CHM 3410 – Problem Set 2
Due date: Friday, September 10th
Do all of the following problems. Show your work.
1) Consider the work involved in the isothermal compression of a gas obeying the van der Waals equation of state.
a) Find a general expression for w, the work, when a gas obeying the van der Waals equation of state is
compressed isothermally and reversibly from an initial volume Vi to a final volume Vf.
b) The general expression for w when an ideal gas undergoes the same process is
w = - nRT ln(Vf/Vi)
(1.1)
Compare the value for work found from your expression in a and the ideal gas result given in b for the case of 1.000
mol of CO2, compressed isothermally and reversibly from an initial volume V i = 25.000 L to a final volume Vf =
1.000 L, at a temperature T = 300.0 K. The van der Waals coefficients for CO 2 are a = 3.610 L2.atm/mol2, b =
0.0429 L/mol.
2) Consider 1.000 mol of nitrogen gas (N2) at an initial temperature T = 300.0 K and an initial pressure p = 1.000
bar. The gas can be assumed ideal, and the constant pressure molar heat capacity of the gas, C p,m, is given by the
expression
Cp,m = a + bT + (c/T2)
a = 28.58 J/mol.K
b = 3.77 x 10-3 J/mol.K2
c = - 0.50 x 105 J.K/mol
Heat is added to the gas reversibly and under conditions of constant pressure until a final temperature T f = 400.0 K is
reached. Find q, w, U, and H for the process.
3) 1.000 mol of molecular oxygen (O2) is contained in a closed system at an initial temperature and pressure T i =
300.0 K and pi = 10.000 atm. The gas is expanded adiabatically and irreversibly against a constant external pressure
pex = pf = 4.000 atm. You may assume for this process that molecular oxygen obeys the ideal gas law, and that the
constant volume molar heat capacity is CV,m = 5/2 R = 20.786 J/mol.K. Find Tf, q, w, U, and H for the process.
4) Heating magnesium carbonate can cause it to decompose. The balanced decomposition reaction is
MgCO3(s)  MgO(s) + CO2(g)
(4.1)
Using the data in Table 2.8 of Atkins find Hrxn for the above process at T = 25.0 C and T = 100.0 C. You may
use the values for Cp,m in the data table and assume that Cp,m for each reactant and product is approximately constant
over the temperature range 25.0 C – 100.0 C.
Also do the following from Atkins:
Exercises
2.3b
A sample consisting of 2.000 mol He is expanded isothermally at 22.0 C from
22.4 dm3 to 44.8 dm3 (a) reversibly, (b) against a constant external pressure equal to the final pressure of the gas, and
(c) freely (against zero external pressure). For the three processes calculate q, w, U, and H.
(Note that 1 dm3 = 1 L, an exact relationship).
2.19b
From the following data, determine Hf for diborane, B2H6(g), at 298. K.
(1)
B2H6(g) + 3 O2(g)  B2O3(s) + 3 H2O(g)
H = - 2036. kJ/mol
(2)
2 B(s) + 3/2 O2(g)  B2O3(s)
H = - 1274. kJ/mol
(3)
H2(g) + 1/2 O2(g)  H2O(g)
H = - 241.8 kJ/mol
(Use the above reactions only in doing this problem).
Solutions.
1)
p=
a) The van der Waals equation of state is
nRT - an2
(V - nb)
V2
Work can be found using the relationship
w = - if pex dV
The process is reversible (so pex = p) and isothermal (so T = constant), and so
w = - if [ nRT/(V - nb) - an2/V2 ] = - [ nRT ln(V - nb) + (an2/V) ]if
or
w = - nRT ln[(Vf - nb)/(Vi - nb)] - an2 [ (1/Vf - 1/Vi) ]
As a check, note that when a = b = 0 we get w = -nRT ln(Vf/Vi), the ideal gas result, as expected.
b) We now consider the value for w for the isothermal reversible compression of one mole of CO 2 from an
initial volume Vi = 25.00 L to a final volume Vf = 1.000 L, at a temperature T = 300.0 K.
Ideal gas result
w = - (1.000 mol) (8.314 J/mol.K) (300.0 K) ln(1.000/25.00) = + 8.03 kJ.
van der Waals result
w = - (1.000 mol) (8.314 J/mol.K) (300.0 K) ln[(1.000 - 0.0429)/(25.000 - 0.0429) ]
- (3.610 L2.atm/mol2) (1.000 mol)2 [(1/1.000 L) - (1/25.00 L) ] (101.3 J/L.atm)
= 8134. J - 351. J = 7.78 kJ (a difference of ~ 3%)
2) The process is reversible and constant pressure, and the gas is ideal. Therefore
U = if nCV,m dT
CV,m = Cp,m - R
q = H = if nCp,m dT
So
H = if n [ a + bT + (c/T 2) ]dT
= n { a (Tf - Ti) + b/2 (Tf2 - Ti2) - c [ (1/Tf) - (1/Ti) ] }
= (1.000 mol) { (28.58 J/mol.K) (400.0 K - 300.0 K) + (3.77 x 10-3 J/mol.K2)/2 [ (400.0 K)2 2
(300.0 K) ]
- ( - 0.50 x 105 J.K/mol) [ (1/400.0 K) - (1/300.0 K) ] }
2858. J + 132.0 J - 41.7 J = + 2948. J
Since CV,m = Cp,m - R, it follows that U = H - nR (Tf – Ti)
So U = 2948. J - (1.000 mol) (8.314 J/mol.K) [ (400.0 K) - (300.0 K) ] = + 2117. J
As noted above, q = H = + 2948. J
Finally, from the first law, w = U - q = [ 2200. J – 3032. J ] = - 832 J
3) The process is adiabatic and irreversible, which means we cannot use the results derived in class, which were for
adiabatic and reversible expansions of ideal gases with constant heat capacity.
Since the process is adiabatic, q = 0
Therefore, U = w
U = if n CV,m dT
But
Since the gas is ideal, and CV,m = constant
U = n CV,m if dT = n CV,m (Tf - Ti)
Also
w = - if pex dV
The expansion is irreversible against a constant external pressure equal to the final pressure, and so
w = - pf if dV = - pf (Vf - Vi)
We face a problem similar to the one we faced when considering an adiabatic reversible expansion, we have one
equation ( U = w ) and two unknowns (T f, Vf). However, we can use the ideal gas law to rewrite Vf in terms of Tf.
U = w
nCV,m (Tf - Ti) = - pf (Vf - Vi)
nCV,mTf - nCV,mTi
=
pfVi - pfVf = pf (nRTi/pi) - nRTf = nR(pf/pi)Ti – nRTf
where we have used pfVf = nRTf and Vi = nRTi/pi in the above.
Rearranging, we get
nCV,mTf + nRTf = nR(pf/pi)Ti + nCV,mTi
nCp,mTf = nR(pf/pi)Ti + nCV,mTi
where we have used Cp,m = CV,m + R
Tf = (1/nCp,m) [ nR(pf/pi)Ti + nCV,mTi ] = (Ti/Cp,m) [ R(pf/pi) + CV,m ]
Now we can put in some numbers. Note that CV,m = 5/2 R = 20.79 J/mol.K ; Cp,m = 7/2 R = 29.10 J/mol.K
Tf = [ (300.0 K)/(29.10 J/mol.K) ] [ (8.314 J/mol.K)(4.0/10.0) + (20.79 J/mol.K) ] = 248.6 K
(NOTE: If we had left everything in terms of R we could show that in this example T f = 29/35 Ti )
To do the rest of the problem we will use results we have derived before, without rederivation.
U = w = nCV,m(Tf - Ti) = (1.000 mol)(20.79 J/mol.K)(248.6 K - 300.0 K) = - 1069. J
H = nCp,m(Tf - Ti) = (1.000 mol)(29.10 J/mol.K)(248.6 K - 300.0 K) = - 1497. J
4) At T = 25.0 C
Hrxn = [Hf(MgO(s)) + Hf(CO2(g)) ] - Hf(MgCO3(s))
= [ (- 601.70) + ( - 393.51) ] - ( - 1095.8) = + 100.6 kJ/mol
At T = 100.0 C, we use the general relationship
Hrxn(T2) = Hrxn(T1) + T1T2 Cp dT
Since we are told to assume that the heat capacities of the products and reactants are constant over the temperature
range of the problem, the above equation simplifies to
Hrxn(T2) = Hrxn(T1) + Cp (T2 - T1)
Cp = [ Cp,m(MgO(s)) + Cp,m(CO2(g)) ] – Cp,m(MgCO3(s))
= [ 37.15 + 37.11 ] – 75.52 = - 1.26 J/mol.K
And so Hrxn(100.0 C) = + 100.6 kJ/mol + (75.0 K) (- 1.26 x 10-3 kJ/mol.K) = 100.5 kJ/mol
(NOTE: This was not a very good reaction to use as an example, as the value for the enthalpy of reaction did not
change by very much. One generally observes larger changes than the above.)
Exercise 2.3 b
We will assume the gas is ideal. Since the processes are isothermal, U = H = 0 for all processes. Except
for the last process (the free expansion) we have encountered the above processes elsewhere, and discussed them in
class, and so will not rederive the results we will use.
Isothermal reversible
w = - nRT ln(Vf/Vi) = - (2.00 mol) (8.314 J/mol.K) (295.2 K) ln(44.8/22.4) = - 3403. J (so q = + 3403. J)
Isothermal irreversible
w = - pf (Vf - Vi) = - (nRT/Vf) (Vf - Vi) = nRT [ 1 – (Vi/Vf) ] = (2.00 mol) (8.314 J/mol.K) (295.2 K) [ 1 22.4/44.8 ]
= - 2454. J (so q = + 2454. J)
Free expansion
w = - if pex dV . But for a free expansion pex = 0, and so w = q = 0
Exercise 2.19b
The formation reaction is
2 B(s) + 3 H2(g)  B2H6(g)
We need some combination of processes that sum to this reaction.
B2O3(s) + 3 H2O(g)  B2H6(g) + 3 O2(g)
H = (-1) ( - 2036. kJ/mol) =
+ 2036. kJ/mol
2 B(s) + 3/2 O2(g)  B2O3(s)
H = (+1) ( - 1274. kJ/mol) =
- 1274. kJ/mol
3 H2(g) + 3/2 O2(g)  3 H2O(g)
_________________________________
H = (+3) (- 241.8 kJ/mol) =
- 725.4 kJ/mol
___________
2 B(s) + 3 H2(g)  B2H6(g)
Hf = +
37. kJ/mol
In the above we have used the fact that reversing the direction of a reaction changes the sign of H, and multiplying
a reaction by a constant value means that H must also be multiplied by the same constant. Both of these are
consequences of enthalpy being a state function.