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