# 普通化學文本分析

In document 國 立 中 央 大 學 (Page 101-111)

## 第 4 章 研究結果分析

### 4.2.4 普通化學文本分析

A09、A22、A23、A26、A34、A58、A59、B01、B02、B04、B06、B07、B08、B10、

B12、D09、G07、G09、G10、I 06、I 07、I 15、K17、L12、L13、L14、M06、M12、

M19、M20、Q25、Q26、Q27、Q28、Q35、Q37 詳細校系編號對照請見表 3.2.18

( u) F ma m

t

= = ∆

Where F represents force , a represents the acceleration , ∆ represents a change in u velocity , and ∆ represents a given length of t time.

Since we assume that the particle has constant mass , we can write

( )

u mu

F m

t t

∆ ∆

= =

∆ ∆

5.6 The Kinetic Molecular Theory of Gases

p.157

5th，p.161

2 32

2

( ) 4 2

2

B mu k T

B

f u m u e

π k T π

 

=  

 

Where

5.6 The Kinetic Molecular Theory of Gases

Maxwell-Boltzmann

distribution(馬克斯威爾-波茲 曼常態分佈曲線)

p.161

u = velocity in m/s

m = mass of a gas particle in kg

kB= Boltzmann’s constant=1.38066 10× 23 J/K T = temperatire in K

The product of f u du represents the fraction ( ) of gas molecules with velocities between u and u+du , where du represents an infinitesimal velocity increment.

p.161

5th，p.308

8.5 Titrations and pH Curves 利用 Titration 滴定法製作 pH 酸鹼值曲線圖。製圖過程 以實驗為基礎，國中數學教授 的坐標平面來描點坐標，當中 並未用微積分的口吻深入探 討曲線的行為。

p.308

Infinite-Step Expansion. If one continues to increase the number of steps , the magnitude of

wn (for an n-step process)

1 n

n i i

i

w P V

=

=

### ∑

Continues to increase

10.2 The Isothermal Expansion and Compression of an Ideal Gas

p.408

Now we consider the limiting case – a process in which Pex is changed by infinitesimally small increments . This case corresponds to use of an infinite number of weights , each differing from the previous one by an infinitesimally small mass . Under these conditions the successive volume changes become

infinitesimally small ( V)d , and the process

10.2 The Isothermal Expansion and Compression of an Ideal Gas

p.408

requires an infinite number of steps . The mathematical operation needed to sum the steps in this instance is the integral

2

1 V V ex

Work =

### ∫

P dV

The diagram corresponding to this process is given in Fig. 10.8.

5th，p.409

10.2 The Isothermal Expansion and Compression of an Ideal Gas

Vi Vi Vi−1

∆ = − ，將其視為膨脹 的體積增量，而 d 視為活塞i 頭移動的對應距離，若 P 代i 表壓力 P 對應第 i 個增量的 值，則活塞頭所受的力為 P i 與活塞頭面積的乘積

2

P riπ 。因此，在第 i 個增量 所作的功為

2

2 2

Vi V

i i i i i

P r d P r P

π π r

π

∆ 

=  = ∆

 

，所以，

1

V

n

i i

i

W P

=

### ∑

∆ 而當分割的範數趨近於零 時，此近似值更佳，故

2

1 V

V P V W =

### ∫

d

p.409

Since PexPgas=P in the reversible expansion , by use of the ideal gas law ,

ex V

PP=nRT

and 2

1 V

rev V

Total work = V

V nRT d w = w =

### ∫

Since n and T are held constant in this experiment ,

10.2 The Isothermal Expansion and Compression of an Ideal Gas

V

P=nRT，再將前一積分是作 變換推導，計算過程中使用到

ln b x a dt

=

t

p.410

2

1 V

rev V

V V w =nRT

### ∫

d

2 1 2

1

(ln V ln V ) ln V

nRT nRT V 

= − =  

  In this specific experiment V2 =4V1 . Therefore ,

rev ln 4 1.4 w =nRT = nRT

And since P1V1=nRT , wrev =1.4 VP1 1 For this particular expansion .

, 0

a b> 。此觀念超過高三下 選修數學 II 的教材內容，並 且需先學習

ln 1 0

d x x

dx = x ， > 。然而 ln 為自然對數，亦非高中數 學有提及的內容。

2

max rev

1

ln V w w nRT V 

= =  

 

2 rev

1

ln V w nRT V 

= −  

 

2

rev rev

1

ln V q w nRT V 

= − =  

 

2

rev rev

1

ln V w q nRT V 

= − = −  

  Or in terms of pressures ,

1

rev rev

2

ln P

w q nRT

P

 

= − = −  

 

10.2 The Isothermal Expansion and Compression of an Ideal Gas

p.449 中也有用到此式來計 算例題。

p.410

p.412 p.449

Infinite-Step Compression . Notice that in compressing the gas isothermally , as the number of steps increases , the work required to compress the gas decreases . If we compress the gas in an infinite number of steps (in which , at all times PexP) , the work required is

10.2 The Isothermal Expansion and Compression of an Ideal Gas

p.411

2

1

V 2

V 1

1

1 1 1

' V ln V V ln V 1.4 V

4V

w P d nRT

nRT P

 

= =  

 

 

=  =

 

### ∫

(In this specific experiment V2 =4V1 .)

B ln S =k Where

kB = Boltzmann’s constant , the gas constant per molecule (R N/ A)

Ω = the number of microstates corresponding to a given state (including both position and energy)

10.3 The Definition of Entropy 運用到自然對數 ln，這些式 子單純地出現 ln，並沒有牽 扯到其他觀念。關於 entropy 有許多 ln 的計算，為避免版 面過長不再贅述

p.414

2

1

ln V S nR V 

∆ =  

  with

2 rev

1

ln V q nRT V 

=  

 

10.3 The Definition of Entropy 運用到自然對數 ln，這些式 子單純地出現 ln，並沒有牽 扯到其他觀念。

p.416

Temperature Dependence of Entropy

For an isothermal process we have seen that the change in entropy is defined by the relationship

qrev

S T

∆ =

We can calculate ∆ for a change in S temperature from T1 to T2 by summing infinitesimal increments in entropy at each temperature T：

dqrev

dS= T

Using integration , we have

2 2

1 2

1 1

rev

V V

T T

T T

S dS dq

T

∆ =

=

### ∫

If the process is carried out at constant pressure , then

rev P

dq =nC dT

for n moles of substance . Thus

10.4 Entropy and Physical Changes

S T

∆ = ，當溫度從 T1T2 的增量十分微小 時，可以利用積分是得到

2 2

1 2

1 1

rev

V V

T T

T T

S dS dq

T

∆ =

=

，而

2 2

1 2

1 1

rev

V V

T T

T P T

dq dT

S nC

T T

=

=

1 2

2

V V

1 P ln S nC T

T

 

⇒ ∆ =  

 

p.416

2 2

1 2

1 1

rev

V V

T T

T P T

dq dT

S nC

T T

=

=

### ∫

assuming that C is constant between P T and 1 T . Performing the integration gives 2

1 2

2

V V

1 P ln S nC T

T

 

∆ =  

 

Similarly , for a process carried out at constant volume

1 2

2

V V V

1

ln T

S nC

T

 

∆ =  

 

1 2

2

V V V

1

ln T

S nC

T

 

∆ =  

 

(1)

J 373 (2.00 mol) 75.3 ln

K mol 323

S    

∆ =    

   

=21.7 J/K

(3)

J 423 (2.00 mol) 36.4 ln

K mol 373

S    

∆ =    

   

=9.16 J/K

10.4 Entropy and Physical Changes

100 C° 、 50 C° 與 150 C° 的 絕對溫度值，但這裡並沒有詳 細說明計算 373

ln 323

 

 

  與 ln 423

373

 

 

  的過程，而這裡計 算 373

ln 323

 

 

 與 423 ln 373

 

 

  的 過程應屬於對數計算的範疇。

p.418

1 2

2

V V

1

ln T

S nC

T

 

∆ =  

  where C is CP or C V

10.8 Entropy Changes in Chemical Reactions

p.427

ln( ) G=G°+RT P

reactants products

G G G

∆ =

### ∑

(10.7)

ln( ) G G° RT Q

∆ = ∆ +

10.10 The Dependence of Free Energy on Pressure

10.11 Free Energy and Equilibrium

p.434

~ p.441 p.475

Concertration 當中也有利用 到 ∆G= ∆G°+RT ln( )Q 來 作計算。

2

2

ln( ) H S

K RT R

° °

−∆ ∆

= +

and

1

1

ln( ) H S

K RT R

° °

−∆ ∆

= +

Subtracting the second equation from the first gives the combined equation：

2

1 2 1

1 1 ln(K ) H

K R T T

° 

=−∆  − 

 

This is called the van’s Hoff equation after the Dutch chemist Jacobus van’s Hoff .

10.11 Free Energy and Equilibrium

p.442

Recall that the energy of an ideal gas depends only on its temperature：E=nC TV

So for an adiabatic process ,

ext V V

E w P nC T

∆ = = − ∆ = ∆

For an infinitesimal adiabatic change ,

ext V V

dE= −P d =nC dT

Assume that the adiabatic expansion or

cpmpression is carried out reversibly . That is , Pext is only infinitesimally different from Pgas

ext gas

(PP ) .

Then ext gas V P =P = nRT

Thus , for a reversible , adiabatic expansion – compression , we have

V ext V gas V V

V dE=nC dT = −P d = −P d = −nRT d

and V V V

nRT d nC dT

− =

10.14 Adiabatic Process 此小節主要在講的是絕熱過 程，此段課文利用理想氣體方 程式為基礎，考慮能量的微小 變化，進而延伸到積分關係 式。

ext V V

E w P nC T

∆ = = − ∆ = ∆

ext V V

dE= −P d =nC dT

ext gas

V P =P =nRT

CV

T dT 轉變為

V V Rd

− ，進而推導到

2 2

1 1

V

V V

1 1

V V

T

C T dT R d

T = −

### ∫ ∫

p.447

which can be rearranged to V V V

C R

dT d

T = − We can derive an expression for a reversible , adiabatic change from V to 1 V and from 2 T 1 to T by summing (integrating) the 2

infinitesimal changes required：

2 2

1 1

V

V V

1 1

V V

T

C T dT R d

T = −

### ∫ ∫

where C is assumed to be independent of V temperature over the interval T to 1 T . 2 Evaluating the integrals gives

2 2 1

V

1 1 2

V V

ln ln ln

V V

C T R R

T

     

= − =

     

     

Taking the antilog of each side we have

V

2 1

1 2

V V

C R

T T

   

  = 

   

Since CP =CV+ , we can write R

V ( V)

2 1

1 2

V V

C CP C

T T

   

  = 

   

or V

( 1) ( 1)

2 1 1

1 2 2

V V

V V

CP

T C

T

γ

     

= =

     

      where

V

CP

γ =C Thus

-1

2 1

-1

1 2

V V T T

γ

= γ or T1V1γ1=T2V2γ1

Using the ideal gas law we can also express this result in terms of pressure . Since in this case ,

1 1 2 2

1 2

V V

P P

T = T then

-1

2 2 2 1

-1

1 1 1 2

V V V V T P

T P

γ

= = γ and P1V1γ =P2V2γ

10.14 Adiabatic Process 這段課文裡繼續作前面積分 的計算，計算的過程中運用到 的是對數的基本觀念與簡單 的四則運算；其中有提及反對 數，但是指對數觀念於高一數 學程度即可理解，

p.447 p.448

11 章為止，第 12 章以後為下學期之授課範圍，由於本篇論文的研究目的為大一上學 期必修課程與高中微積分的關聯性，因此文本分析作到第 11 章為止。

In document 國 立 中 央 大 學 (Page 101-111)