Third year thermodynamics Exam questions and Solutions

WORKED OUT PROBLEMS




E. van Steen
April 2016

,Preamble
Worked-out problems serve as a guide to problem solving (and in this particular case dealing with
thermodynamic problems). The problems listed here may give you some guidance in how to solve
the problems. Students are urged to seriously attempt problems by themselves before going over
the answers. The answers may contain some errors, but the methodology applied will be correct.

The following types of problems are listed:

1. Heating of a single component
2. Compression of a single component
3. Expansion of a single component
4. Single component phase equilibrium
5. Fugacity
6. Vapour-liquid equilibrium
7. Liquid-liquid equilibrium
8. Solid-liquid equilibrium
9. Chemical equilibrium

, 1. Heating/cooling of a single component

Calculate the enthalpy change of nitrogen as it undergoes a process from -50°C and 2 MPa to 40°C
and 6 MPa. Assume that:
(a) Nitrogen can be described by the ideal gas equation
𝑅𝑇 𝑎
(b) Nitrogen can be described by the following equation of state: 𝑉 = 𝑝 − 𝑅𝑇 + 𝑏 with a =
0.01386 Pa·m6/mol2 and b = 3.86410-5 m3/mol
Data: 𝐶𝑃,𝑁2 = 27.318 + 0.623 ∙ 10−2 ∙ 𝑇 − 0.095 ∙ 10−5 ∙ 𝑇 2
For an ideal gas, the enthalpy is only a function of temperature (and not of pressure):
𝑇2 =313𝐾
∆𝐻 𝐼𝐺 = ∫ 𝐶𝑃,𝑁2 ∙ 𝑑𝑇
𝑇1 =223𝐾
0.623 ∙ 10−2 0.095 ∙ 10−5
= 27.318 ∙ (𝑇2 − 𝑇1 ) + ∙ (𝑇2 2 − 𝑇1 2 ) − ∙ (𝑇2 3 − 𝑇1 3 )
2 3
𝑘𝐽
Substituting all numbers: ∆𝐻 𝐼𝐺 = 2.60 ∙ 𝑚𝑜𝑙

For the real gas: ∆𝐻 = 𝐻(𝑇2 , 𝑝2 ) − 𝐻(𝑇1 , 𝑝1 )
𝜕𝑉
𝑑𝐻 = 𝐶𝑝 ∙ 𝑑𝑇 + (𝑉 − 𝑇 ∙ ( ) ) ∙ 𝑑𝑝
𝜕𝑇 𝑝
𝜕𝑉 𝑅𝑇 𝑎 𝑅 𝑎 2∙𝑎
(𝑉 − 𝑇 ∙ ( ) ) = − +𝑏−𝑇∙( + 2
)=− +𝑏
𝜕𝑇 𝑝 𝑝 𝑅𝑇 𝑝 𝑅𝑇 𝑅𝑇
Pathway (T1,p1)  ideal gas state (T1,p=0)  ideal gas state (T2,p=0)  final state (T2,p2)

Pathway A: (T1,p1)  ideal gas state (T1,p=0)
𝑇1 =233𝐾 𝑝=0𝑀𝑃𝑎
2∙𝑎
∆𝐻𝐴 = ∫ 𝐶𝑝 ∙ 𝑑𝑇 + ∫ (− + 𝑏) ∙ 𝑑𝑝
𝑇1 =223𝐾 𝑝1 =2𝑀𝑃𝑎 𝑅𝑇
2 ∙ 0.01386 𝐽
= 0 + (− + 3.864 ∙ 10−5 ) ∙ (0 − 2 ∙ 106 ) = −47.47 ∙
8.314 ∙ 223 𝑚𝑜𝑙
Pathway B: ideal gas state (T1,p=0)  ideal gas state (T2,p=0)
𝑇 =313𝐾 𝑝 =0𝑀𝑃𝑎 2∙𝑎 𝐽
∆𝐻𝐵 = ∫𝑇 2=223𝐾 𝐶𝑝 ∙ 𝑑𝑇 + ∫𝑝=0𝑀𝑃𝑎
2
(− 𝑅𝑇 + 𝑏) ∙ 𝑑𝑝 = 2602.69 ∙ 𝑚𝑜𝑙 (from above)
1
𝑇1 =313𝐾 𝑝2 =6𝑀𝑃𝑎
2∙𝑎
∆𝐻𝐶 = ∫ 𝐶𝑝 ∙ 𝑑𝑇 + ∫ (− + 𝑏) ∙ 𝑑𝑝
𝑇2 =313𝐾 𝑝=0𝑀𝑃𝑎 𝑅𝑇
2 ∙ 0.01386 𝐽
= 0 + (− + 3.864 ∙ 10−5 ) ∙ 6 ∙ 106 = 147.93 ∙
8.314 ∙ 313 𝑚𝑜𝑙
𝑘𝐽
The overall change in enthalpy is given by: ∆𝐻 = ∆𝐻𝐴 + ∆𝐻𝐵 + ∆𝐻𝐶 = 2.72 ∙ 𝑚𝑜𝑙

, A stream of pure ethylene is to be cooled down from 250oC to 25oC in a single-pass, counter-flow
heat exchanger. The gas enters the heat exchanger at 25 bar and a volumetric flow of 3 m3/min. The
𝐽
specific heat of ethylene as an ideal gas is given as 𝑐𝑃 𝐼𝐺 (𝑚𝑜𝑙∙𝐾) = 17.8 + 0.09 ∙ 𝑇(𝐾). Determine
the molar flow rate of ethylene entering the heat exchanger and the amount of heat transferred if
(a) ethylene is an ideal gas.
𝑅𝑇 𝑎
(b) ethylene obeys the van der Waals equation of state as 𝑝 = 𝑉−𝑏 − 𝑉 2 with a = 0.4614
(Pa.m6)/mol2 and b = 5.823.10-5 m3/mol.
𝑝
For an ideal gas, the molar flow rate is given by the ideal gas law 𝑛̇ = 𝑅𝑇 ∙ 𝑉̇. Hence, the molar flow
25∙105 𝑚𝑜𝑙
rate for the ideal gas is law 𝑛̇ 𝑖𝑛 = 8.314∙523.15 ∙ 3 = 1724 ∙ 𝑚𝑖𝑛

The amount of heat transferred in the heat exchange is given by the energy balance:
0 = 𝐻̇𝑖𝑛 − 𝐻̇𝑜𝑢𝑡 + 𝑄̇
Substituting the (rather trivial) molar balance (𝑛̇ 𝑖𝑛 = 𝑛̇ 𝑜𝑢𝑡 ) 
0 = 𝑛̇ 𝑖𝑛 ∙ (𝐻𝑖𝑛 − 𝐻𝑜𝑢𝑡 ) + 𝑄̇
or 𝑄̇ = 𝑛̇ 𝑖𝑛 ∙ (𝐻𝑜𝑢𝑡 − 𝐻𝑖𝑛 )
The heat exchange process takes place isobarically and thus the change in the enthalpy is given by:
298.15
𝑄̇ = 𝑛̇ 𝑖𝑛 ∙ (𝐻𝑜𝑢𝑡 − 𝐻𝑖𝑛 ) = 𝑛̇ 𝑖𝑛 ∙ ∫523.15 𝐶𝑃 ∙ 𝑑𝑇
Since the gas is an ideal gas, we can use the ideal gas heat capacity:
298.15 1724 𝑚𝑜𝑙 0.09
𝑄̇ = 𝑛̇ 𝑖𝑛 ∙ ∫523.15
𝑐𝑃 𝐼𝐺 ∙ 𝑑𝑇 = ∙ ∙ (17.8 ∙ (298.15 − 523.15) + ∙ (298.152 −
60 𝑠 2
𝑘𝐽
523.152 )) = 354 ∙ 𝑠


If ethylene is a gas obeying the van der Waals equation of state has to be used to determine the
molar flow rate. The easiest way is to transform the van der Waals equation into:
𝑍3 + 𝛼 ∙ 𝑍2 + 𝛽 ∙ 𝑍 + 𝛾 = 0
𝑎∙𝑝 𝑏∙𝑝
With 𝛼 = −1 − 𝐵; 𝛽 = 𝐴; 𝛾 = −𝐴 ∙ 𝐵 𝑎𝑛𝑑 𝐴 = (𝑅𝑇)2 ; 𝐵 = 𝑅𝑇
𝑁∙𝑚4 𝑁 𝑚3 𝑁
0.4614 ∙ ∙25∙105 ∙ 2 5.823∙10−5 ∙ ∙25∙105 ∙ 2
𝑚𝑜𝑙2 𝑚𝑜𝑙
𝐴= 𝑁𝑚
𝑚
2 = 0.0610 𝐵= 𝑁𝑚
𝑚
= 0.0335
(8.314∙ ∙523.15𝐾) 8.314∙ ∙523.15𝐾
𝑚𝑜𝑙∙𝐾 𝑚𝑜𝑙∙𝐾

Hence, the equation to be solved is 𝑍 3 − 1.0335 ∙ 𝑍 2 + 0.0610 ∙ 𝑍 − 0.00204 = 0 . This equation
can be solved iteratively using Newton-Raphson:
𝑍𝑜𝑙𝑑 3 − 1.0335 ∙ 𝑍𝑜𝑙𝑑 2 + 0.0610 ∙ 𝑍𝑜𝑙𝑑 + −0.00204
𝑍𝑛𝑒𝑤 = 𝑍𝑜𝑙𝑑 −
3 ∙ 𝑍𝑜𝑙𝑑 2 − 2 ∙ 1.0335 ∙ 𝑍𝑜𝑙𝑑 + 0.0610
starting with Z=1:
Zold Znew
1 0.974384
0.974384 0.972961
0.972961 0.972957
0.972957 0.972957

1 𝑝 𝑛̇ 𝐼𝐺 𝑚𝑜𝑙
Hence, Z=0.9730  𝑛̇ = 𝑍 ∙ 𝑅𝑇 ∙ 𝑉̇ = 𝑍 = 1772.3 ∙ 𝑚𝑖𝑛 (slightly larger seeing that the attractive
force will result in a more dense gas phase than the ideal case)
298.15
The amount of heat transferred is given by: or 𝑄̇ = 𝑛̇ 𝑖𝑛 ∙ (𝐻𝑜𝑢𝑡 − 𝐻𝑖𝑛 ) = 𝑛̇ 𝑖𝑛 ∙ ∫523.15 𝐶𝑃 ∙ 𝑑𝑇. The
gas is no longer an ideal gas (although seeing the value for Z still quite close to the ideal gas state),
we cannot put the ideal gas heat capacity in this equation. We can reformulate the energy balance,
as