Chem 340 – 1st Hour Exam
Friday, Oct. 4, 2013, 2 - 2:50 –(on to 3:50) PM, 250 BSB
Closed book exam, only pencils and calculators permitted. No Computers, no sheets,
just you. Put all of your work (and answers!!) in the answer book. “Floating” answers
without work receive little credit. If you get stuck, move to the next problem and come
back, indicate how to find your answers. Useful information is attached. Good Luck!!
1. (12 pts) We did a number of problems using isothermal expansion coefficient,  or ,
and compressibility, , often assuming an ideal gas
a. evaluating the expansion coefficient or compressibility is more difficult for a van der
Waals gas, P = RT/(Vm-b) - a/Vm2. However the ratio of / is easier. Evaluate /
in terms of a,b,Vm . (Hint: use cyclic rule to get a nice form, one partial derivative).
b. Now evaluate/ for an ideal gas.
c. Briefly discuss the physical meaning of the difference in your answer to (a) and (b).
2. (18 pts) Five liters of nitrogen gas initially at 27 C and 1 bar are heated to 127 C
Assume it is ideal gas and its molar heat capacity is given by:
CP,m = (27.0 + 5.90x10-3 K-1 T) J K-1 mol-1
a. Assuming constant Pressure, calculate the values for w, q, U and H.
b. Now calculate w, q, U and H for a constant Volume change T=27 C  T=127 C
3. (14 pts) A sample of liquid benzene weighing 0.78 g was burned
in a bomb calorimeter, consisting of a sealed metal “bomb”
containing the benzene and excess O2, surrounded by a water
bath, all in a vessel that is isolated from the environment.
Assume the effective heat capacity of the calorimeter is 6.0 kJ/K.
C6H6 MW=78. The combustion reaction is:
C6H6 (l) + 15/2 O2 (g)  6 CO2 (g) + 3H2O (l)
The temperature of the bath changed from 25.0 C to 30.5 C.
a. For this reaction calculate w, q, U
b. Use this to calculate Hof for benzene, assume U ≈ Horxn for the above and:
C(s) + O2 (g)  CO2 (g)
Hfo (CO2) = -394 kJ/mol
H2(g) + ½ O2(g)  H2O (l)
Hfo (H2O) = -286 kJ/mol
4. (20 pt) One mole of an ideal gas is expanded at T=300 K from V1 = 20 L to V2 = 40 L.
Assume CVm = 3/2 R.
Calculate w, q, U and H if:
a. This is done reversibly and isothermally
b. Again isothermally against constant pressure equal to the final equilibrium pressure.
c. The same volume expansion is done adiabatically and reversibly from T1 = 300 K
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5. (8 pt) Consider the decomposition of glucose to form lactic acid which makes your
muscles ache after a run (in the lactate form).
C6H12O6 (s)  2 CH3CHOHCOOH (s)
values at
T = 298K
glucose
ΔHof (kJ/mol)
Cp (J/molK)
-1274
218
lactic acid
-694
128
What is ΔHorxn for glucose lactic acid at body temperature, 37 C?
6. (20 pt) Conceptual questions, brief answers, 3-4 sentences (choose only 5 – five!)
a. Briefly explain the difference between an ideal (or perfect) gas and a real gas
b. Briefly explain the difference in the virial form and the van der Waals equations of
state for a real gas
c. Briefly explain the difference in internal energy, U, and enthalpy, H.
d. Briefly explain the difference in CP and CV (not just definitions or V and P!) and
why CP > CV
e. If H depends on both T and P, express this using a complete differential form?
f. Show how compressibility and the isothermal expansion coefficient are intrinsic
or extrinsic variables (you can use their definitions)
g. Briefly explain how or why I can use ∫CVdT = U for a process that is not constant
volume, and what parameters must I know to do this? (this is more general, but
you may assume ideal gas if that helps you)
h. Briefly explain why we use Hfo values for reactants and products to determine
the enthalpy change Hrxno without knowing the absolute values of enthalpy for
reactants or products
i. Briefly explain how entropy, dS, can be a state function if it depends on dq which
is an inexact differential, i.e. path dependent and not a state function
j. Briefly explain the physical reasons most gases cool on expansion (high to low
pressure change) due to the Joule-Thomson effect.
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Extra information for Chem 340 Exams:
Alternates (non-SI): cm = 10-2 m, g = 10-3 kg, C = K – 273.15
cm3 = 10-3 L = 10-6 m3, Torr = atm/760 = 133.32 Pa, atm = 1.013 bar
Transport
Temperature (T), activity/concentration (a)
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w = ∫ RT.d(ln a)
J
Equations:
C = ᵭq/dT CV = (U/T)V
Cp = (H/T)p ideal: CPm - CVm = R
Adiabatic, reversible change: Ti/Tf= (Vf/Vi)-1 where  = CPm/CVm = 1+ R/CVm
Ti/Tf= (Vf/Vi)1/c  Tic Vi = TfcVf where c = CVm/R
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