CHM 3400 – Problem Set 2
Due date: Monday, January 30th
Do all of the following problems. Show your work.
“Quite ordinary people can be good at science. To say this is not to depricate science but to appreciate ordinary
people.” – Peter Medawar
1) The coefficient of thermal expansion of a substance () measures the change in the volume of a substance as
temperature is increased under conditions of constant pressure. It is defined by the relationship
 = (1/V) (V/T)p
(1.1)
Find an expression for  for the following substances.
a) An ideal gas (pV = nRT).
b) A substance whose equation of state is given by the expression
V = V0 (1 + aT + bT2)
(1.2)
where a, b, and V0 are constants.
2) Consider 1.00 mol of argon (Ar), an ideal monatomic gas (CV,m = 12.47 J/mol.K) at an initial temperature and
pressure T = 300.0 K and p = 5.00 atm. Find q, w, U, and H for each of the following processes.
a) The gas is expanded isothermally and reversibly to a final pressure p f = 2.00 atm.
b) The gas is expanded isothermally and irreversibly against a constant pressure equal to the final pressure,
pex = pf = 2.00 atm.
c) The gas is heated reversibly at constant pressure to a final temperature T f = 360.0 K.
3) A substance obeys the following equation of state.
p = nRT ( 1 + (B/V) )
(3.1)
V
where B is a constant.
a) Find a general expression for w, the work, when a sample of the above substance is compressed
reversibly and isothermally from an initial volume Vi to a final volume Vf.
b) Find q and U for the above process, or briefly explain why we lack the information need to find them.
4) The constant pressure molar heat capacity of a substance is often written as
Cp,m = a + bT + c/T2
(4.1)
where a, b, and c are constants.
a) Using the above expression and the data in Table 2.2, p 59 of Atkins, find q when the temperature of 1.00
mol of graphite (C(s)) is changed at constant pressure from an initial value T i = 300.0 K to a final value T f = 400.0
K.
b) Find w, U, and H for the above process, or briefly explain why we lack the information to find them.
5) Over a limited temperature range the molar internal energy for a substance is given by the expression
Um(T) = a + bT + cT2
(5.1)
where a, b, and c are constants. Starting with eq 5.1 find an expression for C V,m, the constant volume molar heat
capacityof the substance.
Solutions.
1)
a) Since pV = nRT
V = nRT/p
Substituting into equn 1.1 gives
 = (1/V) (V/T)p = (1/V) /T)p (nRT/p)
= (1/V) (nR/p) = nR
pV
But for an ideal gas pV = nRT. If we substitute into the denominator, we get
 = nR = 1
nRT
T
b) V = V0 (1 + aT + bT2)
So
 = (1/V) (V/T)p = (1/V) /T)p [V0 (1 + aT + bT2) ]
= V0 (a + 2bT)
V
If we substitute for V, we get
=
2)
V0 (a + 2bT)
V0 (1 + aT + bT2)
=
(a + 2bT)
(1 + aT + bT 2)
a) The process is isothermal and the gas is ideal, and so U = H = 0
w = - if pex dV
Since the process is reversible, pex = p. Since the gas is ideal, p = nRT/V. Substituting, we get
w = - if (nRT/V) dV
But n and T are constant, and so nRT can be taken outside the integral, and so
w = - nRT if (1/V) dV = - nRT ln(Vf/Vi)
Finally, since the process is isothermal and the gas is ideal
piVi = pfVf, and so (pi/pf) = (Vf/Vi). Therefore
w = - nRT ln(pi/pf) = - (1.000 mol) (8.314 J/mol.K) (300.0 K) ln(5.00/2.00) = - 2285. J
Finally, since U = q + w = 0, q = - w = + 2285. J
b) The process is isothermal and the gas is ideal, and so U = H = 0
w = - if pex dV
The process is irreversible against a constant pressure equal to the final pressure, and so p ex = pf = const. We can
take pex outside the integral, to get
w = - pex if dV = - pex (Vf – Vi)
We may find Vi and Vf from the ideal gas law
Vi = nRT = (1.00 mol) (0.08206 L.atm/mol.K) (300.0 K) = 4.924 L
pi
(5.00 atm)
Vf = nRT = (1.00 mol) (0.08206 L.atm/mol.K) (300.0 K) = 12.309 L
pf
(2.00 atm)
and so
w = - (2.00 atm) (12.309 L – 4.924 L) = - 14.77 L.atm 101.3 J = - 1496. J
1 L.atm
Finally, since U = q + w = 0, q = - w = + 1496. J
c) The gas is ideal, and so
U = if nCV,m dT
H = if nCp,m dT
Since the gas is ideal
Cp,m – CV,m = R
Cp,m = CV,m + R = 12.47 J/mol.K + 8.314 J/mol.K = 20.78 J/mol.K
Since both CV,m and Cp,m are constant, we can take them outside the above integrals, to get
U = if nCV,m dT = nCV,m if dT = nCV,m (Tf – Ti)
= (1.00 mol) (12.47 J/mol.K) (360.0 K – 300.0 K) = + 748. J
H = if nCp,m dT = nCp,m if dT = nCp,m (Tf – Ti)
= (1.00 mol) (20.78 J/mol.K) (360.0 K – 300.0 K) = + 1247. J
The process is carried out at constant pressure, and so q = H = + 1247. J
Finally, since U = q + w, w = U – q = (748. J) – (1247. J) = - 499. J
NOTE: We could find w directly using w = - pex if dV = - pex (Vf – Vi), since we have sufficient
information to find both Vi and Vf.
3)
a) The process is reversible, and so pex = p.
So
w = - if pex dV = - if p dV = - if (nRT/V) [ 1 + (B/V) ]dV
Since the process is isothermal, T = const. We can take a factor of nRT outside of the integral, to get
w = - nRT if (1/V) [ 1 + (B/V) ] dV
= - nRT if [ (1/V) + (B/V2) ] dV
= - nRT { ln(V) – B (1/V) }if
= - nRT { ln(Vf/Vi) – B [ (1/Vf) – (1/Vi) ] }
b) From the first law U = q + w. We have an expression for w, so if we could find either U or q we
could use the first law to find the missing quantity. However, we do not have sufficient information to find either U
or q, and so neither of them can be found.
4)
a) The heating is reversible and at constant pressure, and so
q = if n Cp,m dT = n if Cp,m dT = n if [ a + bT + (c/T2) ]dT
= n { a (Tf – Ti) + (b/2) (Tf2 – Ti2) – c [ (1/Tf) – (1/Ti) ] }
Using the information from the problem and from Table 2.2, we get
q = (1.00 mol) { (16.86 J/mol.K) (400.0 K – 300.0 K) + (4.77 x 10-3 J/mol.K2)/2 [ (400.0 K)2 – (300.0 K)2 ]
- (- 8.54 x 105 J.K/mol) [ (1/400.0 K) – (1/300.0 K) ]
= (1.00 mol) { (1686. J/mol.K) + (167.0 J/mol.K) – 711.7 J/mol.K) }
= + 1141. J
b) Since the process is carried out at constant pressure, H = q = + 1141. J.
From the first law U = q + w. We have an expression for q, so if we could find either U or w we could use the
first law to find the missing quantity. However, we do not have sufficient information to find either U or w, and so
neither of them can be found.
NOTE: While the above is true, we know that the volume occupied by a solid does not change very much
with temperature. Since pressure is constant for the above process
w = - if pex dV = - pex (Vf – Vi)  0, and so U  q = 1141. J.
5) As we showed in class, CV = (U/T)V, and so
CV,m = (Um/T)V = /T)V (a + bT + cT2) = b + 2cT