Learning Center
Plans & pricing Sign in
Sign Out
Get this document free

Slide 1 - Cardinal Newman


									Kinetic Molecular Theory of
 Know the points of KMT points p 115
    1. gases are mainly empty space
      2. gas molecules are in constant straight
 line motion until they collide with something
        gas pressure = caused by the collisions of gas
        molecules with the sides of the container
    3. collision are totally elastic
               no energy is lost
  Description of Gases

Need 4 variables to describe a gas
  Amount of gas present = moles
Volume and Amount of Gas
 Vol = space the gas occupies
     Units = L, mL or cm3
       1 mL = 1 cm3
     Gases fill any container
       diffusion
     Gas vol can change
       Gases are compressible

 Amount of a gas is measured in
     Avogadro's # of particles
     If given grams convert to moles
Use only Kelvins in this chapter!!!
Units ºC or Kelvins
  ºC + 273 = K
Temp is a meas of average kinetic
energy of particles
  Ek = ½ mass x (velocity)2
   temp   movement      kinetic energy
Temp tells us the direction of
energy flow

  Pressure is caused by gas molecules
colliding with the sides of a container
  Pressure is measured with a meter
      Used to meas press of an open sample of gas
      Used to meas the press in a closed system

Measures the pressure of the
 based on the height of a column of
 mercury that air will support
 Measured in mm of mercury
   Mm Hg

Measures the pressure in a closed
system by using a mercury plug
between the gas sample and the
The press of the atmosphere is
The press of the gas is related to
the press of the atmosphere
    Units of Pressure

mm Hg (also called torr)
  atmospheric pressure ~ 760 mm Hg
1 atmosphere = 760 mm Hg
101.3 Kilopascal (KPa) = 760 mm Hg
  1.0 pascal = 1 newton/m2
    Basic SI unit of pressure
Be able to convert from one unit to
Ideal Gas Law: PV = nRT

The ratio of the (pressure x vol) of
a contained sample of a gas, to the
(number of moles x pressure) is a
R = PV      P = press   V = vol
    nT      n = mole    T = temp
R has specific units
    L x atm       the 4 variables must be
  mol x Kelvins   measured in these units
R = 0.0821 L atm/mol K
                Gases Laws
Charles’ Law = vol and temp
  Vi = Vf = k           direct
  Ti Tf             determine 0 Kelvins
Boyle’s Law = vol and pressure
  PiVi = PfVf = k       indirect

Avogadro’ Law = vol and moles
  Vi = Vf = k           direct
  ni n f
Guy-Lusaac’s Law = press and temp
  Ti = Tf = k           direct
  Pi Pf

 inverse      direct

 direct       direct
         PV = nRT

4 types of problem:
  Calcs of P, V, n, or T
  Initial and final state problems
  Molar mass determinations
  Density determinations
           Calcs of P, V, n, or T

Given 3 of the 4 variables deter the other
  PV = nRT (R = 0.0821 Latm/molK)
  A 2.50 g sample of SF6 is placed in a 500.0 mL
  vessel at 83ºC. What pressure in mm Hg is
  developed? (make a table of given variables)
n = 2.50 g SF6 x 1 mol SF6 = 0.0171 mol SF6
                 146.1 g SF6
V = 500.0 mL x    1L       = 0.5000 L
                 1000 mL
T = 83ºC + 273 = 356 K

P = nRT = (0.0171 mol)(0.0821 L atm/mol)(356 K) = 1.00 atm
     V                  (0.5000 L)
 Initial and final state problems
  Given a change in conditions (P, V, n, or
  T) of a gas, determine the new
  conditions (P, V, n, or T)
  Units are not as important (except T)
    Units must be consistent
  PiVi = Ri = Rf = PfVf
  niTi              nfTf
Cancel out any values that remain constant.
  In the following example n is constant
  ni = nf and Ti = Tf they cancel out
  PiVi = Ri = Rf = PfVf
  niTi              nfTf
  Find the volume of a gas at 800.0
  mm Hg and 40.0ºC, if it's vol at
  720.0 mm Hg and 15.0ºC is 6.84 L?
       Make a table: initial         final
             P     800 mm Hg       720 mm Hg
 n is held
 constant    T     313 K           288 K
             V      Vi             6.84 L
  PiVi = PfVf    Vi = PfVfTi
  niTi nfTf            Tf Pi
Vi = PfVfTi = (720 mmHg)(6.84 L)(313 K) = 6.69 L
     nfTfPi      (288 K)(800 mmHg)
 Calculation of Molar Mass
    Mol = grams
Subbing into Ideal gas law
   PV = nRT    PV = gRT
Therefore: mm = gRT
If you know the grams of a gas at a
specific temp, press, and vol you
can determine it’s molar mass
     Determination of Density
  Density = mass/vol
  PV = nRT      n = mol = g/mm
  PV = gRT or g = P(mm) = density
       mm      V     RT
Example: Determine the density, in g/L, of ammonia (NH3),
  at 97ºC and 755 mm Hg?
  P = (755/760 atm) = 0.993 atm
  mm = (17.0 g/mol)
  T = (97ºC + 273) = 370 K

D = Pmm = (0.993 atm)(17.0 g/mol)
     RT    (0.0821 L atm/mol K)(370 K)
  = 0.556 g/L
                       Gas Stoichiometry

    What is the volume of CO2 produced at 37.0˚C and 1.00
    atm when 5.60 g of glucose are used up in the reaction:

    C6H12O6 (s) + 6O2 (g)               6CO2 (g) + 6H2O (l)
g C6H12O6          mol C6H12O6          mol CO2      V CO2

                   1 mol C6H12O6     6 mol CO2
5.60 g C6H12O6 x                 x               = 0.187 mol CO2
                   180 g C6H12O6   1 mol C6H12O6

                   0.187 mol x 0.0821       x 310.15 K
          nRT                         mol•K
     V=       =                                          = 4.76 L
           P                     1.00 atm
                   Gas Stoichiomerty
PROBLEM:    A laboratory method for reducing a metal oxide is to heat it
            with H2. The pure metal and H2O are products. What
            volume of H2 at 765 torr and 225˚C is needed to form 35.5g
            of Cu from copper (II) oxide?

PLAN:   Since this problem requires stoichiometry and the gas laws, we
        have to write a balanced equation, use the moles of Cu to
        calculate mols and then volume of H2 gas.
                SOLUTION:      CuO(s) + H2(g)              Cu(s) + H2O(g)

                                   mol Cu       1 mol H2
                       35.5g Cu                             = 0.559mol H2
                                  63.55g Cu     1 mol Cu
        PV = nRT
                                                L atm
                     0.559 mol H2 x 0.0821              x    498K
        V = nRT                                 mol K
                                                                       = 22.6L
             P                      765 atm
Dalton's Law of Partial Pressure

Two ways of looking at this law
  1 - The total pressure of a mixture of
  gases is equal to the sum of the
  pressures of the individual gases if
  alone in the same volume
  Ptotal = Pgas1 + Pgas2 + Pgas3 …
  Used a lot for gases collected under
    Water vapor pressure must be accounted
    for in the collected gas sample
    Ptotal = Pgas a - Pwater
  Gases Collected over H2O
A 152 mL sample of O2 gas is collected under
water at 25°C (vpH2O= 23.76 mm). If the O2
exerts a pressure of 758 mm Hg determine:
a. the partial pressure of the O2
   Ptotal = PH2O + PO2
   Ptotal - PH2O = PO2
   758 - 23.76 = 734 mm Hg
b. the grams of O2 collected
nH2 = PH2V = (734 / 760 atm)(0.152 L) = 0.00600 mol O2
       RT (0.0821 L atm/mol K)(298K)
    = 0.00600 mol O2 x 32.0 g = 0.192 g O2
Dalton's Law of Partial Pressure

2-The pressure of the components of a gas
mixture is relative to their percents or mole
  Mol frac =    mole gas A
              total moles gases
I know it's a decimal, but it's called mole fraction!!!

  Mol fraction is the same as percent
  PgasA = Ptotal x MFgasA or
  PgasA = Ptotal x PercentgasA(as a decimal)
Dalton's Law of Partial Pressure
A gas sample contains 2.0 mol O2, 1.0
mol N2, and 3.0 mol He. Determine
  The mol fraction of each gas.
    MFO2 = 2.0/6.0 = 0.33
    MFN2 = 1.0/6.0 = 0.17
    MFHe = 3.0/6.0 = 0.50
  The partial press of each gas if the
  total press on the system = 950 mm Hg
    PPO2 = 950 x 0.33 = 314 mm Hg
    PPO2 = 950 x 0.17 = 162 mm Hg
    PPO2 = 950 x 0.50 = 475 mm Hg
  Dalton's Law of Partial Pressure
A sample of natural gas contains 8.24 moles of CH4,
0.421 moles of C2H6, and 0.116 moles of C3H8. If
the total pressure of the gases is 1.37 atm, what is
the partial pressure of propane (C3H8)?

Pi = Xi PT     PT = 1.37 atm

Xpropane =                            = 0.0132
           8.24 + 0.421 + 0.116

Ppropane = 0.0132 x 1.37 atm = 0.0181 atm
  Velocity of Gas Molecules

Root mean speed = u

 U = 3RT ½
Thermodynamics = diff value for R
 R = 8.31 x 103 g m2 / s2 mol K
   Calc the root mean speed, u, of N2 at 25°C
      u = (3)(8.31 x 103 g m2 / s2 mol K)(298K) ½
                    (28.0 g / mol)
         = 515 m/s
Graham's Law of Effusion
Effusion = movement of gas
through a small opening or pore
Lighter gases effuse faster than
heavier gases at the same T and P
 ua = ratea = mmb   ½   = timeb
 ub rateb     mma         timea
Determination of Molar Masses

An unknown gas effuses 1.66 times
slower than CO2., Determine the
molar mass of the gas
rateCO2 = mmx ½
ratex     mmCO2

1.66 =     mmx      ½

         44.0 g/mol

mmx = 121 g/mol
         Rates of Effusion
Consider H2 and O2 gases at the same
temp and pressure. Which gas
effuses faster? How much faster?
  H2 will effuse faster as it is lighter
rateH2 =    mmO2   ½   =   32.0 g/mol ½
rateO2      mmH2           2.0 g/mol

rateH2   = 4.0 H2 will effuse 4.0 times faster
       Rates of Effusion

Hydrogen gas, H2 effuses 4.5 times
faster than gas X. Determine the molar
mass of gas X.
rateH2 =    mmX    ½

rateX       mmH2
 4.5    =     X    ½
 1.0         2.0

 20.25 x 2.0 = X = 40.5 g/mol
Boltzmann’s Distribution
      Real vs. Ideal Gases
Ideal gases obey the gas laws
  PV/nT = constant
  Occurs at room conditions of T and P
Gases don't always obey the gas laws
  PV/nT is not a constant
  Called "real gas"
  Occurs under extreme conditions
Gases act in an ideal or real manner
  Depends on conditions
       Real vs. Ideal Gases

Under extreme conditions the gases don't
follow the gas laws
  Low temp
    Gas molecules stay by each other long enough to be
    attached together
      – V (and PV/RT) is too small
  At high pressure (200-400 atm)
    Gas molecules close enough to attract each other
      – V (and PV/RT) is too small
  Extreme press (above 400 atm)
    Gas molecules squeezed together and e- clouds repel
      – V (and PV/RT) too large
  Van der Waal's Equation

Amends the IGL for real behavior

 a and b are constants for each diff gas
   a corrects for attractive forces
     – Occurs at low T or high P
   b corrects for V of gas molecules
     – Occurs at extreme P

To top