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 CV,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, ΔU, and ΔH:
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 pex = 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 pex = 1
bar to a final pressure of p = 1 bar.
2
4. (16 pts) The standard specific internal energy of combustion of crystalline fullerene
C60 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:
 n2 
nRT
p=
− a 2 
V − nb  V 
Derive and expression for dp in terms of dT and dV. Remember that if a function z depends
on two variables x and y, its differential can be expressed as
€
 ∂z 
 ∂z 
dz =   dx +   dy
 ∂x  y
 ∂y  x
€
6. (16 pts) The standard molar entropy of NH3(g) is 192.45 J K-1 mol-1 at 298 K, and its
heat capacity is given by the equation Cp,m = a + bT + c/T2, 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.
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, ΔS < 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, ΔSsur > 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 = nRT1
(pV)2 = nRT2
Since n = 1 and T = t(°C) + a, we can write
(pV)1 = R(t1 + a)
(pV)2 = R(t2 + 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-1 (°C)-1 = 8.308 J K-1 mol-1
a = 22.98 / R = 22.98 / 0.082 L atm mol-1 (°C)-1 = 280.2°C
1
3. (16 pts) Consider the expansion of 0.500 mol of a perfect monoatomic gas with CV,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, ΔU, and ΔH:
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 pex = 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 pex = 1
bar to a final pressure of p = 1 bar.
(a) isothermal, reversible: w = -nRT ln(Vf/Vi). For constant T, piVi = pfVf Vf/Vi = pi/pf
w = -nRT ln(pi/pf) = (0.5 mol) × (8.314 J K-1 mol-1) × (300 K ) ln(3.25) = -1470 J
ΔU = 0
q = -w = 1470 J
ΔT = 0
ΔH = ΔU + nRΔT = 0
(b)
isothermal, against constant pressure: w = -pexΔV
We can find Vi and Vf from the perfect gas law:
Vi = nRTi/pi = (0.5 mol) × (8.314 J K-1 mol-1) × (300 K ) / 3.25×105 Pa =
3.837×10-3 m3 = 3.837 L
Vf = nRTf/pf = (0.5 mol) × (8.314 J K-1 mol-1) × (300 K ) / 1.0×105 Pa =
12.47×10-3 m3 = 12.47 L
w = -(1.0×105 Pa)( 12.47×10-3 m3 - 3.837×10-3 m3) = -863 J
ΔU = 0
q = -w = -863 J
ΔT = 0
ΔH = ΔU + nRΔT = 0
(c)
For reversible adiabatic expansion,
piViγ = pfVfγ
γ = Cp,m/CV,m
For a perfect monoatmic gas, CV,m = (3/2)R, γ = 5/3
Vf = (pi/pf)1/γ Vi = (3.25 bar / 1.0 bar)3/5 × 3.837 L = 7.783 L
Tf = (Vi/Vf)1/c Ti = (Vi/Vf)γ-1 Ti = (3.837 L/7.783 L)2/3 × (300 K) = 187.23 K
ΔT = -112.77 K
w = nCV,mΔT = -703 J
q=0
ΔU = w + q = -703 J
ΔH = ΔU + nRΔT = -1172 J
(d)
For adiabatic expansion against a constant external pressure:
wad = -pexΔV = CVΔT
-pex(Vf - Vi) = CV(Tf - Ti)
On the other hand, pfVf = nRTf
 nRT f

Vf =
− pex 
− Vi  = nCV ,m (T f − Ti )
p
pf
 f

p V + nCV ,m Ti
Solving this equation for Tf gives
T f = ex i
p nR
nCV ,m + ex
pf
€
nRTf
€
Since pf = pex = 1 bar, the formula can simplified as
€
2
Tf =
pexVi /n + CV ,m Ti
CV ,m + R
Tf = {(1 bar × 3.837 L / 0.500 mol) + (3/2)×(0.0831447 L bar K-1 mol-1)×(300 K)} /
{3/2)×(0.0831447 L bar K-1 mol-1) + (0.0831447 L bar K-1 mol-1)} = 217 K
ΔT = -83 K
w = nCV,mΔT = -519 J
q=0
ΔU = w + q = -519 J
ΔH = ΔU + nRΔT = -865 J
€
4. (16 pts) The standard specific internal energy of combustion of crystalline fullerene
C60 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.
ΔcUspecific = –36.0334 kJ g-1
ΔcUø = ΔcUspecific × M(C60) = (–36.0334 kJ g-1) × 720.6 g mol-1 = -25965.67 kJ mol-1
ΔcHø = ΔcUø + ΔngRT
Combustion of C60: C60(s) + 60 O2(g) → 60 CO2(g)
Δng = 0
ΔcHø = ΔcUø = -25965.67 kJ mol-1
If one writes the reaction enthalpy for the combustion reaction,
ΔcHø = 60×ΔfHø(CO2,g) – 60×ΔfHø(O2,g) – ΔfHø(C60,s)
ΔfHø(C60,s) = 60×ΔfHø(CO2,g) – 60×ΔfHø(O2,g) – ΔcHø
ΔfHø(C60,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:
 n2 
nRT
p=
− a 2 
V − nb  V 
Derive and expression for dp in terms of dT and dV. Remember that if a function z depends
on two variables x and y, its differential can be expressed as
€
 ∂z 
 ∂z 
dz =   dx +   dy
 ∂x  y
 ∂y  x
 ∂p 
 ∂p 
dp =   dT +   dV
∂T V
 ∂V T
 ∂p €

 ∂p 
nR
nRT
2an2
+ 3
  =
  =−
2
∂T V V − nb
∂V T
(V − nb) V

€
nR
nRT
2an2 
dp =
dT + −
+ 3 dV
2
 (V − nb)
V − nb
V 
€
€
3
€
6. (16 pts) The standard molar entropy of NH3(g) is 192.45 J K-1 mol-1 at 298 K, and its
heat capacity is given by the equation Cp,m = a + bT + c/T2, 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.
2
Cp
T f a + bT + c /T
Sm (T f ) = Sm (T f ) + ∫
dT = Sm (T f ) + ∫
dT
Ti T
Ti
T
 Tf 
c 1
1 
Sm (T f ) = Sm (T f ) + a ln  + b(T f − Ti ) −  2 − 2 
2  T f Ti 
 Ti 
Tf
€
Sm(373 K) = 201 J K-1 mol-1
Sm(773 K) = 233 J K-1 mol-1
€
4
Midterm Exam I
CHM 3410, Dr. Mebel, Fall 2006
1. (20 pts.) What mass of N2 gas is present in a 50-liter container at 400 K under 20 atm
of N2 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, ΔU, and ΔH in each case.
1
3. (20 pts) A 1:3 mixture of CO and H2 is passed through a catalyst to produce methane at
500 K.
CO(g) + 3 H2(g) → CH4(g) + H2O(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 ΔS if 1 mol of liquid water is heated from 0°C to 100°C under
constant pressure if Cp,m = 75.291 J K-1 mol-1.
(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. The boiling temperature is 100°C and the enthalpy of
vaporization is 40.6563 kJ mol-1. Calculate ΔS for the transformation
H2O(solid, 0°C) → H2O(gas, 100°C)
2
5. (20 pts) Derive an expression for the internal pressure,
Berthelot equation of state,
π T , of a gas that obeys the
RT
a
−
Vm − b TVm2
Use the fact that π T is related to p, V, and€
T by the following formula:
 ∂p 
πT = 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
N2(g) + 3 H2O(l) → 2 NH3(aq) + 3/2 O2(g)
a. Calculate the standard reaction enthalpy and standard reaction internal energy change
for such biosynthetic fixation of nitrogen at T = 298 K. For NH3(aq), ammonia dissolved
in aqueous solution, use ΔfH∅ = -80.3 kJ mol-1.
b. In some bacteria, glycine is produced from ammonia by the reaction
NH3(g) + 2 CH4(g) + 5/2 O2(g) → NH2CH2COOH(s) + 3 H2O(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, ΔfH∅ = -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 Cp,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 ΔU, ΔH, and
ΔS for this process.
4
5. (20 pts) Derive expressions for the internal pressure,
π T , of (a) an ideal gas and (b) van
π T is related to p, V, and T by the following formula:
 ∂p 
πT = T  − p
 ∂T V
€
Show that your result for van der Waals gas will tend to the result for the ideal gas as the
volume increases and
€ the attractive interaction between molecules decreases.
der Waals gas using the fact that
€
5