Download Chemistry 3510: Physical Chemistry Midterm Exam 1 19 February 2007 Name:

Chemistry 3510: Physical Chemistry
Midterm Exam 1
19 February 2007
CH3510 - F2008
Questions from Old and
Sample Exams
Name:
Score:
Problem 1
/25
2
/25
3
/25
4
/25
5
/25
TOTAL
/100
NB: Do any four (4) of the problems. Indicate which ones should be
graded.
Possibly Useful Information
dU = dq
¯ + dw
¯ + µdn
dw
¯
= −pdV
dq
¯ = ncX dT
dS ≥ dq/T
¯
dH = dU + d(pV )
dG = dH − d(T S)
dA = dU − d(T S)
!
!
!
!
∂p
∂T
∂V
∂T
=−
=
∂V S,n
∂S V,n
∂p S,n
∂S p,n
∂S
∂V
1=
∂x
∂y
!
∂p
∂T
=
T,n
!
j
∂y
∂z
!
j
!
∂z
∂x
V,n
!
∂S
∂p
!
−1 =
j
T,n
∂V
=−
∂T
∂x
∂y
!
Tc
ε = 1−
Th
−2
R ∼ 8 × 10 Lbar/(molK) ∼ 8J/(molK)
P V = nRT
3
dU = nRdT
2
!
γ
Pi
Vj
=
Pj
Vi
z
∂y
∂z
!
!
x
p,n
∂z
∂x
!
y
1J ∼ 10−2 L bar
5
3
5
dH = nRdT Cv = R Cp = R
2
2
2
!
(γ−1)
Vj
Ti
Cp
=
γ=
Tj
Vi
Cv
No programmable calculators. Extra sheets of paper are available if needed.
Reminder: Show all work!
1
Chemistry 3500: Physical Chemistry
Mideterm Exam 1
SAMPLE
Name:
Score:
Problem 1
/25
2
/25
3
/25
4
/25
5
/25
TOTAL
/100
NB: Do any four (4) of the problems.
Possibly Useful Information
PV
Pi
Pj
!
=
Vj
Vi
= nRT!
(γ−1)
Ti
Vj
Cp
γ=
=
Tj
Vi
Cv
!
!
!
∂x
∂y
∂z
−1 =
∂y z ∂z x ∂x y
γ
1 =
∂x
∂y
!
j
∂y
∂z
!
j
∂z
∂x
3
5
Cv = R C p = R
2
2
!
j
¯ v − dw
¯
dU = dq
¯ p + V dp
¯
dH = dq
dS = dq/T
Tc
ε = 1−
Th
R ∼ 8 × 10−2 Lbar/(molK) ∼ 8J/(molK)
No programmable calculators. Extra sheets of paper are available if needed.
Reminder: Show all work!
1
Chemistry 3500: Physical Chemistry
Midterm Exam 1
18 October 2005
Name:
Score:
Problem 1
/25
2
/25
3
/25
4
/25
5
/25
TOTAL
/100
NB: Do any four (4) of the problems.
Please indicate which ones should be
graded.
Possibly Useful Information
¯ v − dw
¯
dU = dq
¯ p + V dp
¯
dH = dq
dS = dq/T
!
!
!
∂x
∂y
∂z
−1 =
∂y z ∂z x ∂x y
!
!
!
∂x
∂y
∂z
1 =
∂y j ∂z j ∂x j
Tc
ε = 1−
Th
R ∼ 8 × 10−2 Lbar/(molK) ∼ 8J/(molK)
P V = nRT
3
dU = nRdT
2
!
γ
Pi
Vj
=
Pj
Vi
5
3
5
dH = nRdT Cv = R Cp = R
2
2
2
!
(γ−1)
Cp
Ti
Vj
=
γ=
Tj
Vi
Cv
No programmable calculators. Extra sheets of paper are available if needed.
Reminder: Show all work!
1
1. Two moles of a monatomic ideal gas are sealed into an expandable vessel with an
initial volume of 0.5L vessel under a pressure of 8 bar and at a temperature of 25K.
This system is then taken through a cycle consisting of the following legs:
(a) an isobaric heating to 12L;
(b) a reversible adiabatic expansion to a pressure of 0.2 bar; and
(c) an isothermal reversible compression back to the initial conditions.
Draw and label the cycle in the state variable space.
Calculate the work, heat, change in internal energy, change in enthalpy, and change in
entropy of both system and surroundings for each of the legs and for the cycle. Fill in
the table provided with these results.
process
(a)
w
q
∆U
(b)
(c)
cycle
2
∆H
∆Ssys
∆Ssurr
3. One mole of an ideal gas initially contained in a 5L vessel under a pressure of 10 bar
at a temperature of 625K is carried through a cycle consisting of the following legs:
(a) an isothermal reversible expansion to a final volume of 25L;
(b) a reversible adiabatic compression to a volume of 5L at a pressure of 4 bar; and
(c) an isochoric reversible heating back to the initial conditions.
Draw and label the cycle in the state variable space.
Calculate the work, heat, change in internal energy, change in enthalpy, and change in
entropy of both system and surroundings for each of the legs and for the cycle. Fill in
the table provided with these results.
process
(a)
w
q
∆U
(b)
(c)
cycle
4
∆H
∆Ssys
∆Ssurr
3. Consider the following process for one mole of an ideal gas initially at a pressure of 1
bar and a temperature of 300K:
(a) an isothermal reversible expansion to 0.1 bar;
(b) an isothermal expansion against a constant pressure of 0.1 bar;
(c) an adiabatic reversible expansion to 0.1 bar.
In each case, calculate the work, heat, change in internal energy, change in enthalpy,
and change in entropy of both system and surroundings. Fill in the table provided
with these results.
process
w
q
∆U
∆H
∆Ssys
∆Ssurr
(a)
(b)
(c)
4
2. Given the definitions of the isothermal compressibility, κT , and the isobaric expansion
coefficient, α,
!
!
1 ∂V
1 ∂V
κT = −
α=
V ∂P T
V ∂T p
(a) express
∂P
∂T
V
in terms of α and κT ; and
(b) show that, starting from the total differential of V,
∂α
∂P
!
3
T
∂κ
=−
∂T
!
p
2. Derive an expression for the constant volume heat capacity, cv , of a gas described by
the equation of state
(p + α) V¯ − β = RT,
starting from the total differential of the internal energy and first law (with only pV
work).
3
5. Show the logical development from the first and second laws of thermodynamics to the
criterion for a system at or seeking equilibrium, (dA)V,T,n ≤ 0.
6
5. Find the difference between the constant pressure and constant volume heat capacities
for a gas obeying the van der Waals equation of state:
p=
RT
a
− 2,
υ−b υ
where a and b are constants.
Compare this difference to that for an ideal gas.
6
4. A block of copper being shaped (m = 20.0g) is flash cooled to 273K from its initial
temperature of 1200K by imersion into a 300L ice water bath at 0 ◦ C.
(a) [5 pts] Is the cooling process reversible or irreversible? Why?
(b) [20 pts] Find the entropy change for the copper in the cooling process, using the
heat capacity for copper over the temperature range:
Cp (T ) ≃ a + b · T + c · T 2 + d · T 3 , where
a = 0.27899J/(g K)
b = 4.42151 · 10−4 J/(g K2 )
c = −4.91784 · 10−7 J/(g K3 )
d = 2.04287 · 10−10 J/(g K4 ).
5
¯
1. Show that the integral of dq/T
around a Carnot cycle, shown below, is zero. Then
show that the integral is zero if the isothermal reversible expansion stage is replaced
by an isothermal irreversible expansion.
p
6
I
isothermal, Th
II
adiabatic
adiabatic
IV
isothermal, Tc
2
III
-
V
5. Devise an engine utilizing four explicit states of an ideal gas to perform 1000J of work
at 50% efficiency.
6
5. For a cooling system based on an ideal gas Carnot cycle connecting reservoirs at 50 ◦ C
and 200◦ C, what amount of work will be needed to remove 1000J of heat?
6
1. The Clausius Inequality states
¯
dS ≥ dq/T.
Starting from the first law of thermodynamics and the fundamental equation for internal energy,
¯ + dw,
¯
T dS − pdV = dq
and considering the change in state of a gas contained in a frictionless and massless
piston and doing work against an external pressure, show that
(a) the inequality holds when pext > p
(b) the inequality holds when pext < p
(c) the equality holds when pext = p
2
4. Calculate the heat evolved by the combustion of octane
C8 H18 (l) + 25
O2 (g) → 8CO2 (g) + 9H2 O(l)
2
at its flash point, 15◦ C, given the following information:
cp (C4 H6 (g))= 80 J/(molK)
cp (C8 H12 (g))= 140 J/(molK)
cp (H2 O(g))= 35 J/(molK)
cp (C8 H18 (g))= 250 J/(molK)
cp (H2 O(l))= 80 J/(molK)
cp (H2 (g))= 30 J/(molK)
cp (CO2 (g))= 40 J/(molK)
cp (O2 (g))= 30 J/(molK)
∆Hf298K (H2 O(l))= -280 kJ/mol
∆Hf298K (H2 O(g))= -240 kJ/mol
∆Hf298K (CO2 (g))= -390 kJ/mol
∆Hf298K (C4 H6 (g))= 110 kJ/mol
C8 H12 (g)+3H2 (g) → C8 H18 (l)
∆Hr298K = −400kJ/mol
∆Hr298K = −180kJ/mol
2C4 H6 (g) → C8 H12 (g)
5
4. Compute the heat produced by the following reaction at 500K
C2 H4 (g) + H2 (g) → C2 H6 (g)
using the following information:
C2 H4 (g) + 3O2 (g) → 2CO2 (g) + 2H2 O(l)
H2 (g) + 21 O2 (g) → H2 O(l)
∆Hr = −1400kJ/mol at 298K
∆Hr = −300kJ/mol at 298K
C2 H6 (l) + 72 O2 (g) → 2CO2 (g) + 3H2 O(l)
∆Hr = −1500 kJ/mol at 298K
cp (H2 C4 (g))= 40 J/(molK)
cp (H2 (g))= 30 J/(molK)
cp (H2 C6 (g))= 50 J/(molK)
cp (H2 C6 (l))= 80 J/(molK)
∆Hv (H2 C6 )= 20 kJ/mol at 400K
∆Hv (H2 C4 )= 14 kJ/mol at 200K
5