MODULE 2

CHE 216
Lecture 8A:
1st
Law of
Thermodynamics
Part 4: Joule-Thomson Effect
D E PA R T M EN T O F C H E M IC A L E N G IN E E R ING
U N IVER SIT Y O F SA N TO TO MA S
M A N IL A , PH IL IPP IN E S

4/30/2020 UST - DEPARTMENT OF CHEMICAL ENGINEERING 1

Background
▪ 1852

▪ Conducted an experiment in
which they pumped gas at a James Prescott William Thomson
steady rate through a lead pipe Joule (Lord Kelvin)
that was cinched to create a
construction.

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 Porous Plug /
Throttle
Apparatus

a) at beginning of experiment,
b) after finishing the experiment.

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Experiment
Starting conditions:

▪Adiabatic, q = 0

▪Gas on high-pressure side, pi at

Ti, occupying volume Vi

▪Gas on low-pressure side, pf at

Tf, occupying volume Vf
▪Upstream and downstream
pressures act as “pistons”
which compress the gas

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Observation
▪When a real gas at a certain pressure expands
adiabatically through a porous plug or a fine hole into
a region of low pressure:

→It is accompanied by cooling

→Hydrogen and Helium gets warmed up.

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Analysis
Isothermal Compression

Gas on left is compressed isothermally by the

upstream piston (pressure is pi and volume Vi becomes
0), so the work is:
wL = -Pi (0 - Vi) = PiVi
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Analysis
Isothermal Expansion

Gas expands isothermally on right of throttle (maybe

different T) against Pf, with work:
wR = - Pf (Vf - 0) = -Pf Vf
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 Initial Analysis
▪ Total work done: w = wR + wL
thus, w = PiVi – PfVf

▪Since the system is under adiabatic conditions (q = 0;

 U = w ):
 U = Uf –Ui = PiVi – PfVf
▪ Rearranging the terms:
Uf + PfVf = Ui + PiVi
(isoenthalpic) Hf = Hi
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The Joule-Thomson Coefficient
What is measured experimentally is:
𝑇2 − 𝑇1
𝜇𝐽𝑇 = lim
∆𝑝 → 0 𝑃2 − 𝑃1

which becomes, taking into account that the

process is isoenthalpic:

∂𝑇
𝜇𝐽𝑇 =
∂𝑃 𝐻
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Implications of 𝜇𝐽𝑇

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Implications of 𝜇𝐽𝑇 Coefficients
▪ Real gases have non-zero J-T coefficients: depends
on gas, pressure, attractive and repulsive
intermolecular interactions & temperature:

∂𝑇
𝜇𝐽𝑇 =
∂𝑃 𝐻
H1 = H2
P1 > P2
T1 ? P2
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 Implications of 𝜇𝐽𝑇 Coefficients
▪ Real gases have non-zero 𝜇𝐽𝑇 coefficients: depends on
gas, pressure, attractive and repulsive intermolecular
interactions & temperature:

∂𝑇
𝜇𝐽𝑇 =
∂𝑃 𝐻
𝑇2 − 𝑇1
𝜇𝐽𝑇 =
For a throttling process (P2 < P1): 𝑃2 − 𝑃1

µJT > 0: is T2 < T1 Write as P1-P2 because this will give us

positive quantity (just for explanation):
µJT < 0: is T2 > T1 𝑇2 = 𝑇1 − 𝜇𝐽𝑇 (𝑃1 − 𝑃2)
µJT = 0: Ideal Gas +
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4/30/2020 13

Example -1
When a Freon is used in
refrigeration was expanded
adiabatically from an internal
pressure of 32 atm and 0 oC to
a final pressure of 1.00 atm,
the temperature fell by 22 K.
Calculate the Joule – Thomson ∂𝑇 ∆𝑇
Coefficient at 0 oC assuming 𝜇𝐽𝑇 = ≈
that it remains constant over ∂𝑃 𝐻 ∆𝑃
this temperature.

−22 𝐾
𝜇𝐽𝑇 = = 0.71 𝐾𝑎𝑡𝑚−1
1−32 𝑎𝑡𝑚

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13
Summary
▪Joule-Thompson expansion is considered an adiabatic process.
▪ In the Joule Thomson experiment, the enthalpy of the gas remains
constant.
▪The Joule-Thomson coefficient can be calculated as the ratio of change
in temperature to change in pressure:
μJT​ ​= δT/δP
▪the Joule-Thompson coefficient can then be used to predict the
direction of temperature change based on the following criteria:
▪ μJT ​= 0 - no temperature change, ideal gases
▪ μJT​ < 0 - gas heats on expansion
▪ μJT​ > 0 - gas cools on expansion

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 MODULE 2
CHE 216
Lecture 8B:
1st
Law of
Thermodynamics
Part 4: Joule-Thomson Effect
D E PA R T M EN T O F C H E M IC A L E N G IN E E R ING
U N IVER SIT Y O F SA N TO TO MA S
M A N IL A , PH IL IPP IN E S

4/30/2020 UST - DEPARTMENT OF CHEMICAL ENGINEERING 15

Further Analysis
Consider enthalpy as a function of pressure and
temperature: H (P,T).
∂𝐻 ∂𝐻
𝑑𝐻 = 𝑇 dP + ∂𝑇 𝑃 𝑑𝑇
∂𝑃
∂𝐻
𝑅𝑒𝑐𝑎𝑙𝑙: 𝑃 = 𝐶𝑝
∂𝑇

∂𝐻
𝑑𝐻 = 𝑇 dP + 𝐶𝑝 𝑑𝑇
∂𝑃
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 Further Analysis
∂𝐻
𝑑𝐻 = 𝑇 dP + 𝐶𝑝 𝑑𝑇
∂𝑃
𝑑𝐻 = 0 ∂ 𝐻
0= 𝑇 dP + 𝐶𝑝 𝑑𝑇
∂𝑃
1 ∂𝐻
0= 𝑇 dP + 𝑑𝑇
𝐶𝑝 ∂𝑃
1 ∂𝐻 𝑑𝑇
0= 𝑇 +
𝐶𝑝 ∂𝑃 𝑑𝑃
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 Further Analysis
1 ∂𝐻 𝑑𝑇
0= 𝑇 +
𝐶𝑝 ∂𝑃 𝑑𝑃
1 ∂𝐻 𝑑𝑇
− 𝑇=
𝐶𝑝 ∂𝑃 𝑑𝑃

Joule-Thomson 1 ∂𝐻 𝑑𝑇 𝜇𝐽𝑇
Coefficient − 𝐶𝑝 ∂𝑃 𝑇=
𝑑𝑃 𝐻
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 Further Analysis
1 ∂𝐻
− 𝑇= 𝜇𝐽𝑇
𝐶𝑝 ∂𝑃

▪ Thermodynamic Relationship:

∂𝐻 ∂𝑉
𝑇= V −T
∂𝑃 ∂𝑇 𝑃

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 Further Analysis
1 ∂𝐻
− 𝑇= 𝜇𝐽𝑇
𝐶𝑝 ∂𝑃
Substitute:
∂𝐻 ∂𝑉
𝑇= V −T 1 ∂𝑉
∂𝑃 ∂𝑇 𝑃 − V −T = 𝜇𝐽𝑇
𝐶𝑝 ∂𝑇 𝑃

∂𝑉
T −𝑉
∂𝑇 𝑃
= 𝜇𝐽𝑇
𝐶𝑝
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 ∂𝐻
= −𝜇𝐽𝑇 𝐶𝑝,𝑚
∂𝑃
Example -2 𝑇

Given that 𝜇 = 0.25 𝐾 𝑎𝑡𝑚−1 ∂𝐻

= −(0.25 𝐾𝑎𝑡𝑚−1 )(29.125 𝐽𝐾 −1 𝑚𝑜𝑙 −1 )
for nitrogen, calculate the ∂𝑃 𝑇
value of its isothermal Joule-
Thomson coefficient. ∂𝐻
= −7.28 𝐽 𝑎𝑡𝑚−1 𝑚𝑜𝑙 −1
Calculate the energy that must ∂𝑃 𝑇
be supplied as heat to
maintain a constant
∆𝐻 /𝑛
temperature when 15.0 moles = −7.28 𝐽 𝑎𝑡𝑚−1 𝑚𝑜𝑙 −1
∆𝑃 𝑇
of N2 flows through a throttle
in an isothermal Joule – ∆𝐻 = −(7.28 𝐽 𝑎𝑡𝑚−1 𝑚𝑜𝑙 −1 ) n ∆𝑃
Thomson experiment and the
pressure drop is 75 atm. ∆𝐻 = − 7.28 𝐽 𝑎𝑡𝑚−1 𝑚𝑜𝑙 −1 (15.0 mol) (−75 atm)
Assume the Cp,m for N2 is
29.125 J K−1 mol−1. ∆𝐻 = 8190 J
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 Why is 𝜇𝐽𝑇 = 0 for Ideal
Gases ?
∂𝑉
T ∂𝑇 −𝑉
𝑃
= 𝜇𝐽𝑇
𝐶𝑝
▪ Rearrange the Equation:
1 ∂𝑉
T − 𝑉 = 𝜇𝐽𝑇
𝐶𝑝 ∂𝑇 𝑃

▪ For One mole of Ideal Gas: PV = RT

∂𝑉 𝑅
=
∂𝑇 𝑃

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 Why is 𝜇𝐽𝑇 = 0 for Ideal
Gases?
▪ Substituting to the equation:

∂𝑉 𝑅 1 𝑅𝑇
= − 𝑉 = 𝜇𝐽𝑇
∂𝑇 𝑃
𝑃 𝐶𝑝 𝑃

𝑅𝑇 1 𝑅𝑇 𝑅𝑇
𝑉= − = 𝜇𝐽𝑇
𝑃 𝐶𝑝 𝑃 𝑃
Joule-Thomson
Coefficient for an 𝟎 = 𝝁𝑱𝑻
Ideal Gas
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 Implications of 𝜇𝐽𝑇 Coefficients
▪ Real gases have non-zero 𝜇𝐽𝑇 coefficients: depends on
gas, pressure, attractive and repulsive intermolecular
interactions & temperature:

∂𝑇
𝜇𝐽𝑇 =
∂𝑃 𝐻 𝑇2 − 𝑇1
𝜇𝐽𝑇 =
For a throttling process (P2 < P1): 𝑃2 − 𝑃1

µJT > 0: is T2 < T1 𝑇2 = 𝑇1 − 0 (𝑃1 − 𝑃2)

µJT < 0: is T2 > T1
µJT = 0: Ideal Gas 𝑇2 = 𝑇1
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Summary
▪Consider that H (P,T) we were able to derive this formula:

1 ∂𝐻
− 𝑇= 𝜇𝐽𝑇
𝐶𝑝 ∂𝑃

▪Using a thermodynamic relationship we were able to show

that:
1 ∂𝑉
T − 𝑉 = 𝜇𝐽𝑇
𝐶𝑝 ∂𝑇 𝑃

▪Joule Thomson coefficient for an ideal gas is 0

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 MODULE 2
CHE 216
Lecture 8C:
1st
Law of
Thermodynamics
Part 4: Joule-Thomson Effect
D E PA R T M EN T O F C H E M IC A L E N G IN E E R ING
U N IVER SIT Y O F SA N TO TO MA S
M A N IL A , PH IL IPP IN E S

4/30/2020 UST - DEPARTMENT OF CHEMICAL ENGINEERING 26

The 𝜇𝐽𝑇 Coefficient for a van
der Waals gas
A model for Real gas:
𝑎𝑛2
𝑃+ 2 𝑉 − 𝑛𝑏 = 𝑛𝑅𝑇
𝑉

The Joule-Thomson Coefficient becomes:

1 2𝑎
𝜇𝐽𝑇 = −𝑏
𝐶𝑝 𝑅𝑇

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 Deriving the 𝜇𝐽𝑇 Coefficient
for a van der Waals gas
𝜕𝑇
By Definition: 𝜇𝐽𝑇 =
𝜕𝑃 𝐻
Using Cyclic Rule: 𝜕𝑇 𝜕𝑃 𝜕𝐻 𝜕𝑇
∂𝑦 ∂𝑥 ∂𝑧 = = −1
= −1 𝜕𝑃 𝐻
𝜕𝐻 𝑇
𝜕𝑇 𝑃
𝜕𝑃 𝐻
∂𝑥 𝑧 ∂𝑧 𝑦 ∂𝑦 𝑥
𝜕𝑇 1
=−
𝜕𝑃 𝜕𝑃 𝜕𝐻
𝐻
𝜕𝐻 𝑇 𝜕𝑇 𝑃
𝜕𝑇 1
=−
𝜕𝑃 𝜕𝑃
𝐻 𝐶𝑃
𝜕𝐻 𝑇
𝜕𝐻
𝜕𝑇 𝜕𝑃 𝑇
=−
𝜕𝑃 𝐻
𝐶𝑃
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 Deriving the 𝜇𝐽𝑇 Coefficient
for a van der Waals gas
∂𝐻 ∂𝑉
Thermodynamic Relationship: 𝑇= V −T
∂𝑃 ∂𝑇 𝑃
Reciprocal Rule:
∂𝑥 1 ∂𝐻 1
𝑧 = 𝑇= V −T
∂𝑇
∂𝑦 ∂𝑦 ∂𝑃
∂𝑥 𝑧 ∂𝑉 𝑃

Van der Waals Equation: 𝑎𝑛2

𝑃+ 2 𝑉 − 𝑛𝑏 = 𝑛𝑅𝑇
𝑉

1 𝑎𝑛2
𝑇= 𝑃+ 2 𝑉 − 𝑛𝑏
𝑛𝑅 𝑉

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 Deriving the 𝜇𝐽𝑇 Coefficient
for a van der Waals gas
Differentiate w.r.t V at 1 𝑎𝑛2
𝑇= 𝑃+ 2 𝑉 − 𝑛𝑏
constant P 𝑛𝑅 𝑉
𝑑 𝑓 + 𝑔 = 𝑑𝑓 + 𝑑𝑔 ∂𝑇 1 2𝑎𝑛2 𝑎𝑛2
𝑑 𝑓 ∙ 𝑔 = 𝑓 ∙ 𝑑𝑔 + 𝑔 ∙ 𝑑𝑓 = − 3 𝑉 − 𝑛𝑏 + 𝑃 +
∂𝑉 𝑛𝑅 𝑉 𝑉
𝑑𝑥 𝑛 𝑑(𝑥) 𝑃
= 𝑛𝑥 𝑛−1 ; =1
𝑑𝑥 𝑑𝑥
∂𝑇 1 2𝑎𝑛2 2𝑎𝑛3 𝑏 𝑎𝑛2
= − 2 + + 𝑃+
∂𝑉 𝑃 𝑛𝑅 𝑉 𝑉3 𝑉
Substitute the van der Waals
equation for the last term: ∂𝑇 1 2𝑎𝑛2 2𝑎𝑛3 𝑏 𝑛𝑅𝑇
= − 2 + 3 +
𝑎𝑛2
∂𝑉 𝑃 𝑛𝑅 𝑉 𝑉 𝑉−𝑛𝑏
𝑃+ 𝑉 − 𝑛𝑏 = 𝑛𝑅𝑇
𝑉
𝑎𝑛2 𝑛𝑅𝑇
𝑃+ =
𝑉 𝑉 − 𝑛𝑏

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 Deriving the 𝜇𝐽𝑇 Coefficient
for a van der Waals gas
The dependence of ∂𝐻 ∂𝐻 ∂𝐻
enthalpy (H) on ≅ +
constants “a” and “b” ∂𝑃 𝑇
∂𝑃 𝑇, 𝑎→0
∂𝑃 𝑇,𝑏→0
are additive:

Evaluate the following ∂𝑇 1 2𝑎𝑛2 2𝑎𝑛3 𝑏 𝑛𝑅𝑇

= − 2 + 3 +
∂𝐻 ∂𝐻
∂𝑉 𝑃 𝑛𝑅 𝑉 𝑉 𝑉−𝑛𝑏
and
∂𝑃 𝑇, 𝑎→0
∂𝑃 𝑇,𝑏→0

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 Deriving the 𝜇𝐽𝑇 Coefficient
for a van der Waals gas
Evaluate the equation ∂𝑇 1 2𝑎𝑛2 2𝑎𝑛3 𝑏 𝑛𝑅𝑇
= − 2 + +
as a → 0 : ∂𝑉 𝑃 𝑛𝑅 𝑉 𝑉3 𝑉−𝑛𝑏

∂𝑇 1 𝑛𝑅𝑇 T
= =
∂𝑉 𝑃
𝑛𝑅 𝑉 − 𝑛𝑏 V − nb

∂𝐻 1
Evaluate the ∂𝐻 𝑇= V −T
∂𝑃 ∂𝑇
∂𝑃 𝑇, 𝑎→0 ∂𝑉 𝑃
Using the following ∂𝐻 1
derived relationship: 𝑇= V −T
T
= V − 𝑉 − 𝑛𝑏
∂𝑃
∂𝐻 1 V − nb
𝑇= V −T
∂𝑃 ∂𝑇 ∂𝐻
∂𝑉 𝑃
∂𝑃 𝑇= V − V + nb = nb
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 Deriving the 𝜇𝐽𝑇 Coefficient
for a van der Waals gas
Evaluate the equation ∂𝑇 1 2𝑎𝑛2 2𝑎𝑛3 𝑏 𝑛𝑅𝑇
= − 2 + 3 +
as b → 0 : ∂𝑉 𝑃 𝑛𝑅 𝑉 𝑉 𝑉−𝑛𝑏

∂𝑇 1 2𝑎𝑛2 𝑛𝑅𝑇
= − 2 +
∂𝑉 𝑃
𝑛𝑅 𝑉 𝑉

Substitute to the derived

∂𝐻 1
thermodynamic relationship: 𝑇= V −T
∂𝑃 1 2𝑎𝑛2 𝑛𝑅𝑇
∂𝐻 1 − 2 +
𝑇= V −T 𝑛𝑅 𝑉 𝑉
∂𝑃 ∂𝑇
∂𝑉 𝑃

∂𝐻 𝑛𝑅𝑇
Simplify the Equation: 𝑇= V −
∂𝑃 2𝑎𝑛2 𝑛𝑅𝑇
− 2 +
𝑉 𝑉

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 Deriving the 𝜇𝐽𝑇 Coefficient
for a van der Waals gas
∂𝐻 𝑛𝑅𝑇 𝑛𝑅𝑇
Factor out V in the 𝑇= V − =V −
denominator: ∂𝑃 2𝑎𝑛2 𝑛𝑅𝑇 1 2𝑎𝑛2
− 2 + 𝑉 − 𝑉 + 𝑛𝑅𝑇
𝑉 𝑉

∂𝐻 𝑛𝑅𝑇𝑉 𝑛𝑅𝑇𝑉
Re-arrange the 𝑇= V − = +𝑉
∂𝑃 2𝑎𝑛2 2𝑎𝑛2
− 𝑉 + 𝑛𝑅𝑇 𝑉 − 𝑛𝑅𝑇
equation:

Your Turn to Show the Steps!

∂𝐻 1
𝑇= V 1− 2𝑎𝑛
∂𝑃 1 − 𝑅𝑇𝑉

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 Deriving the 𝜇𝐽𝑇 Coefficient
for a van der Waals gas
Approximation 2𝑎𝑛 2
∂𝐻 1 1−
Note: 2an/RTV is = V 1 − ≅ V 1 − 𝑅𝑇𝑉
small quantity ∂𝑃 𝑇 1−
2𝑎𝑛
1−
2𝑎𝑛
𝑅𝑇𝑉 𝑅𝑇𝑉

∂𝐻 2𝑎𝑛 2𝑎𝑛
𝑇≅ 𝑉 1− 1+ =−
Simplify the
equation: ∂𝑃 𝑅𝑇𝑉 𝑅𝑇

∂𝐻 2𝑎𝑛
=−
∂𝑃 𝑇, 𝑏→0
𝑅𝑇

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 Deriving the 𝜇𝐽𝑇 Coefficient
for a van der Waals gas
Recall: ∂𝐻 ∂𝐻 ∂𝐻
≅ +
∂𝑃 𝑇
∂𝑃 𝑇, 𝑎→0
∂𝑃 𝑇,𝑏→0

Substituting the ∂𝐻 2𝑎𝑛

values we derived: ≅ 𝑛𝑏 −
∂𝑃 𝑇
𝑅𝑇

𝜕𝐻
−
Recall: 𝜕𝑇 −1 𝜕𝑃 𝑇
= =
𝜕𝑃 𝜕𝑃 𝜕𝐻 𝐶𝑝
𝐻
𝜕𝐻 𝑇 𝜕𝑇 𝑃

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Deriving the 𝜇𝐽𝑇 Coefficient
for a van der Waals gas
Substituting the
2𝑎𝑛
𝜕𝑇 − 𝑛𝑏 − 1 2 𝑎𝑛
values we derived: = 𝑅𝑇 = −𝑛𝑏 +
𝜕𝑃 𝐻
𝐶𝑃 𝐶𝑝 𝑅𝑇

Finally! 𝝏𝑻 𝟏 𝟐 𝒂𝒏
𝝁𝑱𝑻 = = −𝒃
𝝏𝑷 𝑯
𝑪𝒑 𝑹𝑻

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Implications of 𝜇𝐽𝑇 Coefficient
for a van der Waals gas
For a Real gas:
1 2𝑎
𝜇𝐽𝑇 = −𝑏
𝐶𝑝 𝑅𝑇

Very Low Temperature: Very High Temperature:

1 2𝑎 −𝑏
𝜇𝐽𝑇 = 𝜇𝐽𝑇 =
𝐶𝑝 𝑅𝑇 𝐶𝑝

A positive Joule-Thomson A negative Joule-Thomson

coefficient means cooling coefficient means heating

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Assessment
1. Joule-Thomson effect is related to:
(A) adiabatic compression
(B) adiabatic expansion
(C) isothermal expansion
(D) isothermal compression

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Assessment
2. When a gas undergoes continuous throttling process by a valve and
its pressure and temperature are plotted, then we get a
a) isotherm
b) isenthalp
c) adiabatic
d) isobar

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Assessment
3. The region inside the inversion curve has ____ Joule-Kelvin coefficient
and the region outside the inversion curve has ____ Joule-Kelvin
coefficient.
a) positive, positive
b) negative, negative
c) negative, positive
d) positive, negative

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Assessment
4. When an ideal gas is made to undergo a Joule-Kelvin expansion, i.e.,
throttling, there is no change in temperature.
a) true
b) false

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Assessment
5. For a gas being throttled, the change in temperature can be
a) positive
b) negative
c) zero
d) all of the mentioned

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Assessment
6. If 𝜇𝐽−𝑇 > 0 , repulsive part of the potential is dominant.
a) True
b) False

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Assessment
7. Joule-Thompson cooling of a gas occurs when the pressure of the gas
drops as it goes through a porous plug separating a high pressure
chamber from one at 1 atm. The process is carried out at constant
enthalpy. The appropriate measure of the cooling is given by
𝜕𝐻
𝜕𝑇 𝜕𝑇 𝜕𝑃 𝑇
show that: =−
𝜕𝑃 𝐻 𝜕𝑃 𝐻 𝐶𝑝

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Assessment
8. A vapour at 22 atm and 5 oC was allowed to expand adiabatically to a
final pressure of 1.00 atm; the temperature fell by 10 K. Calculate the
Joule-Thomson coefficient, , at 5oC, assuming it remains constant over
this temperature range.

Answer: 0.48 K atm-1

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Assessment
9. Given that 𝜇 = 1.11 𝐾 𝑎𝑡𝑚−1 carbon dioxide, calculate the value of
its isothermal Joule-Thomson coefficient. Calculate the energy that
must be supplied as heat to maintain a constant temperature when 12.0
moles of CO2 flows through a throttle in an isothermal Joule –Thomson
experiment and the pressure drop is 55 atm. The Cp,m for CO2 is 37.11
JK-1mol-1.

Answer: ∆𝐻 = 27.2 𝑥 103 𝐽

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Assessment
10. A sample consisting of 1.00 mol of a van der Waals gas is compressed
from 20.00 dm3 to 10.0 dm3 at 300 K. In this process , 20.2 kJ of work is
done on the gas . Given that
2𝑎
−𝑏
𝜇𝐽−𝑇 = 𝑅𝑇
𝐶𝑝,𝑚
The Cp,m = 38.4 JK-1mol-1, a = 3.60 dm6 atm mol-2, and b = 0.044 dm3 mol-1.
Calculate the H for the process.

Answer: -30.5 J mol-1

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