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) DH = -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)

DT = (65-25) = 40K

q = mass * specific heat * DT (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 DHrxn for the combustion of 235g of propane (C3H8) under conditions of STP (be sure to provide balanced equation(s)) (12 points)

H2O(l) DH0f = -286 kJ/mol

CO2(g) DH0f = -394 kJ/mol

C3H8(g) DH0f = -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) DH = -286 kJ

4H2(g) + 2O2(g) ® 4H2O(l) DH = (4 * -286 kJ) = -1144 kJ (2 points)

Carbon dioxide:

C(s) + O2(g) ® CO2(g) DH = -394 kJ

3C(s) + 3O2(g) ® 3CO2(g) DH = (3 * -394 kJ) = -1,182 kJ (2 points)

Propane:

3C(s) + 4H2(g) ® C3H8(g) DH = -104 kJ

C3H8(g) ® 3C(s) + 4H2(g) DH = +104 kJ (2 points)

Oxygen:

5O2(g) ® 5O2(g) DH = 0 kJ

Summary:

4H2(g) + 2O2(g) ® 4H2O(l) DH = (4 * -286 kJ) = -1144 kJ

3C(s) + 3O2(g) ® 3CO2(g) DH = (3 * -394 kJ) = -1,182 kJ

C3H8(g) ® 3C(s) + 4H2(g) DH = +104 kJ

5O2(g) ® 5O2(g) DH = 0 kJ

C3H8(g) + 5O2(g) ® 3CO2(g) + 4H2O(l) DH = -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:

r = 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