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Exercise 1: Thermodynamics,
General:
• Like always, these sample solutions are only possible ways to solve the questions, your
way might be equaly good or better ( I might also use one of your solutions, if I
consider them better than mine)...
• If you find mistakes, PLEASE tell me.
• It might occur that I didn't understand your genious way and marked it as wrong, if
you can explain it to me and convince me, you get the credits.
• If you don't understand the solutions or what's wrong, don't hesitate to ask.
• Try to make your derivation/calculation way as clear as possible. It is sometimes very
hard for me to follow what you did.
• The correct use of units is very important. Keeping track of the units helps to check if
things are reasonable. I got some answers with just an unit at the end, where it is not
clear to me how you got there. Furthermore it is hard to follow for me, what you did
wrong then. If I get answers in future, where units are missing (and I mean also when
you put in the paramters, not just in the end), I will reduce the total credits of the
exercise by one.
Question 1.1
We are looking for F(V, P). Apply ”General Procedure”:
dF =M dV N dP replace dV by V  dT −V  dP (see table)
dF =M V  dT −V  dP N dP=M V  dT  N −V dP
relationship F T , P  dF =−S P V dT P V  dP
−S
compare coefficients: M V =− S P V   M =
−P
V
−S
−
PV = N −M V =N −
−P  N =
S

V
S

dF =−
P dV − S dP

V
if we consider an ideal, monoatomic gas, we can replace  and  by 1/T and 1 /P respectively and get:
S
T
S
V
dF =−P  1dV − S dP=−P  1 dV − S dP
V
P
V
R
Exercise 1: Thermodynamics, Question 1.2
Given:V 1=12 l ,V 2=6 l , P 1=1 atm , P 2=10 atm , T 1=273K , R=0,082 l⋅atm/ mol⋅K
Apply ideal gas law for monoatomic gas :
PV = RT , =1/ T , =1/ P , C P=5 /2 R , C V =C P −PV =C P −R=3/2 R
a) U =U  P , V ?
dU =MdP NdV replace dV =V  dT −V  dP
 dU =M −NV  dP NV  dT
dU T , P=C
 P −T  dP
P −PV  dT V


CV
0
equate like terms :C V =N V   N =C V
T
,
V
M −NV =0  M =C V
T
P
T
T
dPC V dV
P
V
Since U is a state function,  U is independent of the path of change. So we can do the integration
in two steps, first an isobaric change of volume and then an isochoric change of pressure or the other
way round. BUT since we keep one variable constant, T has to change dependently on V or P ,
respectively in the two intermediate steps.(this is an important concept, therefore you didn't get
PV
full credits, if you kept T as a constant) Therefore we replace it by T =
and C V =3/ 2 R or,
R
PV
if you include n, you have to do that also in c V , i.e. T =
and C V =3/2 nR!! ( n is cancelled out then)
nR
dU =C V
dU =3/2 V dP3/2 P dV
1. isochoric change of pressure: dV =0, V 1=12l
10atm
 U V =1 atm∫
atm
3/ 2V dP=[3/ 2V 1 P ]10
1 atm =3/2⋅12 l⋅9 atm=162 l⋅atm
6l
2. isobaric change of volume: U P =12 l∫ 3 /2 P dV =[3/2 P 2 V ]612l l=3 /2⋅10 atm⋅−6 l=−90 l⋅atm
 U tot = U V  U P =162 l⋅atm−90 l⋅atm=72 l⋅atm=7295 J
Exercise 1: Thermodynamics Question 1.3
Version I (as in question)
1000
atm
B
A
1 atm
100 l
20 l
Version II (more reasonable that you compress and get a higher pressure)
1000
atm
B
A
1 atm
20 l
100 l
Apply ideal gas law for monoatomic gas :
PV = RT , =1/ T , =1/ P , C P=5 /2 R ,
H =H  P ,V ?
dH =MdP NdV replace dV =V  dT −V  dP
 dH =M −NV dP  NV  dT
dH T , P=V
1−T  dPC P dT

0
equate like terms :C P =N V   N =
dH =C P
CPT
,
V
M −NV =0  M =C P
T
P
T
T
dPC P dV
P
V
T is not constant during the intermediate steps. Therefore we replace it by T =
if you include n you have to do that also in c P , i.e. T =
PV
and C P=5 /2 Ror,
R
PV
and c P =5/2 nR!! ( n is cancelled out then):
nR
dH =5 /2 V dP5/ 2 P dV
a) 1. isobaric change of volume and 2. isochoric change of pressure:
V2
P2
 U tot =5/2 P 1 V ∫ dV 5/ 2V 2 P ∫ dP=5/ 2 P 1 [V ]VV 5/ 2V 2 [ P ]PP
1
1
2
2
1
1
I. P 1=1 atm , P 2=1000 atm ,V 1=20l , V 2=100l ,
 U tot =5/2⋅1 atm⋅100l−20l5/ 2⋅100l⋅1000 atm−1 atm≈250 000 l atm≈25 MJ
II. P1=1 atm , P 2=1000 atm , V 1=100l ,V 2=20l ,
 U tot =5/2⋅1 atm⋅ 20l−100l5/ 2⋅20l⋅1000 atm−1 atm≈50 000 l atm≈5 MJ
b) 1.isochoric change of pressure and 2.isobaric change of volume:
P2
V2
 U tot =5/2 V 1 P ∫ dP5 /2 P 2 V ∫ dV =5/ 2V 1 [P ]PP 5/2 P 2 [V ]VV
1
1
2
2
1
1
I. P 1=1 atm , P 2=1000 atm ,V 1=20l , V 2=100l ,
 U tot =5/2⋅20l⋅1000 atm−1 atm 5/2⋅1000 atm⋅100l−20l≈250 000 l atm≈25 MJ
II. P1=1 atm , P 2=1000 atm , V 1=100l ,V 2=20l ,
 U tot =5/2⋅20l⋅1000 atm−1 atm5 /2⋅1000 atm⋅20l−100l≈50 000 l atm≈5 MJ
Exercise 1: Thermodynamics, Question 1.4
m=500 g of equimolar (i.e. X 1= X 2=0.5 ) solution from SnCl 4 and CCl 4,
T =500K , a SnCl =0.52 ,  G mix=−7.03 kJ
molar masses M SnCl =118,70 g /mol4⋅35,45 g /mol=260,50 g /mol
M ClCl =12,01 g / mol4⋅35,45 g / mol=153,81 g /mol
4
4
4
 Gk =n RT ln a k
 G mix=ntot R T ∑ X k ln a k
m
n tot =
 M SnCl M CCl /2
you have to divide the sum of molar masses by two, since you still have two seperate molecules.
m
 G mix=
R⋅T⋅1/2⋅ln a SnCl ln aCCl 
M SnCl M CCl /2
 M SnCl M CCl 
260,5 g / mol 153,81 g /mol 
ln a CCl =
⋅G mix −ln a SnCl =−
⋅7030 J −ln 0.52=−0.747
m RT
500g 8,314 J /molK 500K
a CCl =0.474
4
4
4
4
4
4
4
4
4
4
4