CHM1045 Name_______________________________

Summer 1999

Dr. Michael Blaber SS#________________________________

Exam #4 100 points total

Friday Aug 6 1999
  1. Name three characteristics of liquids that distinguish them from gases (6 points)
  1. Liquids have a defined volume, gases expand to fill the volume of their container (have no defined shape or volume)
  2. Gases are highly compressible, liquids are not (liquids are a condensed phase)
  3. The balance between attractive forces and kinetic energy of molecules in liquids are such that neighboring molecules are held in close proximity; in gases the kinetic energy in relationship to the attractive forces is such that neighboring molecules separate from one another
  4. Gases form homogenous mixtures regardless of the chemical properties of the component molecules; chemically distinct liquids may not mix (e.g. oil and water)
  1. A submarine at the bottom of the ocean experiences a pressure of 3.44 x 107 N/m2 due to the water above. Assuming that 1 m3 of seawater has a mass of 1x103 kg, and gravitational force is 9.78 m/s2, how deep is the submarine? (10 points)

    2. P = F/A = 3.44 x 107 N/m2

    F = (mass * acceleration), therefore 3.44 x 107 N = mass * 9.78 m/s2

    Mass = 3.44 x 107 N/9.78 m/s2 = 3.52 x 106 N s2/m

    Since 1 N = 1 kg m/s2 , mass = 3.52 x 106 kg

    This is the mass for a column of water with a 1m2 cross section

    Since 1 m3 of seawater has a mass of 1 x 103 kg, this means that there is 1 x 103 kg for each meter in height for a column of water with a cross section area of 1m2. Thus, the column of water would be 3.52 x 103 meters high, and this is the depth of the submarine.

    Or F = (1x103Kg)*(9.78m/s2) or 9.78x103N for each 1 m depth.

    -3 for calculation error. 2 points each for correctly used P=F/A and F=mass*acc formulas
  2. What is the pressure, in atmospheres, of the gas in the following open-tube manometer? Assume that atmospheric pressure is 99.3 kPa. (5 points)

    3. 99.3 kPa (1 atm / 101 kPa) = 0.983 atm (1point)

    53 mm Hg (1 atm / 760 mm Hg) = 0.0697 atm (1 point)

    Pressure = 0.983 + 0.0697 = 1.05 atm (3 points)

     

  3. A 4.56 liter sample of gas at 60° C and 0.893 atm of pressure is subsequently compressed to 0.127 liters. In response to being compressed, the temperature increases to 156° C. What is the pressure of the gas after this compression? (5 points)

    4. P1V1/T1 = P2V2/T2

    P2 = (T2/V2) * (P1V1/T1) = (273+156 K)/(0.127 L) * (0.893 atm * 4.56 L)/(273+60 K)

    P2 = 3378 K/L * 0.0122 atm L/K = 41.2 atm

    (-2 for wrong math)

  4. A weather balloon holds 423 L of gas at standard temperature and pressure. How many moles of gas are in the balloon? (5 points)

    5. PV = nRT, therefore, n = PV/RT = (1 atm * 423L)/(0.0821 L atm/mol K * 273 K)

    n = 18.9 moles

    (-2 for wrong math)

  5. What is the volume of a 0.523 mol sample of gas at 18.6 atm pressure at 15° C? (5 points)

    6. PV = nRT

    V = nRT/P = 0.523 mol * 0.0821 L atm /mol K * (273+15) / 18.6 atm

    V = 0.665 L

    (-2 for wrong math)
  6. What is the density of Nitrous oxide (N2O) gas at 1.53 atm and 102°C? (5 points)

    7. N2O = (2*14) + 16 = 44 g/mol

    d = PM/RT = (1.53 atm * 44 g/mol)/(0.0821 L atm/mol K * (273+102 K))

    d = 2.19 g/L

    (-2 for wrong math)
  7. An artificial atmosphere for the planet Mars consists of 80% carbon dioxide (CO2), 15% methane (CH4), and 5% Nitrogen (N2). 10g of this gas mixture is placed in a 100L container at 4°C. What is the partial pressure of each gas? (6 points) What is the mole fraction of each compound in the artificial atmosphere? (6 points)

    8. Carbon dioxide (CO2): (10g * 0.80)(1 mol/44g) = 0.182 mol

    Methane (CH4): (10g * 0.15)(1 mol/16g) = 0.0938 mol

    Nitrogen(N2): (10g * 0.05)(1 mol/28g) = 0.0179 mol

    PV = nRT

    P = nRT/V

    P(CO2) = 0.182 mol * 0.0821 mol L/atm K * (273+4K) / 100L = 0.0414 atm

    P(CH4) = 0.0938 mol * 0.0821 mol L/atm K * (273+4K) / 100L = 0.0213 atm

    P(N2) = 0.0179 mol * 0.0821 mol L/atm K * (273+4K) / 100L = 0.00407 atm

    Mole fraction CO2 = 0.0414/(0.0414+0.0213+0.00407) = 0.620

    Mole fraction CH4 = 0.0213/(0.0414+0.0213+0.00407) = 0.319

    Mole fraction N2 = 0.00407/(0.0414+0.0213+0.00407) = 0.061

     

  8. A space ship contains a hydrogen tank with a volume of 5.2 x 106 L. How many grams of hydrogen will the tank hold if it can be pressurized to 1000 kPa at 22°C? (5 points). If the space ship burned all this hydrogen in the presence of oxygen, how many grams of water would be produced? (5 points)

    9. 1.0 x 103 kPa (1 atm/101 kPa) = 9.90 atm

    PV = nRT

    n = PV/RT = 9.90 atm * 5.2 x 106L / 0.0821 L atm/mol K * (273+22)K = 2.13 x 106 moles

    2.13 x 106 moles (2.02 g/mol) = 4.30 x 106 g or 4.30 x 103 kg (2 points if moles correct)

    2H2 + O2 -> 2H2O (2 points)

    2.13 x 106 moles H2 * (2 mol H2O/2 mol H2) = 2.13 x 106 moles H2O

    2.13 x 106 moles H2O * (18g / mol) = 38.8 x 106 g H2O or 38.8 x 103 kg
  9. A box contains 4 molecules of diatomic oxygen. Their individual velocities are 4.7, 3.8, 14.9 and 9.6 m/s. What is the average kinetic energy (in Joules) of these gaseous molecules? (5 points)

    10. RMS velocity = ((4.72+3.82+14.92+9.62)/4)1/2 = 9.36 m/s (2.5 points)

    Ek = ½ * (32 amu)*(1.66 x 10-24 g/amu) * (9.36 m/s)2

    Ek = 2.33 x 10-21 g m2/s2 = 2.33 x 10-24 kg m2/s2 = 2.33 x 10-24 J (2.5 points)

     

     

  10. What is the rate of diffusion for a molecule of water compared to a molecule of carbon dioxide? (5 points)

    11. r1/r2 = (M2/M1)1/2 (2 points)

    rH2O/rCO2 = (44 g/mol /18 g/mol)1/2 = 1.56

    a water molecule diffuses 1.56 times faster than carbon dioxide

     

     

  11. Van der Waals constants for diatomic hydrogen are a = 0.244 L2 atm/mol2 and b= 0.0266 L/mol. Calculate the pressure exerted by 30.2 g of hydrogen gas at 12° C in a container with a volume of 1.2 L (5 points)

    12. T = 273 + 12 = 285 K

    30.2g H2 * 1 mol/2.02g = 15.0 mol (2 points for correct van der Waals equation)
  12. What is the enthalpy change associated with the heating of 117g of H2O from 235K to 385K? (10 points)

    13. 117 g * (1mol/18g) = 6.5 moles

    heating of ice from 235K to 273K = 2.09 J/gK * (38K) * 117g = 9.29 kJ (2 points)

    fusing ice to water = 6.01 kJ/mol * 6.5 moles = 39.1 kJ (2 points)

    heating water from 273K to 373K = 4.18 J/gK * (100K) * 117g = 48.9 kJ (2 points)

    vaporizing liquid water to gas = 40.67 kJ/mol * (6.5 mol) = 264 kJ (2 points)

    heating water vapor from 373 to 385K = 1.84 J/g K * (12K) * 117g = 2.58 kJ (2 points)

    total = 9.29 + 39.1 + 48.9 + 264 + 2.58 kJ = 364 kJ
  13. Identify the changes in processes indicated in the following phase diagram (6 points)

    14. 1. Deposition

    2. Condensation

    3. Freezing

     

  14. Name and describe the four types of intermolecular forces (5 points) What type of intermolecular force probably contributes most to the freezing of water? (1 Point)

Ion-dipole: attractive force between an ion and the (oppositely) partially charged end of a dipolar molecule

Dipole-dipole: attractive force between neutral polar molecules. Electrostatic attraction between the oppositely charged partial charges of the dipoles

London dispersion forces: attractive force between nonpolar molecules. Arises from transient dipoles that can induce dipoles in neighboring molecules.

Hydrogen bonds: special kind of dipole-dipole interaction between hydrogen atom in a polar bond and an unshared electron pair in a nearby electronegative atom.

(Note: ionic bonds are not part of the list of intermolecular forces)

Hydrogen bonds

 

 

 

Useful tables and Constants

Acceleration due to Earth Gravity: 9.78 m/s2

1 Pa = 1 N/m2

1 N = 1 kg m/s2

1 atm = 760 torr = 101 kPa = 760 mm Hg

Specific Heats:

Ice: 2.09 J/g K

Water: 4.18 J/g K

Steam: 1.84 J/g K

Heat of fusion of Water: 6.01 kJ/mol

Heat of vaporization of Water: 40.67 kJ/mol