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
PhysicsNTHU
MFTai-戴明鳳
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.
PhysicsNTHU
MFTai-戴明鳳
Principle (原理)
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)
1  dQ 
1  dU 
cV  
  

n  dT V n  dT V
1 dQ
U
V
cp  ( ) p  (
) p  p( ) p
n dT
T
T
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)
PhysicsNTHU
R  kNA = 8.31 J/mol-K = 氣體常數 (Gas constant)
MFTai-戴明鳳
Derive the  value of idea gas
Ideal gases pV = nRT  pdV + Vdp = nRdT
Constant Pressure dp = 0 (定壓時) dQ = dU + pdV = dU + nRdT
1  dQ 
1  dU 
1 d ( PV ) p
cp  
  
 
n  dT  p n  dT  p n dT
 cP = cv + R

1  dQ 
1 d (nRT ) p 1  dQ 
 
 
 
 R
n  dT  p n
dT
n  dT  p
 = cp/cv = (cV +R)/cv
c p  cv  R
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
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定壓比熱及定容比熱比值:   cp/cV ~ 7/5
MFTai-戴明鳳
(19 – 11)
Internal Energy of an ideal Gas
理想氣體的內能
Internal energy of an ideal gas
3nRT
Eint 
2
Consider a monatomic gas such as He, Ar, or Kr. In this case the internal energy
Eint of the gas is the sum of the translational kinetic energies of the contituent atoms
The average translational kinetic energy of a single atom is given by the equation:
3kT
K avg 
A gas sample of n moles contains N  nN A atoms. The internal
2
nN A 3kT 3nRT
energy of the gas Eint  NK avg 

2
2
The equation above it expressing the following important result:
The internal energy Eint of an ideal gas is a function of gas temperature
only; it does not depend on any other parameter.
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MFTai-戴明鳳
Molar specific heat CV at constant volume
Consider n moles of an ideal gas at pressure p and
temperature T . The gas volume is fixed at V .
These parameters define the initial state of the gas.
A small heat quantity Q is added from the reservoir
that changes the temperature to T  T and the
pressure to p  p and brings the system to its
final state. The heat Q  nCV T The constant
Eint  nCV T
CV 
3R
2
CV is called the molar specific heat at constant
volume. From the first law of thermodynamics
we have: Q  Eint  W . W  pV  0
Thus Q  Eint  nCV T  CV =
Eint
nT
3nRT
3nRT
3R
 Eint 
 CV 
2
2
2
We can write the internal energy of the
gas in the following form: Eint  nCV T
Eint 
 Eint  nCV T
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MFTai-戴明鳳
Molar specific heat C p at constant pressure
We assume that we add a heat amount Q to
the gas and change its temperature from T
to T  T and its volume from V to V  V
while keeping the pressure constant at p
The heat Q  nC p T The constant C p is called
molar specific heat at constant pressure. The
first law of thermodynamics gives: Q  W  Eint
 nC p T  pV  nCV T
Using the law of ideal gases pV  nRT we get:
pV  nRT  nC p T  nRT  nC p T
Thus: C p  CV  R
C p  CV  R
PhysicsNTHU
MFTai-戴明鳳
Degrees of freedom and molar specific heats
f =3
f =5
3R
agrees with the experimental
2
data of monatomic gases but fails for diatomic and polyatomic
gases. The reason is that for diatomic and polyatomic molecules
The equation CV 
have more complex motions that the simple translational motion
we assummed for monatomic gases. The formar can have rotational
motion about 2 orthogonal axes and also oscillatory motion about
f =6
the equilibrium position. To account for these effects Maxwell
introduced the theorem of equipartition of energy that states:
Every type of molecule has f degrees of freedom which defined
as independent ways in which the molecule can store energy.
Each degree of freedom has an average energy kT / 2
per molecule or RT / 2 per mole.
fR
2
f is equal to 3 for monatomic gases, f  5 for diatomic gases,
PhysicsNTHU
and f  6 for polyatomic gases.
The corresponding molar specific heat is: CV 
fR
CV 
2
MFTai-戴明鳳
Cv = 3/2R = 12.5 J/mol.K
Cp = 5/2R = 20.8 J/mol.K
CP 5R / 2
 

 1.67
CV 3R / 2

Theoretical values of CV,
CP, and  are in excellent
agreement for
monatomic gases.
PhysicsNTHU
MFTai-戴明鳳
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.
PhysicsNTHU
MFTai-戴明鳳

pV
i i  p f Vf

Adiabatic expansion of an ideal gas
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.
Such a process is indicated on the p - V diagram of fig.b.
by the red line. The gas starts at an initial pressure pi and
TV
i i
 1
initial volume Vi . The corresponding final parameters
 Tf V f
 1 are p and V . The process is described by the equation:
f
f

piVi  p f V f

Here the constant  
Cp
CV
Using the ideal gas law we can get the equation:
TV
i i
 1
 Tf V f
 1
 1
V
 T f  Ti i  1
Vf
If V f  Vi we have adiabatic expansion and T f  Ti
If V f  Vi we have adiabatic compression and T f  Ti
PhysicsNTHU
MFTai-戴明鳳
Adiabatic expansion dQ = 0
(絕熱膨脹)
For an ideal gas with n =1
pV  RT  pdV  Vdp  RdT
 dQ  0
Cv
dU  Cv dT 
( pdV  Vdp)
R
Cv
Cv  R
Vdp  
pdV
R
R
dp
dV
 
p
V
ln p   ln V  A

pV
i i  p f Vf

TV
i i
 1
 Tf V f
 1
PV   A '
 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.
PhysicsNTHU
MFTai-戴明鳳
Free expansion
In a free expansion a gas of initial volume Vi and initial
pressure pi is allowed to expand in an empty container
so that the final volume is V f and the final pressure p f
Ti  T f
piVi  p f V f
In a free expansion Q  0 because the gas container is insulated. Furthermore
since the expansion takes place in vacuum the net work W  0
The first low of thermodynamics predicts that Eint  0
Since the gas is assumed to be ideal there is no change in temperature
Ti  T f
Using the law of ideal gases we get the following equation
which connects the initial with the final state of the gas:
piVi  p f V f
(19 – 16)
PhysicsNTHU
MFTai-戴明鳳
Block Diagram of ClementDesorms’ Experiment
1
2
P1
T1 = To
V1
P2 = Po
T2
V2

1 1
(1) PV  P2V2
3
P3
T3 = To
V3 = V2

(2) P1V1 = P3V2
p0
B
p
A
h
(1) Adiabatic expansion
(2) Isothermal expansion
PhysicsNTHU
MFTai-戴明鳳
Clement-Desorms’
Configuration
克雷孟-德梭姆
實驗裝置
 球型玻璃瓶(spherical glassware)
 銅蓋(Cupper Tip, 可沿瓶口水平方向滑動
而移開或滑入原位)
 橡皮管 (rubber tube)
 玻璃氣閥 (glass valve)
 打氣球 (gas charging ball)
 U-型軟管(U-shape soft tube, 內裝測
氣體壓力的液體)
PhysicsNTHU
MFTai-戴明鳳
Clement-Desorms’ Experiment Configuration
克雷孟-德梭姆方法的實驗裝置
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
PhysicsNTHU
MFTai-戴明鳳
(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
p0
T3 = T0 (熱平衡)
測量壓力差高度h3，得 p3 = p0 + h3dg
B
A
(A) p1V1 = n1RT0
(T1 = T0)
h
p
= (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)
PhysicsNTHU
or  = h3/(h1 - h3)
(與理論值  = 7/5 = 1.4 比較)
MFTai-戴明鳳
 Value derived in Classical
Mechanics

1 1
p V  p2V2

h1 , h3  10 - 20 cm and p  1 g/cm 3
 p1V1  p3V2 
 1
1
p
p0  76 cmHg
h3 g 2
h1 g 2
(
)  1, (
)  1.
p0
p0

p3

po
p1  p0  h1 g
p3  p0  h3 g
( p0  h1 g )
 1
1
 ( p0  h3 g )
p0
h3 g 
h1 g  1
(1 
)  (1 
)
p0
p0
h3 g
h1 g
1  (  1)
 1 
p0
p0
h2

h1  h3
PhysicsNTHU
MFTai-戴明鳳
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 = nCvT for constant volume processes
– Q = nCpT 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
PhysicsNTHU
MFTai-戴明鳳
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 = nCvT, applies to all ideal
gases, not just monatomic ones.
PhysicsNTHU
MFTai-戴明鳳
Specific Heat of Monatomic Gases

At constant volume, Q = Eint = nCvT
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
CP 5R / 2
 

 1.67
CV 3R / 2

Theoretical values of CV, CP, and  are in excellent
PhysicsNTHU
agreement for monatomic gases or any ideal
gas.
MFTai-戴明鳳
PhysicsNTHU
MFTai-戴明鳳
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
PhysicsNTHU
MFTai-戴明鳳
例題
柴油引擎汽缸

Adiabatic
compression
process
 Vi 
 800 


Pf  Pi    1.00
  37.6 atm
 60.0 
 Vf 
PiVi PfVf

(for ideal gas)
Ti
Tf
1.40
PhysicsNTHU
MFTai-戴明鳳
Equipartition of Energy
for Complex Molecules

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
PhysicsNTHU
MFTai-戴明鳳
Equipartition of Energy

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
 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
PhysicsNTHU
specific heat.
MFTai-戴明鳳
Agreement with Experiment
-Molar specific heat of a diatomic gas (acts like a
monatomic gas) is a function of T.
at high T
CV = 7/2R
at about room T
Cv = 5/2R
at low T
Cv = 3/2R
PhysicsNTHU
MFTai-戴明鳳
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 energy level diagram of the


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.
rotational and vibrational states of
a diatomic molecule.
PhysicsNTHU
MFTai-戴明鳳
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)
PhysicsNTHU
 = W/Qi = 1 – To/Ti (Si = So, 理想熱機/ideal heat engine) MFTai-戴明鳳