CHM 3410 – Problem Set 2
Due date: Friday, September 11th
Do all of the following problems. Show your work.
“Thermodynamics is a funny subject. The first time you go through it, you don't understand it at all. The second time you go through it, you think you understand it, except for one or two small points. The third time you go through it, you know you don't understand it, but by that time you are so used to it, it doesn't bother you anymore.” - Arnold Sommerfeld
1) The coefficient of thermal expansion, a, measures the change in volume of a substance when the temperature of the substance is changed under conditions of constant pressure. The coefficient is defined as
a = (1/V) (¶V/¶T)p
Find an expression for a for the following
a) An ideal gas ( pVm = RT )
b) A van der Waals gas (p = RT/(Vm - b) - a/Vm2 ). This one is a bit tricky. Show that your result for a van der Waals gas reduces to the result for an ideal gas by setting a = b = 0. This serves as a check of your answer.
c) A solid or liquid obeying the equation of state
Vm = Vm* + a (T - T*) + b(T - T*)2
where we assume a constant pressure p = 1.000 bar. T* represents a reference temperature, Vm* is the molar volume of the substance at the reference temperature, and a and b are temperature independent constants.
2) Consider one mole of an ideal monatomic gas (recall that for a monatomic gas CV,m = 3R/2 = 12.472 J/molK, Cp,m = 5R/2 = 20.786 J/mol.K). The gas is initially at a pressure p = 4.000 atm and a temperature T = 400.0 K. Find q, w, DU, and DH for each of the following processes:
a) An isothermal reversible compression of the gas to a final pressure pf = 10.000 atm.
b) An reversible cooling of the gas from Ti = 400.0 K to Tf = 300.0 K, at a constant pressure p = 4.00 atm.
c) An adiabatic reversible compression of the gas to a final pressure pf = 10.000 atm.
d) An adiabatic irreversible expansion of the gas against a constant pressure pex = pf = 1.000 atm.
3) Find a general expression for w (work) when 1.000 mol of a gas obeying the van der Waals equation of state undergoes an isothermal reversible expansion from an initial volume Vi to a final volume Vf. Show that your answer reduces to the result for an ideal gas ( w = - nRT ln(Vf/Vi) ) by setting a = b = 0. This serves as a check of your answer.
4) An unsaturated hydrocarbon contains at least one carbon-carbon double or triple bond. Such compounds can be saturated by reaction with hydrogen(H2), a process called hydrogenation.
Using the data in Table 2.5 and 2.7 of Atkins, find the enthalpy change for the hydrogenation of ethene, propene, and acetylene (ethyne) at T = 25.0 °C. For acetylene also find the enthalpy of hydorgenation at T = 100.0 °C. You may assume that the heat capacities of the reactants and products are approximately independent of temperature over the range T = 25.0 °C to T = 100.0 °C.
Also do the following from Atkins:
Exercises 2.18 a Cyclopropane and propene both have the molecular formula C3H6. The book has an error here, since to do this problem you will need to use the enthalpy of formation for H2O() and not H2O(g). Note that both propene and cyclopropane are gases for standard conditions and T = 25.0 °C. You may check your answer against the value for DH°f(C3H6(g)) that is given in Table 2.5 of Atkins.
Problems 2.10 (The problem is asking you to fit the combustion data to an equation of the form
DH°c = k Mn
where M is the molecular weight of the alkane, and k and n are constants found by carrying out a best fit to the data. Think about a way of manipulating the above equation to give you a relationship from which k and n can easily be found)
2.14 Dy = dysprosium, a rare earth (lanthanide) element.
Solutions
1) As discussed in class, we note that a = (1/V) (¶V/¶T)p = (1/Vm) (¶Vm/¶T)p
a) For an ideal gas, pVm = RT, and so Vm = RT/p
(¶Vm/¶T)p = ¶/¶T)p RT/p = R/p
and so a = 1 R But pVm = RT, so
Vm p
a = R = 1
RT T
b) For a van der Waals gas, p = RT - a
(Vm - b) Vm2
It is difficult to solve this equation for Vm, so instead we will solve for T, and then make use of the fact that
(¶Vm/¶T)p = 1/(¶T/¶Vm)p
to find the partial derivative we need.
p = RT - a Add a/Vm2 to both sides
(Vm - b) Vm2
p + a = RT Multiply both sides by (Vm - b)/R
Vm2 (Vm - b)
T = p (Vm - b) + a (Vm - b)
R R Vm2
So (¶T/¶Vm)p = ¶/¶Vm)p p (Vm - b) + a (Vm - b)
R R Vm2
= p + a - 2a (Vm - b)
R R Vm2 R Vm3
Before inverting this it is convenient to place everything over a common denominator, so
(¶T/¶Vm)p = pVm3 + a Vm - 2a (Vm - b) = [ pVm3 + a Vm - 2a (Vm - b) ]
RVm3 RVm3 RVm3 R Vm3
So (¶Vm/¶T)p = 1/(¶T/¶Vm)p = R Vm3
[ pVm3 + a Vm - 2a (Vm - b) ]
So a = (1/Vm) (¶Vm/¶T)p = 1 R Vm3 = R Vm2
Vm [ pVm3 + a Vm - 2a (Vm - b) ] [ pVm3 - a Vm + 2ab ]
This is not a very simple result (!). One way we can test this result is to see what happens if we set a = b = 0. We would expect in that case to get the result previously obtained for an ideal gas (since the van der Waals equation reduces to the ideal gas law when a = b = 0).
So, if a = b = 0, then
a = R Vm2 = R = R = 1
p Vm3 p Vm RT T
which is the result previously obtained for an ideal gas. This doesn't prove we have done everything correctly, but it does show that the result checks out when we set a = b = 0.
c) For this empirical equation of state, Vm = Vm* + a (T - T*) + b(T - T*)2
so (¶Vm/¶T)p = ¶/¶T)p Vm* + a (T - T*) + b(T - T*)2 = a + 2b (T - T*)
and so a = 1 [ a + 2b (T - T*) ].
Vm
We can get rid of Vm by substituting using the original equation to get
a = [ a + 2b (T - T*) ]
[ Vm* + a (T - T*) + b(T - T*)2 ]
2) a) The gas is ideal and the process is isothermal, and so DU = DH = 0.
The process is reversible and the gas is ideal, pex = p = nRT
V
w = - òif pex dV = - òif p dV = - òif nRT dV = - nRT òif dV = - nRT ln(Vf/Vi)
V V
Since this is an ideal gas and an isothermal process, we know (from Boyle's law) that (Vf/Vi) = (pi/pf), so
w = - nRT ln(Vf/Vi) = - nRT ln(pi/pf) = + nRT ln(pf/pi)
= (1.000 mol) (8.3145 J/mol.K) (400.0 K) ln(10.000/4.000)
= + 3047. J
Finally, since DU = q + w = 0 , q = - w = - 3047. J
b) The process is carried out at constant pressure, and the gas is ideal, and so
DH = òif n Cp,m dT DU = òif n CV,m dT
Since the gas is an ideal monatomic gas, Cp,m is constant, and so may be taken outside the integral, to give
DH = òif n Cp,m dT = n Cp,m òif dT = n Cp,m (Tf - Ti) DU = òif n CV,m dT = n CV,m òif dT = n CV,m (Tf - Ti)
So
DH = (1.000 mol) (20.786 J/mol.K) (300.0 K - 400.0 K) DU = (1.000 mol) (12.472 J/mol.K) (300.0 K - 400.0 K)
DH = - 2079. J DU = - 1247. J
The process is constant pressure, and so q = DH = - 2079. J
Finally, from the first law
DU = q + w , and so w = DU - q = ( - 1247. J) - (- 2079. J) = + 832. J
c) The process is adiabatic, and so q = 0
For an adiabatic reversible expansion or compression of an ideal monatomic gas, we may say (as proved in class)
pi Vig = pf Vfg g = Cp,m/CV,m= (5R/2)/(3R/2) = 5/3
While we could use the above equation (since we know pi and pf and have enough information to calculate Vi from the ideal gas law) it is convenient to rewrite this equation in terms of p and T.
Vi = nRTi Vf = nRTf
pi pf
If we use these relationships to substitute for Vi and Vf, and cancel the common factor (nR)g, we get
pi Tig = pf Tfg Combine the pressure terms
pig pfg
pi(1-g) Tig = pf(1-g) Vfg Take both sides to the 1/g power
pi(1-g)/g Ti = pf(1-g)/g Tf Solve for Tf
Tf = Ti (pi/pf)(1-g)/g
Now, (1-g)/g = (1 - 5/3) / (5/3) = - (2/5) = - 0.400
So Tf = (400.0 K) (4.000/10.000)-0.400 = 577.1 K
We may use the expressions previous found (in part b) for DU and DH
DU = n CV,m (Tf - Ti) = (1.000 mol) (12.472 J/mol.K) (577.1 K - 400.0 K) = + 2209. J
DH = n Cp,m (Tf - Ti) = (1.000 mol) (20.786 J/mol.K) (577.1 K - 400.0 K) = + 3681. J
Finally, from the first law, DU = q + w. Since q = 0, w = DU = + 2209. J
d) The process is adiabatic, and so q = 0
The process is irreversible against a constant external pressure pex = pf, so
w = - òif pex dV = - pex òif dV = - pf (Vf - Vi)
The gas is ideal and monatomic, and so (since CV,m is constant)
DU = òif n CV,m dT = n CV,m òif dT = n CV,m (Tf - Ti)
Since q = 0, it follows that DU = w, and so
n CV,m (Tf - Ti) = - pf (Vf - Vi) = pf Vi - pf Vf If we multiply the first term on the right by pi/pi, then
n CV,m (Tf - Ti) = (pf/pi) (pi Vi) - (pf Vf) But pi Vi = nRTi, pf Vf = nRTf, so
n CV,m (Tf - Ti) = n CV,m Tf - n CV,m Ti
= (pf/pi) nRTi - nRTf Divide both sides by n
CV,m Tf - CV,m Ti = (pf/pi) RTi - RTf Collect terms in Tf on the left, and in Ti on the right
CV,m Tf + R Tf = (pf/pi) RTi + CV,m Ti Recall that CV,m + R = Cp,m, then
Cp,m Tf = [ (pf/pi) R + CV,m ] Ti
Tf = [ (pf/pi) R + CV,m ] Ti = [ (1.000/4.000) (8.3145 J/mol.K) + (12.472 J/mol.K) ] (400.0 K) = 280.0 K
Cp,m (20.786 J/mol.K)
As we did in part c, we may use our previous result from part b to find DU and DH
DU = n CV,m (Tf - Ti) = (1.000 mol) (12.472 J/mol.K) (280.0 K - 400.0 K) = - 1497. J
DH = n Cp,m (Tf - Ti) = (1.000 mol) (20.786 J/mol.K) (280.0 K - 400.0 K) = - 2494. J
Finally, since q = 0 it follows from the first law that w = DU = - 1497. J
3) For a van der Waals gas p = nRT - a n2
(V - nb) V2
In general w = - òif pex dV. Since the process is reversible, pex = p, so
w = - òif [ (nRT)/(V - nb) ] - (an2/V2) dV
The process is isothermal, and so T is constant. If we do the integral, we get
w = - nRT ln[ (Vf - nb)/(Vi - nb) ] - an2 [ (1/Vf) - (1/Vi) ]
We may manipulate the terms on the right to get rid of the - signs, to get
w = nRT ln [(Vi - nb)/(Vf - nb) ] + an2 [ (1/Vi) - (1/Vf) ]
As a check, if we set a = b = 0, we get
w = nRT ln(Vi/Vf) = - nRT ln(Vf/Vi) , the ideal gas result. That doesn't prove our result is correct, but does show it is consistent with what we expect.
4) The reactions of interest are
ethene C2H4(g) + H2(g) ® C2H6(g)
propene C3H6(g) + H2(g) ® C3H8(g)
acetylene C2H2(g) + 2 H2(g) ® C2H6(g)
The data we need are in Table 2.5 of Atkins (recall that DH°f = 0 for an element in its standard state).
Compound DH°f (kJ/mol) Cp,m (J/mol.K)
C2H2(g) + 226.73 43.93
C2H4(g) + 52.26 43.56
C2H6(g) - 84.68 52.63
C3H6(g) + 20.42 63.89
C3H8(g) - 103.85 73.5
H2(g) 0.0 28.824
The two equations we need are as follows ;
At T = 25.0 °C DH°rxn = [ S (DH°f (products)) ] - [ S (DH°f (reactants)) ]
At T = 100.0 °C DH°f(T2) = DH°f (T1) + òT1T2 DCp dT
@ DH°f (T1) + DCp (T2 - T1)
ethene C2H4(g) + H2(g) ® C2H6(g)
T = 25.0 °C ( - 84.68) - ( + 52.26) = - 136.94 kJ/mol
T = 100.0 °C ( - 136.94) + [ ( 52.63) - (43.56 + 28.824) ] x 10-3 ( 100.0 - 25.0 ) = - 138.42 kJ/mol
propene C3H6(g) + H2(g) ® C3H8(g)
T = 25.0 °C ( - 103.85) - ( + 20.42) = - 124.27 kJ/mol
T = 100.0 °C ( - 124.27) + [ ( 73.5) - (63.89 + 28.824) ] x 10-3 ( 100.0 - 25.0 ) = - 125.71 kJ/mol
acetylene C2H2(g) + 2 H2(g) ® C2H6(g)
T = 25.0 °C ( - 84.68) - ( + 226.73) = - 311.41 kJ/mol
T = 100.0 °C ( - 311.41) + [ ( 52.63) - (43.93 + 2 (28.824)) ] x 10-3 ( 100.0 - 25.0 ) = - 315.08 kJ/mol
Exercise 2.18 a
The combustion reaction is
C3H6(g) + 9/2 O2(g) ® 3 CO2(g) + 3 H2O() DH°c = - 2091. kJ/mol
But DH°c = [ 3 (DH°f(CO2(g))) + 3 (DH°f(H2O( ))) ] - ( DH°f(C3H6(g)) )
DH°f(C3H6(g)) = [ 3 (DH°f(CO2(g))) + 3 (DH°f(H2O( ))) ] - DH°c
= [ 3 ( - 393.51) + 3 (- 285.83) ] - ( - 2091. ) = + 53. kJ/mol
Note the value from Table 2.5 is + 53.30 kJ/mol, which agrees to within the uncertainly of the calculation.
The isomerization reaction is
cyclopropane ® propene
DH°iso = DH°f(propene) - DH°f(cyclopropane)
= (20.42) - (53.30) = - 32.88 kJ/mol
Problem 2.10
We are fitting data to the equation DH°c = k Mn where M = molecular mass
If we take the logarithm of both sides of this equation, we get
ln (DH°c) = ln(k) + n ln(M)
If we plot ln (DH°c) vs ln(M) we expect a straight line with slope = n and intercept = ln(k). Data are given below, from Table 2.5 of Atkins
Compound M(g/mol) ln(M) DH°c (kJ/mol) ln (| DH°c |)
CH4(g) 16.04 2.775 - 890. 6.791
C2H6(g) 30.07 3.404 - 1560. 7.352
C3H8(g) 44.10 3.786 - 2220. 7.705
C4H10(g) 58.13 4.063 - 2878. 7.965
C5H12(g) 72.15 4.279 - 3537. 8.171
C6H14() 86.18 4.456 - 4163. 8.334
C7H16() 100.21 4.607 - 4817. * 8.480
C8H18() 114.23 4.738 - 5471. 8.607
* Calculated from the data in the table, as no value is given in the table for DH°c
The data are plotted below.
Based on the data, I get for the best fitting line
ln (| DH°c |) = 4.206 + 0.927 ln(M)
For C10H22() (M = 142.28 g/mol), the predicted values are
ln(| DH°c |) = 8.802
DH°c = - 6648. kJ/mol (experimental value is DH°c = - 6830. kJ/mol)
Severl comments. For consistency we really should only use data for n-alkanes (straight chain alkanes and not branched alkanes) and for initial states that are all in the same phase (so there is no contribution from the enthalpy for the phase transition). I have done the first of these, but the last three compounds in the fitting are liquids, while the first five are gases. The experimental value for C7H16() was calculated from the enthalpy of formation data in Atkins. There is no value for DH°c in Atkins for C10H22(), and so the experimental value quoted above is from another textbook.