Download ACS_Thermodynamics_Exam_1981

Here are 10 questions on Thermodynamics from the 1981 ACS Physical Chemistry Exam
ACS EXAMINATION
PHYSICAL CHEMISTRY
GRADUATE LEVEL PLACEMENT
1981
DATA THAT MAY BE USEFUL FOR SOME QUESTIONS
Speed of light = 2.9979x108 m sec-1 = 2.9979x1010 cm sec-1
Universal Gas Constant = 8.3143 J K-1 mol-1 = 1.9872 cal K-1mol-1
= 82.056 cm3 atm K-1mol-1
Faraday Constant = 96, 487 coulomb equiv-1
Avogadro Number = 6.0222x1023 molecules (gram-mole)-1
Boltzmann Constant = 1.3806x10-23J K-1 = 1.3806x10-16 erg K-1
Planck’s Constant = 6.626x10-34 J sec = 6.626x10-27 erg sec
THERMODYNAMICS
1. The internal energy, E, of a system is a function of mass, volume, and temperature. The
differential of E for a constant mass is
a) dE = dV + dT
b) dE = VdP – SdT
c) dE = (E / V)T dV + (E / T)V dT
d) dE = E (dV / dT) + E (T / V)
2. According to the Laws of Thermodynamics, differentials of the following are exact:
a)
work and enthalpy
b)
energy and work
c)
entropy and heat
d)
energy and entropy
3. An “exact differential,” df, as the term is ordinarily used in Thermodynamics,
satisfies the condition
a)
 df = 0
B
b)
 df
= fA - fB
A
c) M / x = N / y where df =  (x,y) dx + N(x,y) dy
d)
that its line integral between two fixed points is dependent on the path.
4. When 0.5 mole of an ideal gas at 0 C is compressed isothermally from 1.00 atm
to 2.00 atm
a)  G = 0
b) no heat is absorbed by the gas
c) the work done on the gas is -787 joules
d)E = 0
Pexternal
5.
P2
d
c
P1
a
b
V1
V2 Vsystem
The following statements refer to changes between the thermodynamic states a, b, c, d of a
system occurring by the paths shown on the PV diagram. Which statement is true?
a) For the process a -> b -> c -> d -> a the work done by the system is zero.
b) For the process a -> b -> c -> d -> a the entropy change of the system is greater than
zero
c) For the process a -> b -> c the work done by the system is - P1(V2 - V1)
d) For the process a -> d the work done by the system is V1 (P2 - P1)
6. The equation (G / P) X, m = V is correct when the system is in internal equilibrium and
=
a)
T
b)
S
c)
H
d)
V
7. Consider the reaction
ZnO (s) + H2 (g) <=> Zn(s) + H2O (g)
ZnO (s)
H2O (g)
Hf, 298 K
-348 kJ mole-1
-242 kJ mole-1
Gf, 298 K
-318 kJ mole-1
-228 kJ mole-1
From these data, estimate the temperature at which the equilibrium constant for this
reaction is approximately unity.
a)
700 K
b)
1300 Kc)
1900 Kd)
2500 K
x
8. The following thermochemical data are for 25C.
H2O(l)
H+(aq)
Hf
C p
- 286 kJ mole-1
+ 75.3 joule mole-1 K-1
0
OH-(aq)
0
- 230 kJ mole-1
+ 148 joule mole-1 K-1
For the self-ionization of water,
H2O (l) <=> H+ (aq) + OH-- (aq)
the value of H at 35C is
9.
10.
a)
58.1 kJ mole-1
b)
56.6 kJ mole-1
c)
55.9 kJ mole-1
d)
53.8 kJ mole-1
Suppose S = p1 ln p1 – p2 ln p2 with p1 + p2 = 1. Which statement follows?
a)
S vanishes for p2 = p1
b)
S vanishes for p2 = - p1
c)
dS / dP2 vanishes for p2 = p1
d)
none of the above
The heat of vaporization of water at 100C is 40.6 kJ mole-1. If one mole of liquid water
is reversibly vaporized at 100C and 1 atm, which of the following statements in
incorrect?
a)
q = 40.6 kJ
b)
= 37.5 kJ
c)
G = 3.1 kJ
d)
S = 109 J K-1
Answers
1. C
a. We know that E(m, V, T) from the problem. Remember back to taking
derivatives of G, U, etc and the form it looked like when done. Note that mass is
used instead of N as we used.
2. D
a. Exact differentials are from state functions. Path functions depend on the process
of the path and so are not exact. In the case of the answers, energy and entropy
are both state functions. Enthalpy is as well, but it is paired with work which is
not.
3. A
a. B and C are close to what we want but are written with the wrong order. Since we
just stated that state functions have exact differentials we should look something
that signifies this. Answer A is saying that a closed path (the loop in the integral
symbol) is zero. In other words the function is independent of path since the
integral equals zero. For instance (using entropy as an example), taking liquid
water and then freezing it, then vaporizing it and then condensing it back to a
liquid would have a net 0 change in entropy (path independent) since it started and
ended in the same place.
4. D
a. Since the problem states that the process is isothermal we know ∆U=0. n is also
constant. However the pressure is changing and the volume is also though not
explicitly stated. We know that
w    PdV
PV  nRT
nRT
dV
V
Vf 
w  nRT ln  
 Vi 
1
V f  Vi
2
w  787 J
Answer C is the opposite of what we want. Answer B would apply for an
adiabatic process not isothermal. Answer D is best because U  E
w  
5. D
a. Work is defined by the area below the curve of a P vs. V graph. D is the correct
process description for the work value. A is wrong because work is a path
function.
6. A
a. Remember that G(T,P,m) since G is something nice because we can measure it in
the lab by measuring temperature and pressure. In this problem x must equal T
because it is the only variable left out of the expression. You have P, and m
accounted for already.
7. C
a. We can find ∆Hrxn and ∆Grxn for standard temperature by using products minus
reactants. Remember that H2(g) and Zn(s) have values of 0 for ∆Hrxn and ∆Grxn.
For each set of ∆H and ∆G for each individual reactant or product we can find the
standard entropy term ∆Srxn using T=298K. We can then find ∆Srxn. Now we can
assume ∆Grxn=0 for the reaction in question because the problem states the
equilibrium constant is in equilibrium. Use the ∆Hrxn and ∆Srxn and solve for the
new T. *Note that we assumed H2(g) and Zn(s) had zero entropy for formation
when in reality they have nonzero entropy.
8. B
a. Cp is given and ∆H. Remember that
H
 Cp
T
H  Cp  T
H  CpT
Again find the new ∆H for 35C and then do products minus reactants. Notice and
1 to 1 stochiometric ratio.
9. A
a. Simple enough. Positive something minus that same thing is zero.
10. C
a. For a phase change ∆H=q so answer A is correct. For a reversible process,
∆s=q/T. Since
H  U  PV
H  U  PV
H  U  nRT
U  37.5kJ
U  E
Note that n=1 because we are going from no moles of vapor to 1 mole of vapor.
Answer C is the correct choice because it is incorrect for the problem. This makes
since because we are at a phase change equilibrium so ∆G=0.