NORTH CAROLINA STATE UNIVERSITY
Department of Chemistry
Name________________________________
CH 331
Physical Chemistry
g = 9.81 m/s2
R = 8.314 J mol-1 K-1 = 0.08206 L atm mol-1 K-1
1 atm = 1.0133 x 105 Nm-2 = 760 Torr
P = P0exp{-Mgh/RT}
Given:
P2 = P1 +
ln
Extra-term Examination
∆ trsH m T2
ln
∆ trsVm T1
,
ln
P2
∆ H
= trs m 1 – 1
P1
T1 T2
R
K2
∆H m 1
=
– 1
K1
R T1 T2
N
∆S = – R Σ xiln xi where N is the number of components
i=1
ρ = mgh
,
P = ρgh , dS ≥
δq
T
,
dw = – PdV
, dH = dU + PdV
Please answer all questions.
1. A. Assuming that an ideal gas has a density of 7 x 10-3 g/cm3 at 6 atm and 300 K calculate its molar
mass (5 points):
Solution: M = ρRT/P = (7.0 kg/m3)(8.31 J/mol-K)(300 K)/(6.079. x 105 Nm-2)
= 0.028 kg/mole = 28 g/mole
Μ(g mol-1) = ___________________
B. What is the molecule likely to be? ________________________________.
C. Calculate the root-mean-square velocity of argon gas at 475 K (3 points).
Solution: sqrt(<u2>) = sqrt(3RT/M) sqrt(3.0(8.31 J/mol-K)(400 K)/0.0399 kg/mol)
= 500 m/s
C. Calculate the molar heat capacity at constant pressure of diatomic oxygen gas at 360 K (2 points).
Solution: CV = 3/2R = (1.5)(8.31 J/mol-K)
= 12.5 J/mol-K
1
2. Calculate the depth at which you would measure a pressure of 960 bars in the ocean at 10 oC assume
that water has a density of 1.0 x 103 gL-1 (5 points).
Solution:
h = P/ρg = (1764 x 105 Pa)/(1.0 x 103 kg/m3)(9.8 m/s2) = 18,000 m
Assuming that g = 10 m/s2 is acceptable for this problem.
h = P/ρg = (1764 x 105 Pa)/(1.0 x 103 kg/m3)(10.0 m/s2) = 17,640 m
h(meters) = ___________________
3. Calculate the entropy of mixing of 0.4 grams of NaCl in 20 milliliters of H2O. (5 points)
Solution: The molar mass of NaCl is (23 + 35.5)g/mol = 58.5 g/mol.
nNaCl = 1 gram/(58.5 gram/mol) = 0.017 mol
10 millilters of H2O is 10 grams.
nH2O = 10 grams/(18 grams/mol) = 0.555 mol
The total number of moles is 0.572 moles.
The mole fractions are xNaCl = 0.017/0.572 = 0.029
The mole fractions are xH2O = 0.555/0.572 = 0.97
The entropy is ∆S = -nR(x1 ln x1 + x2 ln x2)
= -(0.572 mol)(8.31 J/mol-K)(0.029 ln 0.029 + 0.97 ln 0.97)
= 0.628 J/K
4. Calculate the free energy of mixing of 4 grams of NaCl in 20 milliliters of H2O at 40 oC. (3 points)
Solution: Calculate the mole fraction of each component.
𝑛𝑛𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 =
4 𝑔𝑔
𝑚𝑚𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁
=
= 0.0683 𝑚𝑚𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜
𝑀𝑀𝑚𝑚 ,𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 58.5 𝑔𝑔/𝑚𝑚𝑚𝑚𝑚𝑚
𝑚𝑚𝐻𝐻2𝑂𝑂
20 𝑔𝑔
=
= 1.111 𝑚𝑚𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜
𝑀𝑀𝑚𝑚 ,𝐻𝐻2𝑂𝑂 18 𝑔𝑔/𝑚𝑚𝑚𝑚𝑚𝑚
𝑛𝑛𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁
0.0683
𝑥𝑥𝑁𝑁𝑎𝑎𝑎𝑎𝑎𝑎 =
=
= 0.0579
𝑛𝑛𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 + 𝑛𝑛𝐻𝐻2𝑂𝑂 1.111 + 0.0683
𝑛𝑛𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁
1.111
𝑥𝑥𝐻𝐻2𝑂𝑂 =
=
= 0.942
𝑛𝑛𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 + 𝑛𝑛𝐻𝐻2𝑂𝑂 1.111 + 0.0683
𝑛𝑛𝐻𝐻2𝑂𝑂 =
∆𝑚𝑚𝑚𝑚𝑚𝑚 𝐺𝐺 = 𝑛𝑛𝑛𝑛𝑛𝑛(𝑥𝑥1 𝑙𝑙𝑙𝑙𝑥𝑥1 + 𝑥𝑥2 𝑙𝑙𝑙𝑙𝑥𝑥2 )
∆𝑚𝑚𝑚𝑚𝑚𝑚 𝐺𝐺 = (1.179)(8.31)(313)((0.0579 ln(0.0579) + (0.942) ln(0.942))) = −678 𝐽𝐽/𝑚𝑚𝑚𝑚𝑚𝑚
2
4. An ideal gas is initially at 100.0 atm and 700 K. Its volume is initially 0.007 L and the gas is
allowed expand to 0.5 L under the following conditions (12 points):
(1.) Pexternal = 0.01 atm
(2.) Pexternal = constant (single-step)
(3.) Pexternal = Pgas (reversible compression)
For each of the above conditions calculate q, and w, for the gas.
q(J)
w(J)
(1.)
__________
___________
(2.)
__________
___________
(3.)
__________
___________
Solution: The initial and final temperature is the same for all three expansions. Note that
this means that they are also isothermal (if we look at the initial and final states) although
only process 3 is at constant temperature throughout the expansion. Since U is a state
function ∆U = 0 for all three processes.
Using the first law ∆U = q + w  q = -w. Therefore, if we calculate the work we will also
be able to find the heat (q = -w).
Process 1 is clearly irreversible w = - Pexternal∆V and since Pexternal = 0.01 we have w = (0.01 atm)(10.0 L – 0.1 L) = - 0.099 L-atm ~ - 10.0 J
(Note: the sign is negative since this is an expansion)
Therefore q = 10 J
Process 2 is w = - Pexternal∆V = -(0.1 atm)(10.0 L – 0.1 L) = - 0.99 L-atm ~ - 100 J.
Therefore, q = 100 J.
We determine the final (external) pressure using P2V2 = P1V1. Here it is 0.1 atm.
Process 3 is reversible and isothermal so that w = -nRT ln(V2/V1). We need n.
n = PV/RT = (10 atm)(0.1 L)/(0.08206 L-atm/mol-K)(300 K) = 0.0406 moles.
w = -nRT ln(V2/V1) = -(0.0406 mol)(8.31 J/mol-K)(300 K) ln(10.0/0.1) = -466 J
Therefore, q = 466 J
q(J)
w(J)
(1.)
____10_____
____-10____
(2.)
___100_____
___-100____
(3.)
___466_____
___-466____
Note: In the practice midterm there was a problem that assumed Pext = 0. Some of you worked one of
the examples like this. This is incorrect and received no credit. The goal of the problem is to
understand conditions and correctly reason the method needed.
3
5. A. Calculate the inlet temperature to a steam turbine which has a thermodynamic efficiency 0.32 and
the outlet temperature is 300 K (3 points).
Solution:
η = (Thot - Tcold)/Thot => Tcold = Thot(1 - η) = 800(0.4) = 320 K
B. How much energy is lost as heat for every kJ of heat expelled into the environment by this engine
(2 points)?
Solution:
η = |w|/q The work obtained is |w| = ηq = 0.6 x 1 kJ = 0.6 kJ per kJ of heat
generated. The heat lost is equal to qlost = q - |w| = 1 - 0.6 kJ = 0.4 kJ.
6. The standard enthalpy of the reaction N2 (g) + 3 H2 (g) → 2 NH3 (g) is – 92.2 kJ/mol of ammonia
and the third law entropies of H2, N2, and NH3 at 298 K are given in the table below. Assuming that
the entropy and enthalpy are dependent on temperature calculate the Gibb’s free energy at 200 K.
The enthalpy of vaporization of ammonia is ∆vapHo = 23.3 kJ/mol and Tb = 239.7 K. (12 points).
Substance
NH3 (l)
NH3 (g)
H2 (g)
N2 (g)
Cp (J/mol-K)
75.3
35.4
27.3
29.1
So (J/mol-K)
111.3
192.4
130.6
191.6
Solution:
∆So(298 K) = 2So(NH3,g) - 3So(H2, g) - So(N2, g) = 2(192.4) – 3(130.6) – 191.6 = -198.6 J/mol-K
The temperature dependence of the enthalpy is:
∆Ho(T2) = ∆Ho(T1) + ∆Cp,liq(T2 – T1)
∆So(T2) = ∆So(T1) + ∆Cp,liqln(T2/T1)
There is no phase change over the specified temperature range so:
∆Cp,liq = 2 Cp (NH3,g) – 3 Cp (H2,g) – Cp (N2,g) = 2(35.4) – 3(27.3) – 29.1 = -40.2 J/mol-K
The entropy change is:
∆So(298 K) = 2So(NH3,g) - 3So(H2, g) - So(N2, g) = 2(192.4) – 3(130.6) – 191.6 = -198.6 J/mol-K
∆Ho(240 K) = - 92,200 + (-40.2)(240 – 298) = - 89,870 J/mol-K = - 89.87 J/mol-K
∆So(240 K) = - 198.6 + (-40.2)ln(240/298) = - 189.9 J/mol-K
∆Go(240 K) = ∆Ho(240 K) - T∆So(240 K) = - 89,870 – 240(-189.9) = - 44,293 J/mol
4
7. The binding constant of a drug at 300 K is 6 x 105 M-1 and at 320 K it is 8 x 104 M-1. Calculate the
standard enthalpy and entropy change for the binding process. (8 points)
Solution: Use the van't Hoff equation.
o
ln K 2 – ln K 1 = – ∆H 1 – 1
R T2 T1
o
K
ln 2 = – ∆H 1 – 1
K1
R T2 T1
K
K
Rln 2
RT1T2ln 2
K
K1
1
∆H o =
=
1 – 1
T2 – T1
T1 T2
Plugging in numbers we obtain:
3
(8.31 J/mol–K)(300 K)(320 K)ln 10 5
10
= – 91.8 kJ/mol
∆H o =
20 K
At 300 K, ∆Go = -RT ln K = -(8.31 J/mol-K)(300 K) ln(105) = -28.7 kJ/mol
∆So = (∆Ho - ∆Go)/T = (-91800 - (-28700)) J/mol /300 K = -210 J/mol-K.
8. The equilibrium constant for the reaction I2 (s) + Br2 (g) = 2 IBr (g) is 0.164 at 25 oC. (a) Calculate
∆rG for this reaction (b) Bromine gas is introduced into a container with excess solid iodine. The
pressure and temperature are held at 0.082 atm and 25 oC. Find the partial pressure of IBr (g) at
equilibrium. (10 points)
Solution:
a. First we can calculate ∆rGo using
∆rGo = -RT ln Keq = (8.31 J/mol-K)(298 K)ln(0.164) = +4477 J/mol
= +4.48 kJ/mol
The reaction quotient can be expressed as:
2
PIBr
Q=
PBr2
Using standard conditions PIBr = PBr2 = 1 atm, so Q = 1 and lnQ = 0.
Since
∆rG = ∆rGo + RT lnQ
we have
∆rG = ∆rGo = +4.48 kJ/mol
b. Make a table
Condition
Initial
Equilibrium
Partial pressure
Br2
0.164
0.164-x
(0.164-x)
/(0.164+x)*0.164
IBr
0
2x
2x
/(0.164+x)*0.164
5
Total
0.164
0.164+x
0.164
2
2x
P2
2
0.164 + x
P
4x 2 P
K eq = IBr =
=
0.164 – x P
0.0268 – x 2
PBr2
0.164 + x
2
0.0268K eq – x K eq + 4P = 0
x=
PIBr =
0.0268K eq
K eq + 4P
=
0.0268(0.164)
= 0.0732
5 0.164
2(0.0732)
2x
P=
0.164 = 0.101 atm
0.164 + x
0.237
9. Calculate the entropy change when 90 grams of boiling water are poured into a stainless steel bowl
weighing 70 grams at 25 oC. Assume that the molar mass of the metal in the bowl is 55 grams/mole.
The heat capacity of water is 75 J/mole-K and that of bowl is 25 J/mole-K. (10 points)
Solution: First we must calculate the equilibrium temperature.
n 1C p1T1 + n 2C p2T2
n 1C p1 + n 2C p2
(10 mol)(75 J/mol–K)373 K + (2 mol)(25 J/mol–K)298 K
=
(10 mol)(75 J/mol–K) + (2 mol)(25 J/mol–K)
= 368.3 K
Teq =
Now we calculate the entropy change of the water and bowl separately.
Teq
= (10 mol)(75 J/mol–K)ln 368.3
T1
373
= – 9.51 J/K
Teq
∆S bowl = n 2C p2
= (2 mol)(25 J/mol–K)ln 368.3
T2
298
= + 10.59 J/K
∆S total = ∆S water + ∆S bowl = + 1.08 J/K
∆S water = n 1C p1
The process is spontaneous. If you get a negative entropy for a heat transfer problem, you know that
you have made a mistake somewhere. Heat transfer from a hotter to a colder body is always
spontaneous.
10. You are asked to design a centrifuge that can reach 10,000 g. Given that radius of rotation is 10 cm,
calculate the required angular frequency, 𝜔𝜔.
Solution:
104 𝑔𝑔 = 𝜔𝜔2 𝑥𝑥
104 𝑔𝑔
104 (9.8 𝑚𝑚𝑠𝑠 −2 )
𝜔𝜔 = �
=�
= 3,130 𝑠𝑠 −1
𝑥𝑥
0.01 𝑚𝑚
6