CHM1045
Fall 2000
Dr. Michael Blaber
Name_______________________________SS#_________________________
Exam #2 100 points total
Friday October 20 2000
1. An aquarium with dimensions 0.5m on each side (i.e. a cube with each side 0.5m long) is
filled with water. What is the pressure, in Pa, that this aquarium exerts on the top of a table? (8
points)
P = F/A
A = 0.5m * 0.5m = 0.25m2 (2 points)
F=m*a
Volume of H2O = (0.5m)3 = 0.125m3
Mass of H2O = 0.125m3 * 1g/1x10-6m3 = 125,000g or 125Kg (2 points)
F = 125Kg * 9.81m/s2 = 1,226 Kg m s-2 (2 points)
P = F/A = 1,226 Kg m s-2/0.25m2 = 4,904 K m-1 s-2 = 4,904 Pa (2 points)
2. Calculate the pressure, in KPa, of the gas sample in the following open-tube mercury
manometer. Assume atmospheric pressure is 745 torr. (6 points)
27mm Hg * (760 torr/760mm Hg) = 27 torr (2 points)
P = (745 torr + 27 torr) = 772 torr (2 points)
P = 772 torr * (101 KPa / 760 torr) = 102.6 KPa (or 103 KPa with 3 digits precision) (2 points)
3. What volume would a 25.3g sample of carbon dioxide occupy under conditions of standard
temperature and pressure? (8 points)
CO2 = (12.0) + (2*16.0) = 44.0 amu = 44.0 g/mole (2 points)
25.3g * (1mole/44.0g) = 0.575 mole (2 points)
PV = nRT (2 points)
V = nRT/P = 0.575moles * (0.0821 L atm/mole K) * 273K / 1atm
V = 12.9 L (4 points)
4. A weather balloon is released from the ground and floats up into the atmosphere. When it
was released it had a volume of 500L, the temperature was 27.5°C and the atmospheric pressure
was noted to be 771 torr. At 50,000 feet altitude the temperature is recorded at -61°C and the
presssure is 27.5 torr. What is the volume of the balloon at this altitude? (8 points)
P1*V1/T1 = P2*V2/T2 (2 points)
P1*V1/T1 = (771 torr * 500L)/(27.5+273K) = 1,283 torr L/K (2 points)
1,283 torr L/K = 27.5 torr * V2/(-61 + 273K) = 0.130 torr/K * V2 (2 points)
9,890 L = V2 (2 points)
5. A certain gas occupies 2.35L at 37°C and 2.09 atm pressure. The sample of gas in question
has a mass of 12.3g. What is the molar mass of the gas? (6 points)
PV = nRT (2 points)
Thus, n = PV/RT = (2.09atm * 2.35L)/(0.0821 L atm/mol K * (37+273))
n = 0.193 moles (2 points)
Molar mass = 12.3g/0.193moles
Molar mass = 63.7g/mole (2 points)
6. What volume of carbon dioxide is produced if 57g of propanol (C3H7OH) is combusted to
completion at 60°C and 1.15 atm pressure (be sure to provide balanced equation)? (10 points)
Propanol is (3*12.0)+(1*16.0)+(8*1.01) = 60.1amu or 60.1g/mole
57g * (1mole/60.1g) = 0.948 moles (2 points)
Balanced equation for combustion:
C3H7OH + 9/2O2  3CO2 + 4H2O (2 points)
Stoichiometry is such that for every mole of propanol consumed, 3 moles of CO2 is produced
0.948moles propanol * (3moles CO2/1mole propanol) = 2.84 moles CO2 (2 points)
n = 2.84 moles
T = (273+60) = 333K
P = 1.15 atm
PV = nRT, so V = nRT/P (2 points) = (2.84moles * 0.0821 L atm/mol K * 333K) / 1.15 atm
V = 67.5L (2 points)
7. A sample of gas contains the following: O2 = 27.5%, N2 = 68.2%, Ar = 4.3% by mass. The
pressure of the gas sample is 2.82 atm. What are the partial pressures of the compounds in this
gas sample? (10 points)
In a representative 100g sample we would have:
O2 = 27.5g * (1 mole/32g) = 0.859 moles
N2 = 68.2g * (1 mole/28g) = 2.44 moles
Ar = 4.3g * (1 mole/39.9g) = 0.108 moles
Total moles = 0.859 + 2.44 + 0.108 = 3.41 moles
Mole fraction O2 = 0.859/3.41 = 0.252
Mole fraction N2 = 2.44/3.41 = 0.716
Mole fraction Ar = 0.108/3.41 = 0.0317
(note: 4 points partial credit for correct mole fraction)
Partial pressure = mole fraction * total pressure (2 points)
PO2 = 0.252 * 2.82 atm = 0.711 atm
PN2 = 0.716 * 2.82 atm = 2.02 atm
PAr = 0.0317 * 2.82 atm = 0.0894 atm
(note: 4 points for correct calc of partial pressures)
8. What is the density of ammonia (NH3) vapor at 0.55 atm and 50°C (8 points)
Ammonia is 14.0 + (3 * 1.01) = 17.0 amu or 17.0 g/mole (2 points)
50°C = (50 + 273) = 323K
d = M*P / R*T (2 points)= (17.0g/mole * 0.55 atm) / (0.0821 L atm/mol K * 323K)
d = 0.353 g/L (4 points)
9. What is the kinetic energy (in Joules) of a 15g bullet with a velocity of 1000 feet per minute,
compared to a 120g bullet with a velocity of 3000 feet per minute? (8 points)
(note: 2 points for knowing E = 1/2 mv2)
Bullet #1:
15g * (1Kg/1000g) = 0.015Kg
1000 feet/min * (12 inches/foot) * (1 meter/39.4 inches) * (1min/60sec) = 5.08 m/s
Ek = 1/2 * (0.015Kg) * (5.08 m/s)2 = 0.194 Kg m2 s-2 = 0.194 Joules (3 points)
Bullet #2:
120g * (1Kg/1000g) = 0.120Kg
3000 feet/min * (12 inches/foot) * (1 meter/39.4 inches) * (1min/60sec) = 15.2 m/s
Ek = 1/2 * (0.120Kg) * (15.2 m/s)2 = 13.9 Kg m2 s-2 = 13.9 Joules (3 points)
10. Given the following equation, calculate the enthalpy change associated with the combustion
of 150g of octane. (8 points)
2C8H18(l) + 25O2(g)  16CO2(g) + 18H2O(l) H = -10.9 kJ
Octane is (12.0*8) + (1.01*18) = 114g/mole (2 points)
150g * (1mole/114g) = 1.32 moles (2 points)
Stoichiometry is such that the combustion of 2moles of octane releases 10.9kJ of energy
1.32 moles octane * (-10.9kJ / 2moles octane) = -7.19 kJ (4 points)
11. A building has a 1,500 pound roof that is made of aluminum. On a hot day in the summer
the temperature of the roof in the morning is observed to be 25°C. In the afternoon, the
temperature is observed to be 65°C. How many Joules of energy has the roof absorbed from the
sun? (8 points)
1,500 pounds * (1Kg/2.2 pounds) * (1000g/Kg) = 682,000 g (2 points)
T = (65-25) = 40K
q = mass * specific heat * T (2 points)
q = 682,000 g * 0.9J g-1 K-1 * 40K
q = 24,552,000 J or 24,552 kJ (4 points)
12. Given the following standard enthalpies of formation, calculate Hrxn for the combustion of
235g of propane (C3H8) under conditions of STP (be sure to provide balanced equation(s)) (12
points)
H2O(l) H0f = -286 kJ/mol
CO2(g) H0f = -394 kJ/mol
C3H8(g) H0f = -104 kJ/mol
Propane = (3*12.0) + (8*1.01) = 44.1 amu or 44.1 g/mole
235g * (1mole/44.1g) = 5.33 moles propane
Balanced combustion equation:
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) (2 points)
Water:
H2(g) + 1/2O2(g)  H2O(l) H = -286 kJ
4H2(g) + 2O2(g)  4H2O(l) H = (4 * -286 kJ) = -1144 kJ (2 points)
Carbon dioxide:
C(s) + O2(g)  CO2(g) H = -394 kJ
3C(s) + 3O2(g)  3CO2(g) H = (3 * -394 kJ) = -1,182 kJ (2 points)
Propane:
3C(s) + 4H2(g)  C3H8(g) H = -104 kJ
C3H8(g)  3C(s) + 4H2(g)H = +104 kJ (2 points)
Oxygen:
5O2(g)  5O2(g) H = 0 kJ
Summary:
4H2(g) + 2O2(g)  4H2O(l) H = (4 * -286 kJ) = -1144 kJ
3C(s) + 3O2(g)  3CO2(g) H = (3 * -394 kJ) = -1,182 kJ
C3H8(g)  3C(s) + 4H2(g)H = +104 kJ
5O2(g)  5O2(g) H = 0 kJ
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) H = -1144 + (-1182) + 104 = -2222 kJ (2 points)
5.33 moles propane * (-2222kJ/1mole propane) = -11,843 kJ (2 points)
Constants, Units, Formulas and Conversions
Acceleration due to gravity:
9.81m/s2
Gas Constant:
0.0821 L atm mol-1 K-1
8.31 J mol-1 K-1
62.4 L torr mol-1 K-1
Force:
1N = 1 kg m s-2
F = mass * acceleration
Energy:
1J = 1N m = 1 kg m2 s-2
Length:
1 m = 39.4 inches
1 mile = 5,280 feet
Mass:
1 kg = 2.2 lb
Pressure:
P = force / area
1Pa = 1N/m2
1Pa = 1Kg/m s2
1atm = 760 torr
1atm = 101 Kpa
760mm Hg = 760 torr
Density:
 = mass / volume
density of H2O = 1g/ml
or 1g/1x10-6m3
Specific Heats:
H2O = 4.18 J/g °C
Al = 0.9 J/g °C