Chapter 9: Gases by Q33otz7

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									Chapter 5: Gases

          Renee Y. Becker
     Valencia Community College
              CHM 1045

                                  1
a) Gas is a large collection of particles moving at random
     throughout a volume
b) Collisions of randomly moving particles with the
walls of the container exert a force per unit area that we
perceive as gas pressure




                                                             2
• Units of pressure: atmosphere (atm)
                 Pa (N/m2, 101,325 Pa = 1 atm)
                        Torr (760 Torr = 1 atm)
                      bar (1.01325 bar = 1 atm)
                  mm Hg (760 mm Hg = 1 atm)
                   lb/in2 (14.696 lb/in2 = 1 atm)
                   in Hg (29.921 in Hg = 1 atm)




                                              3
                               Boyle’s Law


• Pressure–Volume Law (Boyle’s Law):




                                         4
                                         Boyle’s Law


• Pressure–Volume Law (Boyle’s Law):
                         1
            Volume 
                     Pressure
• The volume of a fixed amount of gas maintained at
  constant temperature is inversely proportional to
  the gas pressure.
• Can use any units for P & V


   V1P1  k1                P1V1 = P2V2
                                                   5
                            Example 1: Boyle’s Law


A sample of argon gas has a volume of
  14.5 L at 1.56 atm of pressure. What
  would the pressure be if the gas was
  compressed to 10.5 L? (at constant
  temperature and moles of gas)




                                                 6
                                Charles’ Law


• Temperature–Volume Law (Charles’ Law):




                                           7
                                         Charles’ Law


• Temperature–Volume Law (Charles’ Law):

                 V  T
• The volume of a fixed amount of gas at constant
  pressure is directly proportional to the Kelvin
  temperature of the gas.
• Use Kelvins for temperature!!

     V1                     V1 = V2
     T1 =k1                 T1   T2
                                                    8
                           Example 2: Charles’ Law


A sample of CO2(g) at 35C has a volume of
  8.56 x10-4 L. What would the resulting
  volume be if we increased the temperature
  to 85C? (at constant moles and
  pressure)




                                                 9
                            Avogadro’s Law


• The Volume–Amount Law (Avogadro’s
  Law):




                                        10
                                        Avogadro’s Law

• The Volume–Amount Law (Avogadro’s Law):

                 V n
• At constant pressure and temperature, the volume
  of a gas is directly proportional to the number of
  moles of the gas present.
• Use any volume and moles

    V1
        k1                 V1 = V2
    n1                      n1    n2                11
                           Example 3: Avogadro’s Law


6.53 moles of O2(g) has a volume of 146 L. If
  we decreased the number of moles of
  oxygen to 3.94 moles what would be the
  resulting volume? (constant pressure and
  temperature)




                                                  12
                                  Combined Gas Law



We can combine Boyle’s and charles’ law to
 come up with the combined gas law


          P  V1   P2  V2
           1
                 
            T1       T2
Use Kelvins for temp, any pressure, any volume


                                                 13
                        Example 4:Combined Gas Law


Oxygen gas is normally sold in 49.0 L steel
 containers at a pressure of 150.0 atm.
 What volume would the gas occupy if the
 pressure was reduced to 1.02 atm and the
 temperature raised from 20oC to 35oC?




                                                 14
                                 Example 5: Gas Laws



An inflated balloon with a volume of 0.55 L at
  sea level, where the pressure is 1.0 atm, is
  allowed to rise to a height of 6.5 km, where
  the pressure is about 0.40 atm. Assuming
  that the temperature remains constant, what
  is the final volume of the balloon?



                                                       15
                                       The Ideal Gas Law


• Ideal gases obey an equation incorporating the laws
  of Charles, Boyle, and Avogadro.


              P V  n  R  T
• 1 mole of an ideal gas occupies 22.414 L at STP
• STP conditions are 273.15 K and 1 atm pressure
• The gas constant R = 0.08206 L·atm·K–1·mol–1
   – P has to be in atm
   – V has to be in L
   – T has to be in K
                                                        16
                              Example 6: Ideal Gas Law


Sulfur hexafluoride (SF6) is a colorless,
  odorless, very unreactive gas. Calculate the
  pressure (in atm) exerted by 1.82 moles of
  the gas in a steel vessel of volume 5.43 L at
  69.5°C.




                                                         17
                         Example 7: Ideal Gas Law


What is the volume (in liters) occupied by
 7.40 g of CO2 at STP?




                                                    18
                                 The Ideal Gas Law


• Density and Molar Mass Calculations:


        mass n  MM P  MM
    d            
       volume    V   R T

• You can calculate the density or molar mass
  (MM) of a gas. The density of a gas is
  usually very low under atmospheric
  conditions.
                                                 19
                             Example 8: Density & MM


What is the molar mass of a gas with a density
  of 1.342 g/L at STP?




                                                       20
                            Example 9: Density & MM



What is the density of uranium hexafluoride,
 UF6, (MM = 352 g/mol) under conditions of
 STP?




                                                 21
                          Example 10: Density & MM



The density of a gaseous compound is 3.38
 g/L at 40°C and 1.97 atm. What is its
 molar mass?




                                                22
Dalton’s Law of Partial Pressures




                                    23
                             Dalton’s Law of Partial Pressures


• In a mixture of gases the total pressure, Ptot, is the
  sum of the partial pressures of the gases:


                       RT
             P total 
                       V
                                     n
• Dalton’s law allows us to work with mixtures of
  gases.
   • T has to be in K
   • V has to be in L                                            24
                            Example 11: Dalton


Exactly 2.0 moles of Ne and 3.0 moles of
 Ar were placed in a 40.0 L container at
 25oC. What are the partial pressures of
 each gas and the total pressure?




                                             25
                              Example 12: Dalton


A sample of natural gas contains 6.25
  moles of methane (CH4), 0.500 moles of
  ethane (C2H6), and 0.100 moles of
  propane (C3H8). If the total pressure of
  the gas is 1.50 atm, what are the partial
  pressures of the gases?




                                               26
                              Dalton’s Law of Partial Pressures


• For a two-component system, the moles of
  components A and B can be represented by
  the mole fractions (XA and XB).
                                          nA
Mole fraction is related to        XA 
the total pressure by:                  nA  nB

 Pi  X i Ptot
                                          nB
                                   XB 
                                        nA  nB
                                   X A  X B 1
                                                                  27
                        Example 13: Mole Fraction


What is the mole fraction of each
 component in a mixture of 12.45 g of
 H2, 60.67 g of N2, and 2.38 g of NH3?




                                                    28
                      Example 14: Partial Pressure


On a humid day in summer, the mole
 fraction of gaseous H2O (water vapor) in
 the air at 25°C can be as high as
 0.0287. Assuming a total pressure of
 0.977 atm, what is the partial pressure
 (in atm) of H2O in the air?



                                                     29
                          Gas Stoichiometry & Example

• In gas stoichiometry, for a constant
  temperature and pressure, volume is
  proportional to moles.


Example: Assuming no change in temperature
  and pressure, calculate the volume of O2 (in
  liters) required for the complete combustion
  of 14.9 L of butane (C4H10):

 2 C4H10(g) + 13 O2(g)  8 CO2(g) + 10 H2O(l)
                                                        30
                                  Example 15:

All of the mole fractions of elements in a
    given compound must add up to?

1.   100
2.   1
3.   50
4.   2


                                                31
                                 Example 16:


Hydrogen gas, H2, can be prepared by
 letting zinc metal react with aqueous
 HCl. How many liters of H2 can be
 prepared at 742 mm Hg and 15oC if
 25.5 g of zinc (MM = 65.4 g/mol) was
 allowed to react?

Zn(s) + 2 HCl(aq)  H2(g) + ZnCl2(aq)


                                           32
                                    Kinetic Molecular Theory

• This theory presents physical properties of gases in
  terms of the motion of individual molecules.

   • Average Kinetic Energy  Kelvin Temperature
   • Gas molecules are points separated by a great
     distance
   • Particle volume is negligible compared to gas
     volume
   • Gas molecules are in rapid random motion
   • Gas collisions are perfectly elastic
   • Gas molecules experience no attraction or
     repulsion                                                 33
Kinetic Molecular Theory




                       34
 • Average Kinetic Energy (KE) is given
   by:
                      U = average speed of a
         1
     KE  mu 2        gas particle
         2            R = 8.314 J/K mol
                      m = mass in kg
     3RT       3RT    MM = molar mass, in
u                   kg/mol
     mNA       MM
                      NA = 6.022 x 1023



                                               35
• The Root–Mean–Square Speed: is a
  measure of the average molecular
  speed.

     3RT                               3RT
 u 
  2
                             u rms 
     MM                                MM
        Taking square root of both
        sides gives the equation




                                             36
                                Example 17:


Calculate the root–mean–square speeds
 of helium atoms and nitrogen molecules
 in m/s at 25°C.




                                              37
                          Kinetic Molecular Theory


• Maxwell speed distribution curves.




                                                 38
                        Graham’s Law


• Diffusion is the
  mixing of different
  gases by random
  molecular motion
  and collision.




                                   39
                         Graham’s Law


• Effusion is when
  gas molecules
  escape without
  collision, through a
  tiny hole into a
  vacuum.




                                    40
                                        Graham’s Law


• Graham’s Law: Rate of effusion is
  proportional to its rms speed, urms.

                                3RT
                Rate u rms   
                                MM

• For two gases at same temperature and
  pressure:
       Rate1         MM 2             MM 2
                              
       Rate2         MM1              MM1
                                                   41
                                  Example 18:



Under the same conditions, an unknown
 gas diffuses 0.644 times as fast as
 sulfur hexafluoride, SF6 (MM = 146
 g/mol). What is the identity of the
 unknown gas if it is also a hexafluoride?




                                                42
                             Example 19: Diffusion



• What are the relative rates of diffusion
  of the three naturally occurring isotopes
  of neon: 20Ne, 21Ne, and 22Ne?




                                                     43
                                 Behavior of Real Gases


• Deviations result from assumptions about
  ideal gases.

  1. Molecules in gaseous state do not exert
     any force, either attractive or repulsive, on
     one another.

  2. Volume of the molecules is negligibly small
     compared with that of the container.


                                                     44
                              Behavior of Real Gases



• At higher pressures, particles are much
  closer together and attractive forces become
  more important than at lower pressures.




                                                  45
                               Behavior of Real Gases


• The volume taken up by gas particles is
  actually less important at lower pressures
  than at higher pressure. As a result, the
  volume at high pressure will be greater than
  the ideal value.




                                                   46
                                 Behavior of Real Gases


• Corrections for non-ideality require van der
  Waals equation.



        n      2
   P  a 2   V – n  b   nRT
           
        V 

                     Excluded
                      Volume
Intermolecular
  Attractions
                                                     47
                     Example 20: Ideal Vs. Van Der Waals


Given that 3.50 moles of NH3 occupy 5.20 L
   at 47°C, calculate the pressure of the
   gas (in atm) using


  (a) the ideal gas equation


  (b) the van der Waals equation. (a = 4.17, b =
      0.0371)


                                                           48
                  Example 21: Ideal Vs. Van Der Waals


Assume that you have 0.500 mol of N2 in
 a volume of 0.600 L at 300 K. Calculate
 the pressure in atmospheres using both
 the ideal gas law and the van der Waals
 equation.

• For N2, a = 1.35 L2·atm mol–2, and b =
  0.0387 L/mol.


                                                        49

								
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