Exam I
CHM 3410, Dr. Mebel, Fall 2005
1. (10 pts) The evolution of life requires the organization of a very large number of molecules into biological cells. Does the formation of living organisms violate the
Second Law of thermodynamics? State your conclusion clearly and present detailed arguments to support it.
2. (16 pts) Suppose that you measured the product pV of 1 mol of a dilute gas and found that pV = 22.98 L atm at 0
°
C and 31.18 L atm at 100
°
C. Assume that the perfect gas law is valid, with T = t (
°
C) + a , and that the value of R is not known. Determine R and a from the measurements provided. Express your result for R in J K -1 mol -1 .
1

3. (16 pts) Consider the expansion of 0.500 mol of a perfect monoatomic gas with C
V ,m
=
(3/2) R . The initial state is described by p = 3.25 bar and T = 300 K. For each of the following conditions calculate the final temperature, q , w ,
Δ
U , and
Δ
H : a) the gas undergoes a reversible isothermal expansion to a final pressure of 1 bar; b) the gas undergoes an isothermal expansion against an external pressure of p ex bar to a final pressure of p = 1 bar; c) the gas undergoes a reversible adiabatic expansion to a final pressure of 1 bar; d) the gas undergoes an adiabatic expansion against an external pressure of bar to a final pressure of p = 1 bar.
p ex
= 1
= 1
2

4. (16 pts) The standard specific internal energy of combustion of crystalline fullerene
C
60
was measured to be –36.0334 kJ g -1 at 298.15 K. Calculate the standard enthalpy of combustion and the standard enthalpy of formation of this fullerene.
5. (16 pts) The pressure of a given amount of a van der Waals gas depends on T and V : p =
V nRT
nb
a
n
V
2
2
Derive and expression for d p in terms of d T and d V . Remember that if a function z depends on two variables x and y , its differential can be expressed as
dz =
 ∂ z

 ∂ x


 y dx +
 ∂ z

 ∂ y


 x dy
€
6. (16 pts) The standard molar entropy of NH
3 heat capacity is given by the equation C p ,m
= a
(g) is 192.45 J K
+ bT + c / T 2
-1 mol
, where a ,
-1 b
at 298 K, and its
, and c can be found in Table 2.2 of Data Section . Calculate the standard molar entropy at (a) 100 ° C and (b)
500
°
C.
3

Exam I
CHM 3410, Dr. Mebel, Fall 2005
1. (10 pts) The evolution of life requires the organization of a very large number of molecules into biological cells. Does the formation of living organisms violate the
Second Law of thermodynamics? State your conclusion clearly and present detailed arguments to support it.
When complex biological molecules are formed in biological cells, the entropy of the system decreases, Δ S < 0. However, this does not mean the Second Law of thermodynamics is violated. The Second Law states that the total entropy of the system and its surroundings should increase in a spontaneous process. If a reaction is exothermic and heat is released into surroundings,
Δ
S sur
> 0, and then the total entropy change can be positive. Thus, the organism releases heat and causes increasing disorder of the surroundings. Also, endothermic biochemical reactions are driven forward by coupling with highly exothermic reactions.
2. (16 pts) Suppose that you measured the product pV of 1 mol of a dilute gas and found that pV = 22.98 L atm at 0
°
C and 31.18 L atm at 100
°
C. Assume that the perfect gas law is valid, with T = t ( ° C) + a , and that the value of R is not known. Determine R and a from the measurements provided. Express your result for R in J K -1 mol -1 .
According to the perfect gas law,
( pV )
1
= nRT
1
( pV )
2
= nRT
2
Since n = 1 and T = t ( ° C) + a , we can write
( pV )
1
= R ( t
1
+ a ) ( pV )
2
= R ( t
2
+ a )
Now, we need to solve these two equations for a and R :
22.98 L atm = (1 mol) Ra 31.18 L atm = (1 mol) R (100 + a )
Subtracting the two equations, we obtain
31.18 – 22.98 = 100 R R = 0.082 L atm mol a = 22.98 / R = 22.98 / 0.082 L atm mol -1
-1 ( ° C) -1
( ° C) -1 = 280.2
° C
= 8.308 J K -1 mol -1
1

3. (16 pts) Consider the expansion of 0.500 mol of a perfect monoatomic gas with C
V ,m
=
(3/2) R . The initial state is described by p = 3.25 bar and T = 300 K. For each of the following conditions calculate the final temperature, q , w , Δ U , and Δ H : a) the gas undergoes a reversible isothermal expansion to a final pressure of 1 bar; b) the gas undergoes an isothermal expansion against an external pressure of p ex
= 1 bar to a final pressure of p = 1 bar; c) the gas undergoes a reversible adiabatic expansion to a final pressure of 1 bar; d) the gas undergoes an adiabatic expansion against an external pressure of p ex bar to a final pressure of p = 1 bar.
= 1
(a) isothermal, reversible: w = nRT ln( V f
/V i
). For constant T , p i
V w = nRT ln( p i
/ p f
) = (0.5 mol) × (8.314 J K -1 mol -1 i
= p f
V f
V f
/V i
= p i
/ p f
) × (300 K ) ln(3.25) = -1470 J
Δ U = 0 q = w = 1470 J Δ T = 0 Δ H = Δ U + nR Δ T = 0
(b) isothermal, against constant pressure: w = p ex
Δ V
3.837
×
We can find V i
and V f
from the perfect gas law:
V i
= nRT i
/p i
= (0.5 mol) × (8.314 J K -1 mol -1 ) × (300 K ) / 3.25
× 10 5 Pa =
10
V f
-3
-3
m
=
3 = 3.837 L nRT f
/p f
= (0.5 mol)
×
(8.314 J K -1 mol -1 )
×
(300 K ) / 1.0
×
10 5 Pa =
12.47
×
10 w
m 3 = 12.47 L
= -(1.0
×
10 5 Pa)( 12.47
×
10 -3 m 3 - 3.837
×
10 -3 m 3 ) = -863 J
Δ U = 0 q = w = -863 J Δ T = 0 Δ H = Δ U + nR Δ T = 0
(c) For reversible adiabatic expansion, p i
V i
γ = p f
V f
γ γ = C p ,m
/ C
V ,m
For a perfect monoatmic gas, C
V
T f f
= ( p
= ( V i i
/
/ p f
V f
) 1/
γ
) 1/ c
V i
V ,m
= (3.25 bar / 1.0 bar)
T i
= ( V i
/ V f
) γ
1
= (3/2) R ,
γ
= 5/3
3/5
×
3.837 L = 7.783 L
T i
= (3.837 L/7.783 L) 2/3
× (300 K) = 187.23 K
Δ T = -112.77 K q = 0 Δ U = w + q w = nC
= -703 J
V ,m
Δ T = -703 J
Δ H = Δ U + nR Δ T = -1172 J
(d) For adiabatic expansion against a constant external pressure: w ad
= p ex
Δ V = C
V
Δ T p ex
( V f
- V i
) = C
V
( T f
- T i
)
On the other hand, p f
V f
= nRT f nRT
V f
= f
− p ex p f
 nRT

 p f f
− V i

 = nC

V , m
( T f
− T i
)
Solving this equation for T f
gives T f
= p ex
V i
Since p nC
V , m
€ f
= p ex
= 1 bar, the formula can simplified as
+ nC
V , m
T i
+ p ex p f nR
2

T f
T f
= p ex
V i
C
/ n
V , m
+
+
C
R
V , m
T i
= {(1 bar × 3.837 L / 0.500 mol) + (3/2) × (0.0831447 L bar K -1
{3/2)
×
(0.0831447 L bar K -1
Δ
T = -83 K
mol -1 ) + (0.0831447 L bar K -1 w = nC
V ,m
Δ
T = -519 J
mol -1 ) × (300 K)} /
mol -1 )} = 217 K q = 0 Δ U = w + q = -519 J Δ H = Δ U + nR Δ T = -865 J
4. (16 pts) The standard specific internal energy of combustion of crystalline fullerene
C
60
was measured to be –36.0334 kJ g -1 at 298.15 K. Calculate the standard enthalpy of combustion and the standard enthalpy of formation of this fullerene.
Δ c
U ø
Δ c
U specific
= –36.0334 kJ g -1
= Δ c
U specific
× M (C
60
) = (–36.0334 kJ g -1 ) × 720.6 g mol -1 = -25965.67 kJ mol -1
Δ c
H ø = Δ c
U ø + Δ n g
RT
Combustion of C
60
: C
60
(s) + 60 O
2
(g) → 60 CO
2
(g)
Δ c
H ø =
Δ c
U ø = -25965.67 kJ mol -1
If one writes the reaction enthalpy for the combustion reaction,
Δ c
H ø = 60 ×Δ f
H ø (CO
2
Δ
Δ f f
H
H
ø
ø
(C
(C
60
60
,s) = 60 ×Δ f
H
,g) – 60
ø (CO
2
×Δ f
H ø (O
,g) – 60 ×Δ
2 f
,g) – Δ
H ø (O
2 f
H ø (C
,g) – Δ
60 c
Δ n
,s)
H ø g
= 0
,s) = 60
×
(-393.51) – 60
×
0 – (-25965.67) = 2355.07 67 kJ mol -1
5. (16 pts) The pressure of a given amount of a van der Waals gas depends on T and V : p
V nRT
nb
a
n
V
2
2
Derive and expression for d p in terms of d T and d V . Remember that if a function z depends on two variables x and y , its differential can be expressed as
dz =
 ∂ z

 ∂ x


 y dx +
 ∂ z

 ∂ y


 x dy
p
T
€
V
€
= dp = dp = nR
V
nb nR
 ∂ p

 ∂ T
V
nb



V dT
p
V
+
T
 ∂ p

 ∂ V
=
dT +
( nRT



T
V
nb )
2 dV nRT
( V
nb )
2
+
+
2 an 2
V
2 an 2
V
3
dV
3
3

€
6. (16 pts) The standard molar entropy of NH
3
(g) is 192.45 J K heat capacity is given by the equation C p ,m
= a + bT + c / T 2
-1 mol -1 at 298 K, and its
, where a , b , and c can be found in Table 2.2 of Data Section . Calculate the standard molar entropy at (a) 100 ° C and (b)
500
°
C.
S m
( ) = S m
( ) +
T
T i f
C p
T
S m
( ) = S m
( ) + a ln
 T

 T i f
S m
(373 K) = 201 J K
S m
(773 K) = 233 J K
-1
-1
mol
mol
-1
-1 dT

 +

= S m
( ) +
T f
T i a + bT + c / T 2
( f
− T i
) − c
2



T
1 f
2
−
T
1 i
T
2


 dT
4

Midterm Exam I
CHM 3410, Dr. Mebel, Fall 2006
1. (20 pts.) What mass of N
2
gas is present in a 50-liter container at 400 K under 20 atm of N
2
pressure if
(a) the gas is perfect;
(b) the gas obeys the van der Waals equation of state?
2. (20 pts.) One mole of argon at 25
°
C and 1 bar pressure is allowed to expand reversibly to a volume of 50 L (a) isothermally and (b) adiabatically. Assuming perfect gas behavior, calculate the final pressure, temperature, q, w, Δ U, and Δ H in each case.
1

3. (20 pts) A 1:3 mixture of CO and H
2
is passed through a catalyst to produce methane at
500 K.
CO(g) + 3 H
2
(g)
→
CH
4
(g) + H
2
O(g)
How much heat is liberated in producing a mole of methane? How does this compare with the heat obtained from combustion of a mole of methane at this temperature?
4. (20 pts.) (a) Calculate Δ S if 1 mol of liquid water is heated from 0 ° C to 100 ° C under constant pressure if C p
, m
= 75.291 J K
(b) The melting temperature of water at the pressure of interest is 0
°
C and the enthalpy of melting is 6.0095 kJ mol -1
-1 mol -1 .
. The boiling temperature is 100 vaporization is 40.6563 kJ mol -1 . Calculate Δ S
°
C and the enthalpy of
for the transformation
H
2
O(solid, 0 ° C) → H
2
O(gas, 100 ° C)
2

5. (20 pts) Derive an expression for the internal pressure,
Berthelot equation of state,
Use the fact that
T
T
, of a gas that obeys the p =
V m
RT
b
a
TV
2 m
is related to p , V , and
by the following formula:
T
= T
p
T
V
p
3

Midterm Exam I
CHM 3410, Dr. Mebel, Fall 2013
1. (20 pts.) Consider the system at T = 298 K, shown in the figure below:
He
2.00 L
1.50 bar
Ne
3.00 L
2.50 bar
Xe
1.00 L
1.00 bar
Initially, the gases are placed in individual compartments separated by walls (barriers).
Assuming ideal gas behavior, calculate the total pressure and the partial pressure of each component if the barriers separating the compartments are removed.
1

2. (20 pts.) Two ideal gas systems undergo reversible expansion under different conditions starting from the same p and V . At the end of the expansion, the two systems have the same volume. The first system has undergone adiabatic expansion and the second has undergone isothermal expansion. Which system will have the lower pressure?
Explain your answer both with and without using equations.
2

3. (20 pts) Assume the hypothetical reaction for fixing nitrogen biologically is
N
2
(g) + 3 H
2
O(l)
→
2 NH
3
(aq) + 3/2 O
2
(g) a. Calculate the standard reaction enthalpy and standard reaction internal energy change for such biosynthetic fixation of nitrogen at T = 298 K. For NH
3
(aq), ammonia dissolved in aqueous solution, use
Δ f
H ∅ = -80.3 kJ mol -1 .
b. In some bacteria, glycine is produced from ammonia by the reaction
NH
3
(g) + 2 CH
4
(g) + 5/2 O
2
(g) → NH
2
CH
2
COOH(s) + 3 H
2
O(l)
Calculate the standard reaction enthalpy and standard reaction internal energy change for the synthesis of glycine from ammonia at T = 298 K. For glycine, Δ f
H ∅ = -537.2 kJ mol -1 .
c. Calculate the standard reaction enthalpy for the synthesis of glycine from nitrogen, water, oxygen, and methane at = 298 K and 348 K. Use C p ,m
(glycine,s) = 99.14 J K -1 mol -1 .
3

4. (20 pts.) 1.75 moles of an ideal monoatomic gas are transformed from an initial state T =
750 K and p = 1.75 bar to a final state T = 350 K and p = 5.25 bar. Calculate
Δ
U ,
Δ
H , and
Δ S for this process.
4

5. (20 pts) Derive expressions for the internal pressure,
T
, of (a) an ideal gas and (b) van der Waals gas using the fact that
T
is related to p , V , and T by the following formula:
T
= T
p
p
T
V
Show that your result for van der Waals gas will tend to the result for the ideal gas as the
5