Chemistry 114
First Hour Exam
Name:____________
(4 points)
Please show all work for partial credit
1. ( 12 points) I had my blood pressure taken on Friday. My systolic blood pressure was
126 (Slightly high). The units of blood pressure are actually mm of Hg.
Convert my systolic blood pressure into units of:
Atm
Pascal
Bar
PSI
2. (12 points) A large hot air balloon has a volume of 2,800 m3. If the temperature of the
air in the balloon is 99oC, and the balloon has risen to an altitude makes the pressure .87
atm...
A.) Calculate the number of moles of air in the balloon
(Helpful hint: there are 1000L in one m3.)
PV=nRT; n=PV/RT
2,800 m3 ×(1000L/m3)=2.8x106L
990C + 273 = 372K
n=(.87atm×2.8x106L)/(.08206LAatm/KAmol ×372K)
n=7.98x104 mol
B.) Assuming air has an average molar mass of 28.8 g/mole, what is the mass of air inside
the balloon?
7.98x104 mol × 28.8g/mol =2.30x106 g
Extra for the curious
2.30x106g ×(2.2046 lbs/1000g) = 5,070 lbs ~2.5 tons!
2
3. (12 points) I have a canister of CO2 gas with a volume of 5L and a pressure of 760 torr.
I have a second canister of N2 gas with a volume of 10L and a pressure of 1 atm. If I
connect the two canisters together and allow the gases to mix....
A. What is the final pressure of the system ? (use any units you want)
Quick and dirty Both gases are at 1 atm so the final pressure must be 1 atm!
B. What is the partial pressure of CO2 in the system?
P1V1/V2 = P2 ; 760 torr = 1 atm
PCO2 = 5L(1atm)/15L = .333 atm
C. What is the mole fraction of CO2 in the system?
÷=Pco2/Ptotal
=.333/1 = .333
4. (12 points) I have an exercise ball with a volume of 73.6 liters.
A. How much work is involved in filling this ball in Spearfish, where the pressure
is .88 atm? Please give your answer in kJ.
Work = -PÄV; ÄV = 73.6-0 = +73.6L
= - .88 atm ×73.6L
= -6.48 lAatm
-6.48 lAatm ×101.325J/LAatm = -6.56 x103J = - 6.56kJ
B. If I released 2 kJ of heat energy when I filled the ball, what is the ÄU for this
process?
ÄU = w + q
release = -2 kJ
=-6.56 +(-2) = -8.56 kJ
3
5. (12 points) Define the following terms:
Extensive variable
A property that depends on the amount of the system.
State Function
A property that depends only on the change between two states, not the path taken to
move between the two state
Endothermic reaction
A reaction that absorbs heat from the surroundings.
Enthalpy
U + PV or qp
Standard State
The form of an element or compound (s,l,g) at 1 bar of pressure and 273K, or a
solution that is at a 1M concentration.
Adiabatic Conditions
No heat exchange between the system and the surroundings
6. (12 points) Given that the reaction
H2(g) + F2(g)6 2HF(g)
has a ÄH of -546.6 kJ/mol
and
2H2(g) + O2(g) 6 2H2O(l) has a ÄH of -571.6 kJ/mol
calculate ÄH for the reaction
2F2(g) + 2H2O(l) 64HF(g) + O2(g)
Let’s double the first reaction:
2H2(g) + 2F2(g)6 4HF(g)
ÄH -1093.2 kJ/mol
Reverse the second reaction
2H2O(l) 6 2H2(g) + O2(g)
ÄH=+571.6 kJ/mol
Sum
2F2(g) + 2H2O(l) 64HF(g) + O2(g) ÄH= -521.6 kJ/mol
4
7. (12 points) How much heat energy will it take to convert 10 g of solid sodium at 80.0oC
to sodium vapor at 900oC given the following information
MP
97.8oC
BP
8830C
ÄH fus 2.6 kJ/mol
ÄH Vap 97.4 kJ/mol
Heat capacity of the solid: 28.2 J/molAK
Heat capacity of the liquid: 30.8 J/molAK
Heat capacity of the gas: 20.8 J/molAK
10g /22.99g/mol = .434 mol
Heat needed to raise solid sodium from 80 to 97.8
ÄT×molar heat capacity of solid×mol
(97.8-80) ×28.2 × .434
Heat to melt sodium
ÄHfus×mol
2.6 kJ/mol×.434=
Heat to raise liquid Na from97.8 to 883
(883-97.8) ×30.8J×.434
Heat to vaporize Liquid Na
97.4 kJ/mol ×.434 mol
Heat to raise Na vapor from883 to 900
(900-883)×20.8×.434=
Total
218J
Or
.218kJ
1,130 J or
1.13kJ
10,500J or 10.5kJ
42,270J or 42.27kJ
153J
Or
54,271J or
.153 kJ
54.271kJ
8. (12 points) In class we discussed several different intermolecular forces. Name and
rank these forces from strongest to weakest, and briefly describe what causes the force.
Charge-Charge interactions, occurs between ions, the strongest intermolecular
force.
Hydrogen bonding - An especially strong dipole/dipole force that occurs between
hydrogen and nitrogen, oxygen and fluorine. Slightly stronger than Dipole/Dipole
interactions.
Dipole/Dipole interactions- Occurs between two polar molecules, essentially a
charge-charge interaction that occurs between the partial charges on two polar molecules.
Intermediate in overall strength.
London Forces - Occurs between two non-polar molecules, the weakest of all
interactions.