Chemistry Department
College of Science and Humanities Studies
Prince Sattam Bin Abdulaziz University
Al-kharj, March 2016
Exam-I- Physical Chemistry II (Chem 3320)
Exam-1-Physical Chemistry II-Solution
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Multiple choice questions (5 pts)
Q1
√
Q2
√
Q3
√
Q4
Q5
√
Boyle’s law states that when the temperature of a perfect gas (PV = nRT) is constant,
thus:
A PV > 0
√ C P1V1 = P2V2
B PV = constant
D concentration
1
2
For a perfect gas, PV = nRT and pV = nMc . Thus, the root mean square speed of the
3
molecules of gas and its temperature T of are related by the following formula
3RT
M 1/2
C
c=(
)
M
3RT
M
3RT 1/2
2
B
√
D
c =
c=(
)
3RT
M
A mean speed is calculated by multiplying each speed by the fraction of molecules that
have that speed, and then adding all the products together. By using Maxwell
A
c2 =
𝑀
3/2
𝑀
2
distribution 𝑓(𝑣) = 4𝜋 (2𝜋𝑅𝑇) 𝑣 2 𝑒 −2𝑅𝑇𝑣 , the mean speed, 𝑐̅ of the molecules in
a gas is:
πM 1/2
RT 1/2
A
C
c̅ = (
)
c̅ = ( )
8RT
πM
1/2
RT 1/2
8RT
B
D
c̅ = ( )
c̅ = (
)
M
πM
The mean free path, λ (lambda), the average distance a molecule travels between
collisions
kT
√
√
C
λ=
λ = c̅rel Δt
σp
T
c̅rel
√
D
λ=
λ=
σp
z
When a gas at a pressure p and temperature T is separated from a vacuum by a small
hole, the rate of escape of its molecules is equal to the rate at which they strike the area
of the hole. Thus, the rate of effusion from a hole of area A0 is
A
B
𝑝𝐴0 𝑁𝐴
(2πMRT)1/2
𝑝
Rate of effusion =
(2πMRT)1/2
Rate of effusion =
C
D
𝑝𝐴0
(2πMRT)1/2
𝑝𝑁𝐴
Rate of effusion =
(2πMRT)1/2
Rate of effusion =
1
Chemistry Department
College of Science and Humanities Studies
Prince Sattam Bin Abdulaziz University
Al-kharj, March 2016
Exam-I- Physical Chemistry II (Chem 3320)
The conductance, G of a sample decreases with its length l and increases with its crosssectional area A. Thus,
Q6
𝐴
𝑙
𝜅𝐴
G=
𝑙
√
𝑙
𝜅𝐴
𝑙
G=
𝐴
G=
A
B
G=
Kohlrausch was able to establish experimentally that limiting molar conductivity Λ0𝑚
can be expressed as the sum of contributions from its individual ions. If the limiting
molar conductivity of the cations is denoted λ+ and that of the anions λ−, then his law of
the independent migration of ions states that
Q7
Λ0𝑚 = 𝜈+ 𝜆+ + 𝜈− 𝜆−
Λ0𝑚 = 𝜈+ 𝜆+ − 𝜈− 𝜆−
A
B
C
D
Λ0𝑚 = 𝜈+ 𝜆+ /𝜈− 𝜆−
Λ0𝑚 = 𝜈+ 𝜆+ ∗ 𝜈− 𝜆−
where ν+ and ν− are the numbers of cations and anions per formula unit of electrolyte
In a solution, the drift speed, s of an ion is defined
Q8
√
𝑍𝑒𝐸
𝑓
𝑠 = 𝑢/𝐸
𝑠=
A
B
If we divide both sides of J = −𝐷
C
√
d𝒩
dx
D
𝐸
𝑓
𝑠 = 𝑢𝐸
𝑠=
by Avogadro’s constant, thereby converting
numbers into amounts (numbers of moles), then Fick’s law becomes
Data: 𝒩 is the number density
Q9
𝒩=
A
√
B
N
V
dn
dV
dc
J = −𝐷
dx
J = −𝐷
C
D
dn
dx
dV
J = −𝐷
dx
J = −𝐷
The Einstein relation between the diffusion coefficient D and the ionic mobility u is:
Q10
√
A
B
RT
u
zF
RT
D=
uzF
D=
C
D
u
RT
u
D=
zF
D=
2
Chemistry Department
College of Science and Humanities Studies
Prince Sattam Bin Abdulaziz University
Al-kharj, March 2016
Exam-I- Physical Chemistry II (Chem 3320)
Exercises (10 pts)
Exercise 1
The best laboratory vacuum pump can generate a vacuum of about 1 nTorr. At 25°C and
assuming that air consists of N2 molecules with a collision diameter of 395 pm.
Calculate:
(1) The mean speed of the molecules?
(2) The mean free path?
(3) The collision frequency in the gas?
Data:
Atomic mass of N = 14.01 g mol-1; R = 8.3145 J mol-1 K-1;
Boltzmann constant k = 1.381 ×10-23 J K-1; 1 atm = 101325 Pa = 760 torr; 1 Pa = 1 J m-3
Solution
(a) By definition, the mean speed is :
8RT 1/2
c̅ = (
)
πM
where, M is the molar mass of molecule in Kg mol-1. Thus,
1/2
8 × 8.3145 × J mol−1 K −1 × 298 K
c̅ = (
)
3.14 × 28 × 10−3 Kg mol−1
= 475 m s −1
(b) By definition, the mean free path is :
λ=
kT
σp
where k is the Boltzmann constant, σ =πd2 is called the collision cross-section of the molecules.
Thus,
kT
1.381 × 10−23 J K −1 × 298 K
λ=
=
= 6.3 × 104 m = 630 km
σp 3.14 × (395 × 10−12 m)2 × 10−9 torr × (101325 Pa)
760 torr
3
Chemistry Department
College of Science and Humanities Studies
Prince Sattam Bin Abdulaziz University
Al-kharj, March 2016
Exam-I- Physical Chemistry II (Chem 3320)
(b) By definition, the collision frequency is :
z =
c̅rel
where c̅rel = 21/2 c̅
λ
Thus,
z =
21/2 c̅ 21/2 × 475 m s−1
=
= 1.1 × 10−2 s−1
λ
6.3 × 104 m
Exercise 2
Calculate the thermal conductivity of argon (CV,m = 12.5 J K−1 mol−1, σ = 0.36 nm2) at room
temperature (20°C).
Data:
Atomic mass of Argon = 40 g mol-1; R = 8.3145 J mol-1 K-1
Solution
By definition,
1
kT
RT
8RT 1/2
𝑝
κ = λc̅ 𝐶𝑉,𝑚 [𝐴] where λ =
=
; c̅ = (
)
and [𝐴] =
3
σp
σpNA
πM
𝑅𝑇
Thus,
CV,m 8RT 1/2
κ=
(
)
3σNA πM
1/2
12.5 J mol−1 K −1
8 × 8.3145 × J mol−1 K −1 × 293 K
κ=
(
)
3 × 0.36 × (10−9 𝑚)2 × 6.02 × 1023 mol−1
3.14 × 40 × 10−3 Kg mol−1
κ = 7.6 × 10−3 J K −1 m−1 s −1 = 7.6 mJ K −1 m−1 s−1
Exercise 3
At 25°C, the limiting molar conductivities of KCl, KNO3, and AgNO3 are 14.99 mS m2 mol−1,
14.50 mS m2 mol−1, and 13.34 mS m2 mol−1, respectively. What is the limiting molar
conductivity of AgCl at this temperature?
Solution
Here, we use the relationship between the liming molar conductivity of an electrolyte, and the
individual limiting molar conductivities for ions:
4
Chemistry Department
College of Science and Humanities Studies
Prince Sattam Bin Abdulaziz University
Al-kharj, March 2016
Exam-I- Physical Chemistry II (Chem 3320)
𝚲𝟎𝒎 = 𝝂+ 𝝀+ + 𝝂− 𝝀−
where λ+ and λ− are the limiting molar conductivities of the cations is and anions, respectively.
ν+ and ν− are the numbers of cations and anions per formula unit of electrolyte.
Thus,
Λ0m (KCl) = λK+ + λCl− = 14.99 mS m2 mol−1
(1)
Λ0m (KNO3 ) = λK+ + λNO−3 = 14.50 mS m2 mol−1
(2)
Λ0m (AgNO3 ) = λAg+ + λNO−3 = 13.34 mS m2 mol−1
(3)
To obtained the Λ0m (AgCl), we sum the above equations:
λK+ + λCl− + λK+ + λNO−3 + λAg+ + λNO−3 = Λ0m (KCl) + Λ0m (KNO3 ) + Λ0m (AgNO3 )
2λK+ + λCl− + 2λNO−3 + λAg+ = Λ0m (KCl) + Λ0m (KNO3 ) + Λ0m (AgNO3 )
(2λK+ + 2λNO−3 ) + (λCl− + λAg+ ) = Λ0m (KCl) + Λ0m (KNO3 ) + Λ0m (AgNO3 )
2(λK+ + λNO−3 ) + (λAg+ + λCl− ) = Λ0m (KCl) + Λ0m (KNO3 ) + Λ0m (AgNO3 )
2Λ0m (KNO3 ) + Λ0m (AgCl) = Λ0m (KCl) + Λ0m (KNO3 ) + Λ0m (AgNO3 )
Thus,
Λ0m (AgCl) = Λ0m (KCl) + Λ0m (KNO3 ) + Λ0m (AgNO3 ) − 2Λ0m (KNO3 )
Λ0m (AgCl) = Λ0m (KCl) − Λ0m (KNO3 ) + Λ0m (AgNO3 )
Λ0m (AgCl) = (14.99 − 14.50 + 13.34 mS m2 mol−1 ) = 13.83 mS m2 mol−1
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