Ch 25 Problem Set 2 Answers
Part A
Work through the following problem, but nothing is to be turned in
The First Law and Ideal Gases
Derive the following expressions for the changes in various thermodynamic first law
quantities for isothermal (T=constant), isobaric (P=constant), isochoric (V=constant) and
adiabatic (q=0) expansions for n=1 moles of an ideal gas.
Calculation of ∆E, ∆H, q and w for expansion of an ideal gas from P1, V1, T1 to P2,
V2, T2 through various processes:
A.
isobaric process (P=constant)
qp = nCp(T2-T1)
wp= -Pext(V2-V1) = -nR(T2-T1)
∆E = n(Cp-R) (T2-T1)
∆H = ∆E + ∆(PV) = n(Cp-R) (T2-T1) + nR (T2-T1) = n Cp (T2-T1)
B,
isochoric process (V=constant)
qv = nCv (T2-T1)
wv = 0
∆E = nCv (T2-T1)
∆H = ∆E + ∆(PV) = nCv (T2-T1) + nR (T2-T1) = n (Cv+R) (T2-T1)
C.
Isothermal (constant temperature) expansion
Since T = constant, then ΔE = 0 (since the energy of an ideal gas depends only on T)
and q = -w.
dw = −Pext dV
= −nRT
dV
V
V2
dV
V
V1
w = −nRT ∫
= −nRT ln
V2
(reversible process)
V1
=-q
∆H = ∆E + ∆(PV) = 0 + 0 = 0
General process from P1,V1,T1 to P2,V2,T2.
Consider the expansion from P1,V1,T1 to P2,V2,T2. The change in E and H can be
calculated for two different routes from P1,V1,T1 to P2,V2,T2:
1
T1,P1 to T1,P2 to T2,P2
∆E(isothermal) + ∆E(isobaric) = 0 + ∆E(isobaric) = nCp(T2-T1)-nR(T2-T1) = n(Cp-R)(T2T1)
∆H=∆E+∆(PV) = n(Cp-R)(T2-T1) + nR(T2-T1) = n(Cp)(T2-T1)
T1,V1 to T1,V2 to T2,V2
∆E(isothermal) + ∆E(isochoric) = 0 + ∆E(isochoric) = nCv(T2-T1)+0 = nCv(T2-T1)
∆H=∆E+∆(PV) = nCv(T2-T1) + nR(T2-T1) = n(Cv+R)(T2-T1)
since ∆E and ∆H are properties, they are path independent, which requires Cp=Cv+R
per mole of ideal gas.
Another useful process:
D.
Adiabatic Expansion
For this process, q=0, so dE = dw
⎛ ∂E ⎞
⎛ ∂E ⎞
dE = ⎜
dV + ⎜
dT
⎟
⎝ ∂V ⎠ T
⎝ ∂T ⎟⎠ V
⎛ ∂E ⎞
=⎜
dT for an ideal gas, since
⎝ ∂T ⎟⎠ V
⎛ ∂E ⎞
⎜⎝
⎟ =0
∂V ⎠ T
dE = CV dT
T2
ΔE = w = ∫ CV dT = CV (T2 − T1 )
T1
this gives the energy change and work as a function of the temperature change.
In terms of the volume or pressure:
2
CV dT = −Pext dV
dT
dV
CV
= −nR
T
V
T
V
CV ln 2 = nR ln 1
T1
V2
nR /CV
T2 ⎛ V1 ⎞
=
T1 ⎜⎝ V2 ⎟⎠
Summary of changes in various thermodynamic first law quantities for isothermal,
isobaric, ischoric and adiabatic expansions for n=1 moles of ideal gas.
Process
isothermal
∆E
0
isobaric
(C
isochoric
adiabatic
p
− R ) (T2 − T1 )
Cv (T2 − T1 )
Cv (T2 − T1 )
∆H
0
q
w
C p (T2 − T1 )
V
RT ln 2
V1
C p (T2 − T1 )
V
−RT ln 2
V1
(Cv + R )(T2 − T1 )
(Cv + R )(T2 − T1 )
Cv (T2 − T1 )
0
T2/T1
1
−R (T2 − T1 )
0
Cv (T2 − T1 )
⎛ V1 ⎞ (
⎜⎝ V ⎟⎠
R
Cv
)
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Part B.1 Probability and Multiplicity
Complete and turn in the following text problems, Chapter 7 (pp 337-339)
Probs. 13, 15, 16, 21, 24, 25 (10 points each)
since the probabilities are exact, in principle they could be given to an arbitrary number
of significant figures.
13. probability p of heads = 2/3 and of tails = 1/3
probability of HHH = (2/3)3 = 0.296296…
probability of TTT = (1/3)3 = 0.037037…
relative probability of HHH/TTT = 23 = 8
15. let N be the number of X's and M be the number of grids
M!
W (M, N ) =
and S = kB lnW
N !( M − N )!
a. M = 6, N = 4, W = 15, S = kBln15 = 2.7 kB
b. M = 6, N = 2, W = 15, S = kBln15 = 2.7 kB
c. M = 6, N = 3, W = 20, S = kBln20 = 3.0 kB
16.
3
W = M !/ (O!X!Y !Z!(M − O − X − Y − Z )!) = 25!/ (4!2!3!2!14!)
= 308, 897, 820, 000
S = kB lnW = 26.5kB
21. Entropy is a state function so it is independent of the nature of the path used by the
system between initial and final states. It is defined, however, in terms of the reversible
heat transfer. For an ideal gas, energy is independent of T, and so qrev = - wrev which
equals 15 kJ in this case (for this problem, wrev = -15 kJ, since the system is doing work
on the surroundings). Therefore S = qrev/T = 15000 J/300K = 50 J K-1.
24.
spaces!
10000!
=
green!yellow!empty! 2000!4000!4000!
S = k B lnW = 10549.2k B Stirling's approx.
S = k B lnW = 10539.9k B "exact'
W=
25.
spaces!
10000!
=
yellow!empty! 6000!4000!
S = k B lnW = 6730.1k B Stirling's approx
S = k B lnW = 6725.3k B exact
W=
This conversion would not occur spontaneously since the entropy decreases.
Problem 1: The probability of disulfide bond formation
1a. designate cysteines in order from 1 to n (ie 1 2 3 4 … n)
cys 1 can form n-1 different disulfides (1-2, 1-3, … 1-n)
the next free cys (of the n-2 left after the first SS) can form n-3 different disulfides
the next free cys (of the n-4 left after the first and second SS) can form n-5 disulfides
total number of unique disulfides = (n-1)(n-3)(n-5)…1 = (n-1)!!
for n = 4, 6, 8, and 10, this formula gives 3, 15, 105 and 945 possible disulfides.
1b. if disulfides formed randomly and were all of equal probability, ~ 1% activity (1/105)
would be expected based on the random probability of forming the correct pairings.
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2a the melting of one gram of ice to liquid water will require
1.00 gm x 6010 J mol-1 / 18.02 gm mol-1 = 333.5 J
The cooling of 1.00 gm of liquid water from 20 ˚C to 0 ˚C generates (75.4/18.02) * 20 =
83.7 J gm-1
Hence 333.5/83.7 gm = 3.98 gm of liquid water can be cooled by 1.00 gm of ice under
these conditions - very close to the rule of thumb in the Wikipedia article.
The overall entropy change for this process is the sum of the entropy changes of the ice
and of the liquid.
For ice ∆S = 333.5/273 = 1.22 J K-1
For water ∆S = C*ln(T2/T1) = (3.98*4.18) *ln(273/293) = -1.18 J K-1,
for a total ∆S = +0.04 J K-1 > 0
2b. ∆S (ice to water, 0˚C) = Q/T = 0.5 * 6010/273 = 11.0 J/K
∆S (water to ice, -10˚C) = Q/T = 0.5 * (-6010)/263 = -11.4 J/K
∆Soverall = 11.0 - 11.4 = -0.4 J/K; the negative entropy change means the process is
not spontaneous.
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