1
P
dS    dU    dV
T 
T 
 ___ 
 ___ 
dS  
___


 ___  dP
___




Take Home Test
1)
 ___ 
 ___ 
 ___ 
dH  
___  
___  ___  


 ___
 ___  ___
 ___ T
 ___  ___
2)
thus:
but since:
then:
dU  w
___
dV
___
2 1
2 ___
___ 
dT   ___ 
dV
1 ___
1 ___
___ dT   ___  
___ ln  ___/___    ___ ln  ___/___ 
dS 

___
1  ___ 
___ 
 ___  ___


___
___  ___  ___

 S 
 S 
dS  
___  

 ___
___
___

 ___

 ___

1  ___ 
 S 
 ___   ___  ___   ___ 

T
 ___


V
1
ST    ___   ___  dP
___
T
P

d ___
___
ST   ___ ln  ___/___   ___ ln  ___/___ 
___
 S 
 ___   ___

P
S ___
   ___  CV  ln  ___/___  but since:
___
ln  ___/___     ___  1 ln  ___/___  ___  C
V
T2  ___ 

T1  ___ 
___ 1
3) If you put 11 for “thermodynamic potential”, you were close, but still wrong. dU is a
change in thermodynamic potential.
_13_closed system
_1_adiabatic system
__equilibrium system
_7_ CV
__Maxwell relation
_10_compressibility factor
_8 and/or 12_state variable
__thermodynamic potential
__Third Law
_3_Enthalpy
__Entropy
__extent of reaction
_4_Affinity
1. does not exchange energy as heat
2. Entropy of universe is increasing
3. Entropy multiplied by absolute temp. during phase change
4. sum of chemical potentials times their stoichiom. coeff.
5. A form of energy that can convert to other forms
6. Above this value the P-V curve is always single valued
7. Energy that changes a mole of substance’s temperature by 1K
8. Nk
9. Measure of particulate motion
10. pV/nRT
11. dU=TdS-PdV
12. V, T
13. Exchanges energy, but not matter
4) How might one measure the molecular mass of a gas?
See Problem 1.2. One way is to weigh a container of known volvume containing a known
pressure of the pure gas at a known temperature, then to evacuate it and reweigh it. The
difference in mass divided by PV/RT would equal the mass per mole i.e. the molecular
mass.
5) The metabolism of a human generates approximately 100 W of heat. Assume a person is
in a well-insulated sleeping tent which contains about 6000.0 L of air (which consists of
diatomic molecules) and that 50% of the heat generated by the person's body is absorbed
by the air in the tent. Assuming no heat losses, what would the increase in the tent's air
temperature be due to heat generated by the person in 30 minutes?
Watt=J/s 30 min = 1800 s (i.e. q=180,000 J, or qair=90,000 J). ΔT= qair /nCp, where n is the
number of moles, i.e. n=(6000 L)/(22.4 mol/L)=268 mol. So, ΔT= 11.5K. (If you forget
Cp, just remember that Cv is the one tied to internal energy, which deals with degrees of
freedom. There are 5 degrees of freedom for a diatomic gas—three for moving and two
for rotating and zero for bending and stretching since that doesn’t happen at room
temperature—so Cv for this would be 5/2RT. Then remember that Cp-Cv=R. So Cp is
7/2R).
6) For the reaction: C6H12O6 + 6O2  6CO2 + 6 H2O, set up an equation relating ΔHf for
all of the components.
ΔHf,products - ΔHf,rxnts = ΔHrxn
6ΔHf,CO2 + 6ΔHf,H2O - ΔHf,C6H12O6 = ΔHrxn
7) What is the maximum work that can be obtained from 500 J of heat supplied to a steam
engine with a high temperature reservoir at 100 ˚C if the condenser is at 0 ˚C.
q=500 J, effic.=1-Tcold/Thot=w/q=1-273/373=0.268 so w=500*0.268=134 J
8) What is the difference between a chemical potential and a thermodynamic potential.
A chemical potential is mu (µ) and a thermodynamic potential is one of the faces on the
thermodynamic octahedron (U, H, G, etc.). The chem. pot. has to do with how much a
particular substance “wants to be somewhere” in comparison to somewhere else. For
example, if the mu of pure liquid water is high relative to the mu of pure gaseous water,
the will be an energy cost (e.g. in Joules) of moving one molecule of water from the gas
phase to the liquid phase. Once the system has reached equilibrium (and the P, V, and T
are constant and no longer changing), the mu’s of all the components are equal to each
other.
9) Bike racers often fill their tires with helium. Why don’t their tire valves cool when they
let the helium escape?
Short answer: Helium is a gas with very little intermolecular attraction energy, so expanding
this gas at room temperature does not require investing thermal energy from the
surroundings. For other gases this would require an energy investment and would be an
endothermic process
10) Which part of the van der Waals equation deals with intermolecular attractive forces?
a (that’s all you need for full credit!)
11) Considering collision of molecules of mass m and average speed vavg with the walls of a
container, show that the pressure p exerted is:
1
p  mnv 2avg in which n=N/V the number of molecules per unit volume
3
Here is the full derivation from Dr. Kondepudi:
Let us assume the average speed of the gas molecules is vavg.
We denote its x, y and z
components of the by vxavg, vyavg, and vzavg. Thus
v 2avg  v2xavg  v2yavg  v 2z avg
(1.6.1)
Because all directions are equivalent, we must have,
v 2x av g  v 2y av g  v 2zavg  (v2avg / 3)
(1.6.2)
The following quantities are necessary for obtaining the expression for pressure:
N = amount of gas in moles
V = gas volume
M = Molar mass of the gas
m = mass of a single molecule = M/NA
n = number of molecules per unit volume = NNA/V
NA= Avogadro number
(1.6.3)
Now we can calculate the pressure by considering the molecular collisions with the wall. In
doing so we will approximate the random motion of molecules with molecules moving with an
average speed vavg. (A rigorous derivation gives the same result.) Consider a layer of a gas, of
thickness x, close to the wall of the container (see Fig. 1.8). When a molecule collides with the
wall, which we assume is perpendicular to the x-axis, the change in momentum of the molecule in
In the layer of thickness x and area A, because of the
the x-direction equals 2mvxavg.
randomness of molecular motion, about half the molecules will be moving towards the wall, the
rest will be moving away from the wall. Hence, in a time t=(x/vx) about half the molecules in
the layer will collide with the wall.
The number of molecules in the layer = (xA) n
The number of molecules colliding with the walls = (xA) n/2
Since each collision imparts a momentum 2mvxavg, in a time t, the total momentum imparted to
the wall = 2mvxavg (xA) n/2.
Thus the average force F on the wall of area A is:
F

Momentum imparted 2mv xavgxA n mv xavg xAn


 mv 2xavg nA .
t
t
2
(x /v xavg )
(1.6.4)
Pressure, p, which is the force per unit area, is thus:
p
F
 mv 2x av gn
A
(1.6.5)
Since the direction x is arbitrary, it is better to write this expression in terms of the average speed
of the molecule rather than its x-component. By using (1.6.2) and the definitions (1.6.3), we can
write the pressure in terms the macroscopic variables M, V and N:
1
1 N
p  mnv 2avg  M v2avg
3
3 V

(1.6.6)
12) What enthalpies could you combine in order to get the heat of aqueous ion formation for
Cl-? What convention does this use?
½H2(g)  H(g)
H(g)  H+(g) + e–
½Cl2(g)  Cl(g)
Cl(g) + e–  Cl–(g)
H+(g) + Cl–(g)  HCl (g)
this part is the ΔHf,HCl
HCl (g)  HCl (aq)
this part is the ΔHsoln,HCl
ΔHf,HCl + ΔHsoln,HCl = ΔHf,HCl(aq)







–

and: H fo Claq
 H fo HClaq  H fo H aq
 H fo HClaq

The convention is that ΔH˚f,(H+aq) =0
13) Draw the carnot engine’s cycle on an S versus T graph. Label the parts of the cycle
according to what is happening. Write an equation for the heat input and another for the
heat output.
Your drawing should be a rectangle on a graph where the y-axis is S and x-axis is T. The
isotherms are the vertical sides, the adiabats are the horizontal sides. The heat input is:
nRTH ln(VA / VB )
and the heat output is:
nRTL ln(VA / VB )
14) What is the entropy of mixing, ΔSmix, for 1 mole of N2 added to 1 mole of O2?
ΔSmix = Ntot R(XalnXa+XblnXb)=11.53 J/K
15) How does Gibbs free energy differ from Enthalpy?
They are both thermodynamic potentials. “Potentials” measure potential energy. Potential
energy is the potential to expend energy at some later point in time. Enthalpy includes
the potential energy associated with both the ability to do work and to heat something at
some later point in time. The Gibbs free energy subtracts that off and only looks at a
system’s ability to do work at some later point in time. Also, they are both
thermodynamic potentials that do not consider PV work