lab. 12 - determine the value of air 實驗 12 ：空氣值的測定

PhysicsNTHUMFTai-戴明鳳

Lab. 12- Determine the value of air

實驗 12 ：空氣值的測定I. Object: Measure the ratio of air using Clement &

Desorms’ method.目的：以克雷孟和德梭姆的方法測量空氣的值

= cp/cv = 定壓比熱及定容比熱比值 ,-- 氣體動力學中一個很重要參數


Adiabatic process ( 絕熱過程 )Broilge Law: at constant T ， an ideal gas with the fixed particles

obeys PV = nRT =constant.But real gas isn’t a good heat conductor, it need time to reach a

new thermal equilibrium.As P and V change too rapidly, e.g., the propagation of sound,

the energies between the parts of gas can not exchange with each other at once.

So it is impossible for the isothermal process ( 等溫過程 ), the realize reaction shall be an adiabatic process.

Adiabatic process: 1. If gas is compressed (expanded), all work done by the

surrounding becomes (dissipate) the internal energy of gas, thus the P and T of gas will simultaneously increase (decrease).

2. The slope of p(V) curve in adiabatic process must steeper than in the isothermal process.


熱力學第一定律 ( 能量守恆 ) ：系統能量 U 之微分 dU(1st thermodynamical law: energy conservation law)

吸 / 放熱 + 作正功 (endothermic/exothermic reaction + do positive work): dU = dQ

+ dW

吸 / 放熱 + 作負功 (endothermic/exothermic reaction + do negative work): dU = dQ –

pdV

摩爾定容比熱 (molar heat capacity at constant volume)

摩爾定壓比熱 (molar heat capacity at constant pressure)

空氣在室溫 (T ~ 300 K) 可視為接近理想氣體 (Air at 300 K can be as idea

gas)

pV = NkT , pV = nRT

N = 粒子數 (Particle number)

k = 1.38 x 10-23 J/K = 波茲曼常數 (Boltzman constant)

n N/NA = 粒子摩爾數 (Molar no.)

NA = 6.02 x 1023 /mol = 亞佛加德常數 (Advogadro constant)

R kNA = 8.31 J/mol-K = 氣體常數 (Gas constant)

Principle ( 原理 )

VVV dT

dU

ndT

dQ

nc

11pppp T

Vp

T

U

dT

dQ

nc )()()(

1


Derive the value of idea gasIdeal gases pV = nRT pdV + Vdp = nRdTConstant Pressure dp = 0 ( 定壓時 ) dQ = dU + pdV = dU +

nRdT

能量均分 (Equipartition of Energy): kT/2 per degree of freedom

空氣主要為雙原子分子 N2 : 78%, O2 : 21% ，在室溫 (T ~ 300 K)

有 3 個自由度的平移， 2 個自由度的轉動，但不會振動。故每一分子共有 5 自由度，每自由度貢獻 kT/2 能量。總能量 Total = U ~ 5nRT/2

定容比熱 : cv ~ 5 R/ 2 定壓比熱 : cp = cv + R ~ 7 R/

2

定壓比熱及定容比熱比值 : cp/cV ~ 7/5

Rcc

RdT

dQ

ndT

nRTd

ndT

dQ

n

dT

PVd

ndT

dU

ndT

dQ

nc

vp

p

p

p

p

ppp

1)(11

)(111 cP = cv + R

= cp/cv = (cV +R)/cv


int

Consider a monatomic gas such as He, Ar, or Kr. In this case the internal energy

of the gas is the sum of the translational kinetic energies of the contituent atomsE

Internal energy of an ideal gas

int

The average translational kinetic energy of a single atom is given by the equation:

3 A gas sample of moles contains atoms. The internal

23 3

energy of the gas 2

avg A

Aavg

kTK n N nN

nN kT nRTE NK

2

The equation above it expressing the following important result:

intThe internal energy E of an ideal gas is a function of gas temperature

only; it does not depend on any other parameter.

int

3

2

nRTE

(19 – 11) Internal Energy of an ideal Gas理想氣體的內能


Consider moles of an ideal gas at pressure and

temperature . The gas volume is fixed at .

These parameters define the initial state of the gas.

A small he

VC

n p

T V

Molar specific heat at constant volume

at quantity is added from the reservoir

that changes the temperature to and the

pressure to and brings the system to its

final state. The heat The constant

is called the V

V

Q

T T

p p

Q nC T

C

molar

int

intint

int int

. From the first law of thermodynamics

we have: . 0

Thus =

3 3 3

2 2 2We can write the internal ener

V V

V

Q E W W p V

EQ E nC T C

n TnRT nR T R

E E C

specific heat at constant

volume

int

int

gy of the

gas in the following form: V

V

E nC T

E nC T

3

2V

RC

int VE nC T


We assume that we add a heat amount to

the gas and change its temperature from

to and its volume from to

while keeping the pressure constant at

pC

Q

T

T T V V V

Molar specific heat at constant pressure

int

The heat The constant is called

molar specific heat at constant pressure. The

first law of thermodynamics gives:

Using the law of ideal gases we get:

p p

p V

p

Q nC T C

Q W E

nC T p V nC T

pV nRT

p

Thus:

p p

p V

V nR T nC T nR T nC T

C C R

p VC C R


f =3

f =5

f =6

2V

fRC


Cv = 3/2R = 12.5 J/mol.K

Cp = 5/2R = 20.8 J/mol.K

Theoretical values of CV, CP, and are in excellent agreement for monatomic gases.

67.12/3

2/5

R

R

C

C

V

P


Molar Specific Heat for an Ideal gas Several processes can change the

temperature of an ideal gas. Since T is the same for each

process, Eint is also the same.

The heat is different for the different paths.

The heat associated with a particular change in temperature is not unique.

Specific heats are frequently defined at two processes:

– Constant-pressure specific heats cp

– Constant-volume specific heats cv

Using the number of moles, n, defines molar specific heats for these processes.


Consider the ideal gas in fig.a. The container is well

insulated. When the gas expands no heat is transferred

to or from the gas. This process is called adiabatic.

Adiabatic expansion of an ideal gas

Such a process is indicated on the - diagram of fig.b.

by the red line. The gas starts at an initial pressure and

initial volume . The corresponding final parameters

are and . The process

i

i

f f

p V

p

V

p V

11

1

1

is described by the equation:

Here the constant

Using the ideal gas law we can get the equation:

If we have adiab

ati

p

V

if i

i i f

i

i

f

f

i

f

f f

pC

C

VT T

V

V V

V p V

TV T V

c expansion and

If we have adiabatic compression and

f i

f i f i

T T

V V T T

i i f fpV p V

1 1i i f fTV T V


Adiabatic expansion dQ = 0 ( 絕熱膨脹 )

'A

A lnln

)(

0

PV

Vp

V

dV

p

dp

pdVR

RCVdp

R

C

VdppdVR

CdTCdU

dQ

RdTVdppdVRTpV

vv

vv

For an ideal gas with n =1

Here both A and A’ are constant values.The slope of p(V) curve in adiabatic process must steeper than in the

isothermal process.One can determine the value from the adiabatic process.

i i f fpV p V 1 1i i f fTV T V


In a free expansion a gas of initial volume and initial

pressure is allowed to expand in an empty container

so that the final volume is and the final pressure

i

i

f f

V

p

V p

Free expansion

int

In a free expansion 0 because the gas container is insulated. Furthermore

since the expansion takes place in vacuum the net work 0

The first low of thermodynamics predicts that 0

Since the gas

Q

W

E

is assumed to be ideal there is no change in temperature

Using the law of ideal gases we get the following equation

which connects the initial with the final state of the gas:

i f

i i f f

T T

pV p V

i i f fpV p V

i fT T

(19 – 16)


Block Diagram of Clement-Desorms’ Experiment

P1

T1 = To

V1

P2 = Po

T2

V2

P3

T3 = To

V3 = V2

1 2 3

(1)

2211 VPVP

(2) P1V1 = P3V2

hA

B

p

p0

(1) Adiabatic expansion(2) Isothermal expansion


① 球型玻璃瓶 (spherical glassware)② 銅蓋 (Cupper Tip, 可沿瓶口水平方向滑

動而移開或滑入原位 )③ 橡皮管 (rubber tube)④ 玻璃氣閥 (glass valve)⑤ 打氣球 (gas charging ball)⑥ U- 型軟管 (U-shape soft tube, 內裝

測氣體壓力的液體 )

Clement-Desorms’ Configuration克雷孟 - 德梭姆

實驗裝置


Initial condition of system:

at room temperature, T0 ~ 300 K

and under ambient atmosphere, p0 = 1 atm

( 系統起始條件 : 室溫時，一大氣壓力的靜態環境下 )

(1) 關 A 銅蓋閥，開 B 玻璃氣閥用打氣球將空氣打入球型玻璃容器中 ,

使系統壓力增加至 p1, 關玻璃氣閥。等幾分鐘 , 讓系統與外界熱平衡 ,

以 U 型開口壓力計測壓力差高度 h1 ，

可得 p1 = p0 + h1d·g

d: 壓力計液體比重 , g: 重力加速度 T1 = T0

Clement-Desorms’ Experiment Configuration克雷孟 - 德梭姆方法的實驗裝置


(2) 開 B 閥 , 讓氣體急速絕熱膨脹排出 , 使容器內外壓力相同 , 關B 閥 .

p2 = p0

p1V1 = p2V2

( 絕熱膨脹 , 外界熱量來不及進入 )

T2 < T1 = T0 ( 膨脹後溫度降低 )

(3) 等幾分鐘 , 讓外界熱量流入使達熱平衡 T3 = T0 ( 熱平衡 )

測量壓力差高度 h3 ，得 p3 = p0 + h3dg

(A) p1V1 = n1RT0 (T1 = T0)

= (n1/n3)p3V3 (T0 = T3)

~ p3V3 (n1 ~ n3)

(B) p1V1 = p0V2

(p2 = p0)

~ p0V3 (V2 ~ V3, 體積由大玻璃瓶控制 )

(A)/(B) p1-1 = p3

/p0 (p1, p3 代入 )

or = h3/(h1 - h3) ( 與理論值 = 7/5 = 1.4 比較 )

hA

B

p

p0


Value derived in Classical Mechanics

)1()1(

)(1

)(

0

31

0

1

300

110

303

101

311

2311

2211

p

gh

p

gh

ghpp

ghp

ghpp

ghpp

p

pp

VpVp

VpVp

o

31

2

0

3

0

1

2

0

32

0

1

0

331

1)1(1

.1)( ,1)(

cmHg 76

g/cm 1 and cm 20-10 ,

hh

h

p

gh

p

gh

p

gh

p

gh

p

phh


Molar Specific Heat Specific heats are frequently defined at two processes:

– Constant-pressure specific heats cp

– Constant-volume specific heats cv

Using the number of moles, n, defines molar specific heats for these processes.

Molar specific heats:

– Q = nCvT for constant volume processes

– Q = nCpT for constant pressure processes

Q (in a constant pressure process) must account for both the increase in internal energy and the transfer of energy out of the system by work.

Q(constant P) > Q(constant V) for given values of n and T


Ideal Monatomic Gas- contains only one atom per molecule

When energy is added to a monatomic gas in a container

with a fixed volume, all of the energy goes into increasing

the translational kinetic energy and temperature of gas.

– There is no other way to store energy in such a gas.

Eint = 3/2nRT, in general, the internal energy of an ideal gas

is a function of T only.

The exact relationship depends on the type of gas

At constant volume, Q = Eint = nCvT, applies to all ideal

gases, not just monatomic ones.


At constant volume, Q = Eint = nCvT

applies to all ideal gases, not just monatomic ones.

Cv = 3/2R = 12.5 J/mol.K

– Be in good agreement with experimental results for monatomic gases.

In a constant pressure process, Eint = Q + W

Cp – Cv = R Cp = 5/2R = 20.8 J/mol.K

– This also applies to any ideal gas Define a Ratio of Molar Specific Heats

Theoretical values of CV, CP, and are in excellent agreement for monatomic gases or any ideal gas.

Specific Heat of Monatomic Gases

67.12/3

2/5

R

R

C

C

V

P


Adiabatic Process for an Ideal Gas

At any time during the process, PV = nRT is valid

– None of the variables alone are constant

– Combinations of the variables may be constant

The pressure and volume of an ideal gas at any time during an adiabatic process are related by PV = constant

All three variables in the ideal gas law (P, V, T) can change during an adiabatic process


例題柴油引擎汽缸

gas) ideal (for

atm 6.370.60

80000.1

40.1

f

ff

i

ii

f

iif

T

VP

T

VP

V

VPP

Adiabatic compression

process


Other contributions to internal energy must be taken into account:

1. Translational motion of the center of mass: 3 degrees of freedom

2. Rotational motion about the various axes: 2 degrees of freedom for x and z axes

To neglect the rotation around the y axis since it is negligible compared to the x and z axes.

Eint = 5/2nRT

CV = 5/2R = 20.8 J/mol.K

Cp = 7/2R = 29.1 J/mol.K

predicts that = 7/5 = 1.40

Equipartition of Energyfor Complex Molecules


The molecule can also vibrate

There is kinetic energy and potential energy associated with the vibrations

This adds two more degrees of freedom

Eint = 7/2nRT, Cv = 7/2R = 29.1 J/mol.K

= 1.29

Equipartition of Energy

This doesn’t agree well with experimental results.

A wide range of temperature needs to be included.

For molecules with more than two atoms, the vibrations are more complex. The number of degrees of freedom is larger

The more degrees of freedom available to a molecule, the more “ways” there are to store energy, and the higher molar specific heat.


Agreement with Experiment-Molar specific heat of a diatomic gas (acts like a

monatomic gas) is a function of T.

at low TCv = 3/2R

at about room TCv = 5/2R

at high TCV = 7/2R


Quantization of Energy Classical mechanics is not sufficient to explain the

results of the various molar specific heats, we must use some quantum mechanics.

The rotational and vibrational energies of a molecule are quantized

The vibrational states are separated by larger energy gaps than are rotational states.

1. At low T, the energy gained during collisions is generally not enough to raise it to the first excited state of either rotation or vibration.

Even though rotation and vibration are classically allowed, they do not occur.

The energy level diagram of the rotational and vibrational states of

a diatomic molecule.


DVD: 自然界的引擎(Engine of Nature)(MU46)

熱力學第一定律 : 能量守恆定律 dU = dQ – dW ( 系統增加之熱能 = 吸收熱量 – 對外作功

(work)) dU = TdS – pdV (S = dQ/T 亂度 (entropy))

熱機 ( 熱引擎 )(heat engine)

高溫 (Tinput) 吸熱 Qi, Si = Qi/Ti (Ti = constant)

低溫 (Toutput) 放熱 Qo, So = Qo/To

對外作功 W = Qi – Q0

效率 (efficiency) e W/Qi = 1 – Qo/Qi

卡諾循環 (Carnot cycle)/ 卡諾熱機 (Carnot engine)

1. 高溫等溫膨脹 (isothermal expansion) Ti = constant, 吸熱 Qi

2. 絕熱膨脹 (adiabatic expansion) dQ = 0 ( 絕熱 ), Ti To ( 降溫 )

3. 低溫等溫壓縮 (isothermal compression), To = constant, 放熱 Q0

4. 絕熱壓縮 (adiabatioc compressiion) dQ = 0, To Ti ( 昇溫 )

= W/Qi = 1 – To/Ti (Si = So, 理想熱機 /ideal heat engine)

lab. 12 - determine the value of air 實驗 12 ：空氣 值的測定

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lab. 12 - determine the value of air 實驗 12 ：空氣值的測定