expansion of a gas from
Vinitial to Vfinal constant
temperature T.
cwhen PdVPw exex ≠= ∫ ,
Heat (熱)
Heat
通常指藉由溫度差異傳遞的能量稱為heat
Heat : q = CΔT
C = Heat Capacity (熱容)
Cs = Specific Heat Capacity = C/m, where m is the mass.
Cm = Molar Heat Capacity = C/n, where n is the amount of
moles.
Heat Capacities
測量熱變化的方法
DTA: Differential Thermal Analysis
DSC : Differential Scanning Calorimetry
Example
A chemical reaction known to release 1.78 kJ of heat
take place in a constant-volume calorimeter containing
0.1L of solution and the temperature rose by 3.65℃.
Next, 0.05L of 0.2M HCl and 0.05L of NaOH were
mixed in the same calorimeter and the temperature
rose by 1.26 ℃. What is the change in the external
energy?
如何計算系統的內能ΔU大小?
0 whereprocess olumeconstant vfor ,
C andonly work expansion for ,
C andonly work expansion for ,
gernalin ,
exp
exp
nonexpexp
=∆=∆
=∆−=+=∆
≠−=+=∆
++=∆
∫
VqU
PVPqwqU
PdVPqwqU
wwqU
exex
exex
為什麼無法直接計算內能U大小?
內能, U, 指的就是系統的所有能量包括所有分子的
動能、位能、鍵能及電子游離能等。準確的得知U
的大小是件非常困難的工作。
U 只跟系統的狀態有關,如 P、V、T,卻與反應
的過程無關。因此只要知道系統的起始狀態及終了
狀態就可以算出反應前後的內能變化,ΔU。
因此在實際的應用上,通常求ΔU比U容易多了。
U of an Ideal Gas System
While a system
has only ideal
gas molecules, U
contains only
kinetics energies.
3RTNonlinear Molecular
Ideal Gas
(5/2)RTLinear Molecular
Ideal Gas
(3/2)RTMonatomic
Molecular Ideal Gas
Internal
Energy U
How to Get U of Ideal Gases
Monatomic Molecular Ideal Gas
Linear MolecularIdeal Gas
Nonlinear MolecularIdeal Gas
State Function(狀態函數)
25℃
60℃
95℃
常見的狀態函數
Intensive Properties:
與系統的大小(質量)無關的物理量
壓力P、溫度T及密度d
莫爾體積Vn及莫爾位能Un
Extensive Properties:
與系統的大小(質量)有關的物理量
體積V、位能PE、質量M及內能U
Example
在恆溫25℃下的理想氣
體依下列兩種不同的步
驟膨脹500cm3至
1000cm3:
外在環境抽真空,請計
算ΔU, q及w
外在環境保持1Pa,請
計算ΔU, q及w
重新檢視First Law
ΔU = q + w
內能是狀態函數,因此只跟系統的起始狀態及
終了狀態有關,與路徑無關,因此記為ΔU
熱不是狀態函數,因此不同路徑有不同的q,因
此不可記為Δq,而記為 q
功不是狀態函數,因此不同路徑有不同的w,因
此不可記為Δq,而記為 w
Example
Suppose that 1 mol of ideal gas molecules at 292K and
3 atm expands from 8L to 20L and a final pressure of
1.2 atm by two different paths. Please calculate ΔU,
q and w for the two paths.
Path A is an isothermal, reversible expansion.
Path B has two parts. In step 1, the gas is cooled at constant
volume until its pressure has fallen to 1.2 atm. In step 2, it
is heated and allowed to expand against a constant pressure
of 1.2 atm until its volume is 20L and T=292K