CHM 3410 – Problem Set 3
Due date: Wednesday, September 12th
Do all of the following problems. Show your work.
“Science walks on two feet, namely theory and experiment.” Robert Millikan
1) There are a number of interesting applications of adiabatic expansions in science. In this problem, we explore one
of these applictions.
Consider the situation pictured below. A dry parcel of air, initially at sea level, rises in the atmosphere.
Because the pressure of the Earth’s atmosphere decreases with increasing altitude, the air parcel will expand and
cool. To a first approximation, the process can be considered an adiabatic reversible expansion (because heat
exchange between the air parcel and the surrounding atmosphere is slow, and so can be neglected).
a) Let
zi = initial altitude = 0.00 km
pi = initial pressure = 1.00 atm
Ti = initial temperature = 290.0 K
zf = final altitude = 1.00 km
pf = final pressure = 0.882 atm
Tf = ?
For air,  = Cp/CV  7/5 = 1.400
Based on this information, find T f, the temperature of the air parcel at z = 1.00 km. (HINT: Begin with the
general relationship that applies to adiabatic reversible processes with constant 
piVi = pfVf
(1.1)
Transform equn 1.1 to an equation involving pressure and temperature, as done in class. Note that meteorologists
call the change in temperature that you find in your calculation the adiabatic lapse rate, defined as the expected rate
of change in temperature of a dry parcel of air as it rises in the atmosphere.)
b) Assume that your air parcel was humid and had a large partial pressure of water vapor. What will
happen to this water vapor as the air parcel moves up in altitude? Use your answer to explain why it rains.
2) Consider the adiabatic irreversible expansion of 1.00 mol of argon gas (Ar, CV,m = 3/2 R), from an initial
temperature Ti = 300.0 K and an initial pressure pi = 2.00 atm, against a constant external pressure equal to the final
pressure of the gas, pex = pf = 1.00 atm. Find q, w, U, and H for the process. (HINT: Since the process is
irreversible, you cannot use equn 1.1 above. In fact, however, this problem is more simple than an adiabatic
reversible expansion.)
3) The standard enthalpy of combustion of anthracene (C14H10(s)) is – 7163. kJ/mol. Based on this, and the data in
the Appendix of Atkins, find the standard enthalpy of formation for solid anthracene.
4) Calculate the standard enthalpy of formation of N2O5(g) using only the following data
(1)
2 NO(g) + O2(g)  2 NO2(g)
H = - 114.1 kJ/mol
(2)
4 NO2(g) + O2(g)  2 N2O5(g)
H = - 110.2 kJ/mol
(3)
N2(g) + O2(g)  2 NO(g)
H = + 180.5 kJ/mol
5) The Haber process, which converts nitrogen and hydrogen into ammonia, is one of the most important reactions in
industrial chemistry. The reaction may be written as
N2(g) + 3 H2(g)  2 NH3(g)
(5.1)
a) How is reaction 5.1 related to the formation reaction for ammonia?
b) Find Hrxn for reaction 5.1 at T = 25. C.
c) Find Hrxn for reaction 5.1 at T = 450. C, and assuming the constant pressure molar heat capacities are
constant. Use the values given in Table 2.8 of the Appendix of Atkins.
d) Because of the large change in temperature in this problem the assumption that the constant pressure
molar heat capacities are constant is likely invalid. Use the more rigorous expression for Cp,m
Cp,m = a + bT + (c/T2)
(5.2)
to find Hrxn for reaction 5.1 at T = 450. C. Values for a, b, and c for the reactant and products are given in Table
2.2 of the Appendix of Atkins.
e) Briefly compare your results from b, c, and d.
6) The coefficient of thermal expansion () and isothermal compressibility (T) can often be measured for real
substances by relatively simple experiments. Because of this, there is interest in finding general equations of state
that can be written in terms of  and T.
a) Write expressions for dV and dp given that V is a function of p and T (V  V(p,T) ) and p is a function of
V and T (p  p(V,T) ).
b) Deduce expressions for d lnV and d ln p in terms of the coefficient of thermal expansion and the
isothermal compressibility. (HINT: Recall that (dx/x) = d(ln x)).
 = (1/V) (V/T)p
(coefficient of thermal expansion)
(6.1)
T = - (1/V) (V/p)T
(isothermal compressibility)
(6.2)
Solutions.
1) For an adiabatic reversible expansion of an ideal gas with constant heat capacity
piVi = pfVf
 = Cp/CV = Cp,m/CV,m
In this problem it is convenient to rewrite the above expression in terms of pressure and temperature. Since
V = nRT/p
pi (nRTi/pi) = pf(nRTf/pf)
Ti = Tf
pi-1 pf-1
Taking the 1/ power of both sides of the above equation gives
Ti =
pi(-1)/
Tf
pf(-1)/
or, solving for Tf
Tf = (pf/pi)(-1)/ Ti
Note that this relationship was derived in class, and so you really did not need to rederive it here.
-1 = (1.400 – 1.) = 0.2857

1.400
and so
Tf = (0.882/1.000)0.2857 (290. K) = 279.8
This temperature drop, approximately - 10.C/km, is called the adiabatic lapse rate, and represents the rate
of cooling of a dry parcel of air as it moves up in altitude, due to adiabatic expansion.
b) The vapor pressure of water decreases as temperature decreases. Therefore, if the above parcel of air
was humid air (had a large partial pressure of water vapor), the cooling it would experience would eventually cause
the partial pressure of water vapor to reach the vapor pressure. At that point, any additional cooling would lead to
formation of either liquid or solid water, leading to precipitation.
2) The expansion is adiabatic, and so q = 0. Therefore, U = w.
The expansion is irreversible, against constant external pressure equal to the final pressure of the gas. The
gas is ideal, and the constant volume molar heat capacity is constant.
So
w = if pex dV = - pf if dV = - pf (Vf – Vi)
U = if nCV,m dT = nCV,m if dT = nCV.m(Tf – Ti)
So
nCV.m(Tf – Ti) = - pf (Vf – Vi)
Since the gas is ideal, V = nRT/p. If we substitute on the right side of the above equation, we get
nCV.m(Tf – Ti) = - pf [ (nRTf/pf) – (nRTi/pi) ]
There is a factor of n that can be removed. If we multiply out both sides of the above equation, we get
CV,mTf – CV,mTi = - RTf + RTi(pf/pi)
CV,mTf + RTf = CV,mTi + R(pf/pi)Ti
(CV,m + R)Tf = [ CV,m + R(pf/pi) ]Ti
Tf = [ CV,m + R(pf/pi) ] Ti
(CV,m + R)
Substituting into the above, and noting that CV,m = 3/2 R, we get
Tf = (3/2 + ½) (300.0 K) = 240.0 K
(3/2 + 1)
U = nCV.m(Tf – Ti) = (1.000 mol) (3/2) (8.3145 J/mol.K) (240.0 K – 300.0 K) = - 748.3 J
And so w = - 748.3 J
Finally, Cp,m = CV,m + R = 3/2 R + R = 5/2 R
So
H = if nCp,m dT = nCp,m if dT = nCp.m(Tf – Ti)
= (1.000 mol) (5/2) (8.3145 J/mol.K) (240.0 K – 300.0 K) = - 1247.2 J
3) The combustion reaction for anthracene is
C14H10(s) + 33/2 O2(g)  14 CO2(g) + 5 H2O()
Hrxn = Hcomb
From Hess’ law
Hcomb = [ 14 Hf(CO2(g)) + 5 Hf(H2O()) ] – [ Hf(C14H10(s)) ]
Hf(C14H10(s)) = [ 14 Hf(CO2(g)) + 5 Hf(H2O()) ] - Hcomb
= [ 14 (- 393.51) + 5 (- 285.83) ] – (- 7163) = + 225. kJ/mol
4) The formation reaction for N2O5(g) is
N2(g) + 5/2 O2(g)  N2O5(g)
It can be obtained as follows
2 NO2(g) + ½ O2(g)  N2O5(g)
H = ½ (- 110.2 kJ/mol)
2 NO(g) + O2(g)  2 NO2(g)
H = (1) (- 114.1 kJ/mol)
N2(g) + O2(g)  2 NO(g)
___________________________
N2(g) + 5/2 O2(g)  N2O5(g)
H = (1) (+ 180.5 kJ/mol)
______________________
Hf(N2O5(g)) = + 11.3 kJ/mol
5)
a) By definition the formation reaction is the reaction that produces one mole of a single product out of
elements in their standard (thermodynamically most stable) state. Reaction 5.1 produces two moles of ammonia out
of elements in their standard state, and so it is equal to two times the formation reaction for ammonia.
b) Based on the answer in part a, at T = 25.0 C
Hrxn = 2 Hf(NH3(g)) = 2 (- 46.11 kJ/mol) = - 92.22 kJ/mol
c) To do this problem we need an expression for Cp for the reaction. Since we are told we may assume
that the values for Cp,m for the reactants and products are all independent of temperature, then
Cp = 2 Cp,m(NH3(g)) – [ Cp,m(N2(g)) + 3 Cp,m(H2(g)) ]
= 2 (35.06) – [ (29.125) + 3 (28.824) ] = - 45.48 J/mol.K = - 45.48 x 10-3 kJ/mol.K
and so at T = 450. C, using the simplified form of Kirchoff’s law that applies when Cp is independent of
temperature
Hrxn (450. C) = Hrxn(25. C) + Cp (450. C – 25. C)
= - 92.22 kJ/mol + (- 45.48 x 10-3 kJ/mol) (425. K) = - 111.55 kJ/mol
Note that in the above calculation we have made use of the fact that the size of a degree is the same in the Centigrade
and Kelvin scale, so a temperature difference of 425. C is the same as a temperature difference of 425. K.
d) If we use
Cp,m = a + bT + (c/T2)
for the temperature dependence of the constant pressure molar heat capacities of the reactants and products, then we
may say
Cp = a + b T + (c/T2)
Substance
a (J/mol.K)
b (J/mol.K2)
c (J.K/mol)
N2(g)
28.58
3.77 x 10-3
- 0.50 x 105
H2(g)
27.28
3.26 x 10-3
0.50 x 105
NH3(g)
29.75
25.1 x 10-3
- 1.55 x 105
a = 2 (29.75) – [ (28.58) + 3 (27.28) ] = - 50.92 J/mol.K
b = 2 (25.1 x 10-3) – [ (3.77 x 10-3) + 3 (3.26 x 10-3) ] = 36.65 x 10-3 J/mol.K2
c = 2 ( - 1.55 x 105) – [ (- 0.50 x 105) + 3 (0.50 x 105) ] = - 4.1 x 105 J.K/mol
Hrxn (Tf) = Hrxn(Ti) + TiTf Cp dT
= Hrxn(Ti) + TiTf [a + b T + (c/T2)]dT
= Hrxn(Ti) + (a) (Tf – Ti) + (b/2) [ Tf2 – Ti2 ] – (c) [ (1/Tf ) - (1/Ti) ]
Ti = 25. C = 298. K
Tf = 450. C = 723. K
Hrxn (450. C) = - 92.22 kJ/mol + (- 50.92 x 10-3 kJ/mol.K) (723. K – 298. K)
+ (36.65 x 10-6 J/mol.K2/2) [ (723. K)2 – (298. K)2 ]
- (- 4.1 x 102 kJ.K/mol) [ (1/723. K) – (1/298. K) ]
- 92.22 + (- 21.641) + (7.952) – (0.809) = - 106.72 kJ/mol
If we assume that the result in part d is closest to the actual result, then not taking into account the
temperature dependence of Hrxn (part b) leads to an error of ~ 14%, and assuming a constant value of heat capacity
(part c) leads to an error of ~ 5%. So for this large temperature change taking the dependence of the heat capacity on
temperature into account is important.
6)
a)
dV = (V/p)T dp + (V/T)p dT
dp = (p/V)T dV + (p/T)V dT
b) If we divide both sides of the expression for dV by V, we get
dV = (1/V) (V/p)T dp + (1/V) (V/T)p dT
V
But
(1/V) (V/p)T = - T
dV = d ln(V)
V
(1/V) (V/T)p = 
Substituting, we get
d ln(V) = - T dp +  dT
If we divide both sides of the expression for dp by p, we get
dp = (1/p) (p/V)T dV + (1/p) (p/T)V dT
p
But
dp = d ln(p)
p
And
(p/V)T =
Also
(p/T)V (T/V)p (V/p)T = -1
1
=
1
= - 1
(V/p)T
V (1/V) (V/p)T
TV
And so (p/T)V = - (V/T)p = - (1/V) (V/T)p = 
(V/p)T
(1/V) (V/p)T
T
Substituting, we get
d ln(p) = (1/p) ( -1/TV) dV + (1/p) (/T) dT
d ln(p) = -
dV +
T (pV)
 dT
T p
Note that the expression for d ln(V) is much simpler than the expression for d ln(p). I have to admit to
being a bit bothered by this, but I don’t see any simpler way of rewriting the expression for d ln(p), and so I guess
that is just the way things work out.