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Chemistry Department Al-kharj, Nov 2016
College of Science and Humanities Studies Exercises- Physical Chemistry I (Chem 3310)
Prince Sattam Bin Abdulaziz University
1
Exercises-Physical Chemistry I (Chem 3310)
Chapter-3- The second and third laws of thermodynamics
Exercise 1
Calculate the change in entropy when 50 kJ of energy is transferred reversibly and isothermally
as heat to a large block of copper at (a) 0°C, (b) 70°C.
Solution
We have:
𝑑𝑆 =𝑑𝑞𝑟𝑒𝑣
𝑇
For a measurable change,
∆𝑆 = ∫𝑑𝑞𝑟𝑒𝑣
𝑇
Process is reversible and isothermal (T=constant), thus
∆𝑆 =1
𝑇∫ 𝑑𝑞𝑟𝑒𝑣 =
𝑞𝑟𝑒𝑣
𝑇
qrev is the energy transferred as heat (qrev = 50 kJ).
(a) The change in entropy at T = 0°C = (0+273.15) K = 273.15 k
We have
∆S =50 × 103
273.15= 1.8 102 J k−1
(a) The change in entropy at T = 70°C = (70+273.15) K = 273.15 k
We have
∆S =50 × 103J
(70 + 273.15)= 1.5 102 J k−1
Chemistry Department Al-kharj, Nov 2016
College of Science and Humanities Studies Exercises- Physical Chemistry I (Chem 3310)
Prince Sattam Bin Abdulaziz University
2
Exercise 2
Calculate the molar entropy of a constant-volume sample of argon at 250 K given that it is 154.84
J K−1
mol−1
at 298 K.
Data: Cp,m = 20.786 J K-1
mol-1
and R = 8.3145 J K-1
mol-1
Solution
At constant-volume, the change in molar entropy when monoatomic gas argon (Ar) is heated
from 250 K to 298 K is (Tf = 298 K and Ti = 250):
𝑆𝑚(𝑇𝑓) = 𝑆𝑚(𝑇𝑖) + 𝐶𝑉,𝑚 ∫𝑑𝑇
𝑇
𝑇𝑓
𝑇𝑖
Thus,
𝑆𝑚(𝑇𝑖) = 𝑆𝑚(𝑇𝑓) + 𝐶𝑉,𝑚 ∫𝑑𝑇
𝑇
𝑇𝑖
𝑇𝑓
= 𝑆𝑚(𝑇𝑓) + 𝐶𝑉,𝑚 ln𝑇𝑖
𝑇𝑓
where CV,m = Cp,m - R
Sm(250) = Sm(298) + CV,m ∫dT
T
298
250
= 154.84 J K−1 mol−1 + (20.786 − 8.3145) J K−1 mol−1 ln250
298
= 152.65 J K−1 mol−1
Chemistry Department Al-kharj, Nov 2016
College of Science and Humanities Studies Exercises- Physical Chemistry I (Chem 3310)
Prince Sattam Bin Abdulaziz University
3
Exercise 3
Calculate ΔS (for the system) when the state of 2.00 mol diatomic perfect gas molecules, for
which Cp,m = (7/2)R, is changed from 25°C and 1.50 atm to 135°C and 7.00 atm. How do you
rationalize the sign of ΔS?
Solution
ΔS of the system has the same value as if the change happened by reversible heating at constant
pressure (step 1), followed by reversible isothermal compression (step 2).
ΔS = ΔS1 + ΔS2
∆𝑆1 = ∫𝑑𝑞𝑟𝑒𝑣
𝑇= 𝑛𝐶𝑝,𝑚 ∫
𝑑𝑇
𝑇
𝑇𝑓
𝑇𝑖= 𝑛𝐶𝑝,𝑚 ln
𝑇𝑓
𝑇𝑖
∆𝑆1 = (2.00 𝑚𝑜𝑙−1) × (7
2) × (8.3145 𝐽𝐾−1𝑚𝑜𝑙−1) × ln (
(135 + 273.15)𝐾
(25 + 273.15)𝐾) = 18.3 𝐽𝑘−1
For the second step (isothermal process)
∆S2 = ∫dqrev
T=
qrev
T
where
qrev = −w = ∫ pdV = nRT lnVf
Vi= nRT ln
pi
pf
Where pi = nRT/Vi and pf = nRT/Vf
Hence,
∆S2 = nRT lnpi
pf= (2.00 mol−1) × (8.3145 JK−1mol−1) ln (
1.5 atm
7.00) = −25.6 Jk−1
Thus,
ΔS = ΔS1 + ΔS2 = (18.3 – 25.6) JK-1
= -7.3 JK-1
The heat lost in step 2 was more that the heat gained in step 1, resulting in a net loss of entropy.
Alternatively, the ordering represented by confining by confining the sample to a smaller volume
in step 2 overcomes the disordering represented by the temperature rise in step 1. A negative
entropy change is allowed for a system as long as an increase in entropy elsewhere results in ΔStot
> 0.
Chemistry Department Al-kharj, Nov 2016
College of Science and Humanities Studies Exercises- Physical Chemistry I (Chem 3310)
Prince Sattam Bin Abdulaziz University
4
Exercise 4
A sample consisting of 2.00 mol of diatomic perfect gas molecules at 250 K is compressed
reversibly and adiabatically until its temperature reaches 300 K. Given that CV,m = 27.5 J K−1
mol−1
, calculate q, w, ΔU, ΔH, and ΔS.
Solution
Chemistry Department Al-kharj, Nov 2016
College of Science and Humanities Studies Exercises- Physical Chemistry I (Chem 3310)
Prince Sattam Bin Abdulaziz University
5
Exercise 5
Calculate ΔH and ΔStot when two iron blocks, each of mass 1.00 kg, one at 200°C and the other
at 25°C, are placed in contact in an isolated container. The specific heat capacity of iron is 0.449
J K−1
g−1
and may be assumed constant over the temperature range involved.
Solution
Chemistry Department Al-kharj, Nov 2016
College of Science and Humanities Studies Exercises- Physical Chemistry I (Chem 3310)
Prince Sattam Bin Abdulaziz University
6
Exercise 6
The enthalpy of vaporization of methanol is 35.27 kJ mol−1
at its normal boiling point of 64.1°C.
Calculate
(a) The entropy of vaporization of methanol at this temperature
(b) The entropy change of the surroundings.
Solution
Chemistry Department Al-kharj, Nov 2016
College of Science and Humanities Studies Exercises- Physical Chemistry I (Chem 3310)
Prince Sattam Bin Abdulaziz University
7
Exercise 7
The heat capacity of chloroform (trichloromethane, CHCl3) in the range 240 K to 330 K is given
by Cp,m /(J K−1
mol−1
) = 91.47 + 7.5 × 10−2
(T/K).
In a particular experiment, 1.00 mol CHCl3 is heated from 273 K to 300 K. Calculate the change
in molar entropy of the sample.
Solution
Chemistry Department Al-kharj, Nov 2016
College of Science and Humanities Studies Exercises- Physical Chemistry I (Chem 3310)
Prince Sattam Bin Abdulaziz University
8
Chemistry Department Al-kharj, Nov 2016
College of Science and Humanities Studies Exercises- Physical Chemistry I (Chem 3310)
Prince Sattam Bin Abdulaziz University
9
Chemistry Department Al-kharj, Nov 2016
College of Science and Humanities Studies Exercises- Physical Chemistry I (Chem 3310)
Prince Sattam Bin Abdulaziz University
10
Chemistry Department Al-kharj, Nov 2016
College of Science and Humanities Studies Exercises- Physical Chemistry I (Chem 3310)
Prince Sattam Bin Abdulaziz University
11
Chemistry Department Al-kharj, Nov 2016
College of Science and Humanities Studies Exercises- Physical Chemistry I (Chem 3310)
Prince Sattam Bin Abdulaziz University
12
Chemistry Department Al-kharj, Nov 2016
College of Science and Humanities Studies Exercises- Physical Chemistry I (Chem 3310)
Prince Sattam Bin Abdulaziz University
13
Chemistry Department Al-kharj, Nov 2016
College of Science and Humanities Studies Exercises- Physical Chemistry I (Chem 3310)
Prince Sattam Bin Abdulaziz University
14
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