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Chapter 11

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Chapter 11 Chem 130


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Lecture Notes and a Chapter 11 Summary with Review Questions
General Chemistry II
Dr. Yang
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This 12 page Bundle was uploaded by n/a on Thursday January 21, 2016. The Bundle belongs to Chem 130 at University of Tennessee - Knoxville taught by Dr. Yang in Spring 2016. Since its upload, it has received 212 views. For similar materials see General Chemistry II in Chemistry at University of Tennessee - Knoxville.


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Date Created: 01/21/16
January  20,  2016   CHAPTER  11:  GASES   1.   KEY  LEARNING  OBJECTI VES   i.   Converting  between  Pressure  Units   ii.   Relating  Volume  and  Pressure:  Boyle's  Law   iii.   Relating  Volume  and  Temperature:  Charles's  Law   iv.   Relating  Volume  and  Moles:  Avogadro's  Law   v.   Determining  P,  V,  n  or  T  Using   the  Ideal  Gas  Law   vi.   Relating  the  Density  of  a  Gas  to  Its  Molar  Mass   vii.   Calculating  the  molar  Mass  of  a  Gas  with  the  Ideal  Gas  Law   viii.   Calculating  Total  Pressure,  Partial  Pressures,  and  Mole  Fractions  of  Gases  in  a   Mixture   ix.   Calculating  the  Root  Mean  Square  Velocity  o f  a  Gas   x.   Calculating  the  Effusion  Rate  or  the  Ration  of  Effusion  Rates  of  Two  Gases   xi.   Relating  the  Amounts  of  Reactants  and  Products  in  Gaseous  Reactions:   Stoichiometry   2.   KEY  CONCEPTS     i.   Gas  Pressure:  force  per  unit  area  resulting  from  colliding  gas  particles   ii.   Simple  Gas  Laws:       a.  Boyle's  Law:  P V =P V 1 1 2 2     b.    Charles's  Law:  V 1T =V 1T 2 2     c.  Avogadro's  Law:  V /n =V 1n 1 2 2   iii.   Ideal  Gas  Law:  PV=nRT       a.  Can  be  used  to  calculate  molar  volume  (22.4  L  at  STP),  density,  and  molar         mass     b.  Derived  from  simple  gas  laws   iv.   Mixtures  of  Gases  and  Partial  Pressures     a.  Gases  act  independently  in  mixtures,  so  the  total  pressure  can  be  found  by         adding  partial  pressures  of  the  individual  gases.     b.  total=P +Pa+P +.b.  c   P a(n RT)aV         b   P b(n RT)/V     c   c   P =(n RT)/V     v.   Gas  Stoichiometry     a.  Values  are  frequently  reported  at  a  specified  pressure  and  temperature.         We  can  convert  quantities  to  amounts  in  moles  using  the  ideal  gas         law  and  then  use  stoichiometric  coefficients  to  determine  amounts  of       other  reactants  or  products.     b.  General  Calculation:  volume  A →amount  A  (in  moles)  →amount  B  (in         moles)  →quantity  of  B  (in  desired  unit)     c.  At  STP  22.4  L=1  mol   vi.   Mole  Fraction     a.  X =n /a a total     b.  P =X P a a total                        vii.            Kinetic  Molecular  Theory       a.  Three  assumptions:  gas  particles  are  negligibly  small;  the  average  kinetic           energy  of  a  gas  particle  is  proportional  to  the  temperature  in  kelvin;           the  collision  of  one  gas  particle  with   another  is  completely  elastic       b.  We  can  derive  the  root  mean  square  ve locity  of  gas  particles  with  the           theory.  The  RMS  shows  that  smaller  gas  particles  are   moving  more           quickly  than  larger  ones  at   a  given  T.       c.  Real  gases  do  not  always  follow  the  assumptions  of  kinetic  molec ular           theory.                                           2   Review  Questions   1.    What  is  a  manometer?  How  does  it  measure  the  pressure  of  a  sample  of  gas?         2.  Explain  how  the  ideal  gas  law  contains  within  it  the  simple  gas  laws  (show  an   example).           3.  When  a  gas  is  collected  over  water,  is  the  gas  pure?  Why  or  why  not?  How  can  the   partial  pressure  of  the  collected  gas  be  determined?           4.  The  pressure  on  top  of  Mount  Everest  (elevation  29,029  ft)  averages  about  235   mmHg.  Convert  this  pressure  to:     a)torr             c)Pa       b)psi             d)  atm       5.  A  sample  of  gas  has  an  initial  volume  of  13.9  L  at  a  pressure  of  1.22  atm.  If  the   sample  is  compressed  to  a  volume  of  10.3  L,  what  is  the  pressure?         6.  A  balloon  contains  0.158  mol  of  gas  and  has  a  volume  of  2.46  L.  If  we  add  0.113   mol  of  gas  to  the  balloon  (constant  T  and  P)  what  is  the  final  volume?         3   7.  A  1.0-­‐‑L  container  of  liquid  nitrogen  is  kept  in  a    closet  measuring  1.0  m  by  1.0  m   by  2.0  m.  Assuming  that  the  container  is  completely  full,  that  the  temperature  is   25.0°C,  and  that  the  atmospheric  pressure  is  1.0  atm,  calculate  the  percent  (by   volume)  of  air  that  is  displaced  if  all  of  the  liquid  nitrogen  evaporates.  (Liquid   nitrogen  has  a  density  of  0.807  g/mL).             8.  What  is  the  pressure  in  a  15.0  L  cylinder  filled  with  32.7  g  of  oxygen  gas  at  a   temperature  of  302  K?           9.  Use  the  molar  volume  of  a  gas  at  STP  to  determine  the  volume  (in  L)  occupied  by   33.6  g  of  neon  at  STP.           10.  A  248-­‐‑mL  gas  sample  has  a  mass  of  38.8  mg  at  a  pressure  of  721  mmHg  and  a   temperature  of  55°C.  What  is  the  molar  mass  of  the  gas?             11.  Use  the  molar  volume  of  a  gas  at  STP  to  calculate  the  density  (in  g/L)  of  nitrogen   gas  at  STP.           4   12.  A  275-­‐‑mL  flask  contains  pure  helium  at  a  pressure  of  752  torr.  A  second  flask   with  a  volume  of  472  mL  contains  pure  argon  at  a  pressure  of  722  torr.  If  the  two   flasks  are  connected  through  a  stopcock  and  the  stopcock  is  opened,  what  is  the   partial  pressure  of  each  gas  and  the  total  pressure?           13.  Oxygen  gas  reacts  with  powdered  aluminum  according  to  the  following  reaction:       4AL(s)  +  3O (g)  →2Al2O (s)   2 3         What  volume  of  O  gas  (in 2L)  measured  at  782  mmHg  and  25°C  completely     reacts  with  53.2  g  Al?           14.  In  a  common  classroom  demonstration,  a  balloon  is  filled  with  air  and   submerged  in  liquid  nitrogen.  The  balloon  contracts  as  the  gases  within  the  balloon   cool.  Suppose  a  balloon  initially  contains  2.95  L  of  air  at  a  temperature  of  25°C  and  a   pressure  of  0.998  atm.  Calculate  the  expected  volume  of  the  balloon  upon  cooling  to   -­‐‑196°C  (the  boiling  point  of  liquid  nitrogen).  When  the  demonstration  is  carried  out,   the  actual  volume  of  the  balloon  decreases  to  0.61  L.  How  does  the  observed  volume   of  the  balloon  compare  to  your  calculated  value?  Explain  the  difference.             15.  Calculate  the  root  mean  square  velocity  and  kinetic  energy  of  CO,  CO ,  and  SO  at   2 3 298  K.  Which  gas  has  the  greatest  velocity?  The  greatest  kinetic  energy?  The   greatest  effusion  rate?           5   Challenge  Problem:  Two  identical  balloons  are  filled  to  the  same  volume,  one  with   air  and  one  with  helium.  The  next  day,  the  volume  of  the  air-­‐‑filled  balloon  has   decreased  by  5.0%.  By  what  percent  has  the  volume  of  the  helium-­‐‑filled  balloon   decreased?  (Assume  that  air  is  four-­‐‑fifths  nitrogen  and  one  fifth  oxygen,  and  that   temperature  did  not  change.)                                                           *Review  Problems  taken  from  Tro  Textbook  Chapter  11  Review       6   Chapter  11:  Gases   CHEM  130  (Yang)  Lecture  January  14,  2016     Collision  of  Gas  Particles  and  Gas  Pressure     -­‐Gasses  are  composed  of  particles  that  are  moving  randomly  and  very  fast  in  their     containers.     -­‐A  gaseous  atom  or  molecule  exerts  a  force  when  it  collides  with:   •   a  surface   •   another  gaseous  atom   -­‐Gas  Pressure:  the  force  exerted  per  unit  area  by  gas  molecules  as  they  strike  the   surfaces  around  them     Pressure  =  force/area  =  F/A   -­‐Gas  Pressure  depends  on:   •   number  of  gas  particles  in  given  volume   •   volume  of  container   •   temperature     Gas  Pressure  Units     -­‐Pascals(Pa):  SI  unit  of  pressure,  1  Pa=1N/m 2   5   -­‐Bar:  1  Bar=10 Pa=100  kPa     -­‐Atmospheres  (atm):  1  atm=101.325  kPa   •   Atmospheres  are  a  measure  of  weight  of  air  per  unit  area.   •   Normal  Atmospheric  Pressure=  Standard  Atmospheric  Pressure  at  sea  level   •   Atmospheres  are  measured  using  a  barometer.   -­‐mmHg:  the  difference  in  the  heights  measured  in  mm(h)  of  2  connected  columns  of   mercury;  also  known  as  torr     Manometer   -­‐This  device  is  used  to  measure  the  difference  in  pressure  between  atmospheric   pressure  and  that  of  a  gas  in  a  vessel.   -­‐If  P =765  mmHg  and  h=60  then:  P =P +H  =  765+60=825  mmHg   atm gas atm   Basic  Properties  of  Gases     -­‐Pressure  (P)   *The  equations  that  express  the     -­‐Volume  (V)   relationship  among  T,  P,  V,  and  n   are  known  as  gas  laws.     -­‐Temperature  (T)     -­‐Amount  of  gas  in  moles  (m)     Simple  Gas  Laws     1.  Boyle's  Law       -­‐Robert  Boyle  (1627-­‐1691)  and  Robert  Hooke  used  a  J-­‐tube  to  measure  the         volume  of  a  sample  of  gas  at  different  pressures.       -­‐Volume  and  Pressure  are  inversely  related  (as  one  increases,  the  other           decreases)   Chapter  11:  Gases       -­‐At  constant  n  and  T,  V  is  inversely  proportional  to  P.       P V1 1 V 2 2     2.  Charles's  Law       -­‐At  constant  n  and  P,  V  increases  linearly  with  T.       -­‐Volume  and  temperature  are  directly  proportional.       -­‐Extrapolation  of  V=0  leads  to  lowest  possible  T,  which       is  the  absolute  zero.       V /1 =V /T1 2 2     3.  Avogadro's  Law       -­‐V  is  directly  proportional  to  the  number  of  gas  molecules  when  P  and  T  are         constant.       V /1 =V /n1 2 2       Ex:  At  constant  T  and  P,  when  volume  is  doubled,  the  number  of  moles  of  a  gas  is                    _______.           Answer:  doubled     Ideal  Gas  Law     -­‐The  simple  gas  laws  relationship  can  be  combined  as:     PV=nRT  where  R  is  a  constant  (0.0821  L*atm/K*mol)     Ex:  A  4.50  L  cylinder  container  contains  He  gas  at  an  unknown  pressure.  it  is  now  connected  to                 a  92.5  L  evacuated  cylinder.  When  the  connecting  valve  between  the  two  cylinders  are     opened,  the  pressure  falls  to  1.40  atm.  What  is  the  pressure  in  the  4.50  L  container?   Given:   V =1.50  L             What  simple  gas  law  relates  V  to  P?   V =  4.50  L  +  92.5  L=  97.0  L           Boyle's  Law  (P V =P V )   2 1 1 2 2 P 2= 1.40  atm         Solve:  Plug  and  chug:  P x  (4.50  L)=1    atm  x  (97.0  L)         Answer:  P =30.2  a1m     Equations  Covered:   **Remember:   P V1 1 V 2 2     V /1 =P /T1 2 2   Always  convert  temperature  to  Kelvin   V /1 =V /n1 2 2     PV=nRT   When  using  constant  R,  make  sure  units     all  match.       Chem  130  Lecture  Notes  1.19.16  and  1.21.16   Ex.  Tennis  balls  are  usually  filled  with  air  or  N  gas  to  a  pressure  above  atmospher2c  pressure  to     increase  their  "bounce".  If  a  particular  tennis  ball  has  a  volume  of  144cm  and  contains   3   0.33g  of  N  gas,  2hat  is  the  pressure  inside  the  ball  at  24°C?     Given:  V=144  cm =  144  mL     3       PV=nRT       T=  24°C  =  297  K         P(144)=(0.33/28)(0.0821)(297)       n=0.33g  (1  mole/28g)         P=2.0  atm     Molar  Volume  at  STP     -­‐For  1,000  mol  of  ideal  gas  at  1.000  atm  and  0.00°C  (273.15  K),  the  V  is  22.4  L  at  STP     -­‐You  can  calculate  this  using  the  ideal  gas  law  and  STP  values  (T=273,  n=1.00  mol,     P=1.00  atm)     -­‐At  constant  T  and  P,  the  gases  with  the  same  volume  have  the  same  moles  and  number      gas  particles.     Ex.  What  is  th  emass  of  16.2  L  of  SF  at  STP?   6   Method  1:  PV=nRT         Method  2:  At  STP,  1  mole  occupies  22.4  L                  (1)(16.2)=n(0.08206)(273)                    16.2  L*(1  mol/22.4  L)=0.723  mol                  n=0.723  mol  SF  (conv6rt  to  g)                  Covert  to  g,  106  g  SF   6 Density  of  a  Gas     -­‐Density=molar  mass/molar  volume     Unit:  g/L     -­‐At  STP:  d =  4.00He  g/mol  =  0.179  g/L                                22.4  L/mol     -­‐Larger  molar  mass=more  dense     -­‐d=m=PM                V        RT     -­‐One  only  needs  to  know  the  molar  mass,  P  and  T  to  calculate  density  of  gas.     *T  increase,  density  decreases        P  increases,  density  increases          Molar  mass  increases,  density  increases     -­‐At  constant  T  and  P,  the  higher  the  molar  mass,  the  denser  the  gas.     Ex.  Which  of  the  following  is  most  dense  at  1.00  atm  and  20°C?       Cl  because  it  has  the  largest  atomic  mass   2 Molar  Mass  of  a  Gas     -­‐molar  mass=  mass  (g)                              mole  (n)     Ex.  Calculate  the  molar  mass  of  a  liquid  which,  when  vaporized,  at  98°C  and  715  mmHg,         yields  121  mL  of  vapor  with  a  mass  of  0.471  g.       T=98°C         Use  PV=nRT  to  calculate  n:       P=715  mmHg       (715/760)(0.121)=n(0.0821)(273+98)       V=  121  mL       n=0.00373       m=  0.471  g               Molar  Mass=0.471  g/  0.00373  mol=126  g/mol     Chem  130  Lecture  Notes  1.19.16  and  1.21.16   Mixtures  of  Gases     -­‐The  pressure  due  to  any  individual  component  in  a  gas  mixture  is  its  partial  pressure     (P )n     -­‐Ideal  Gas  Law  can  be  used  for  each  individual  gas.  P =n (RT/V)   n n   -­‐Dalton's  Law  of  Partial  Pressure:  P =P +P +P +....   total a b c •   Ideal  Gas  Law  can  be  used  with  a  mixture:  P total =n total (RT/V)   Mole  Fraction:  X a     -­‐The  number  of  moles  of  a  component  in  a  mixture  divided  by  the  total  number  of       moles  in  the  mixture  is  the  mole  fraction.     -­‐X =n /n       P =X (P )   a a total a a total   -­‐The  ration  of  the  partial  pressure  a  single  gas  contributes  and  total  pressure  is  equal  to     mole  fraction.     Ex.  The  air  contains  78%  N  by 2volume.  What  is  the  partial  pressure  of  N  in  a  sample 2of         air  at  1.0  atm?       Given:  composition  of  dry  air       P =N278(1.00  atm)=0.78  atm   Ex.  In  a  2.16  L  container  at  323  K  are  placed  3.00  g  He  and  26.0  g  Ne:     a)  What  is  the  pressure  of  each  gas?       P =He RT)/V=(He00  g/4.0  g)(0.0821)(323  K)/(21.6  L)=1.58  atm     b)  What  is  the  total  pressure?       P total    PHeP = Ne    +  1.58  =  2.50  atm     c)  Mole  fraction  of  Ne?       X =1.58  atm/2.50  atm   He Ex.  Consider  an  apparatus  connecting  two  containers  with  a  closed  valve.  One  container     contains  N  at 2  2.0  L,  1.0  atm,  and  25°C.  The  second  container  contains  O  at  3.0  L,  2.0 2    atm,  and  25°C.  When  the  valve  between  the  two  containers  is  opened,  what  is  the     partial  pressure  of  N ?  O ? 2   2   P V1 1 V  2 2   For  N : 2  5.0  L(P)=1.0  atm  (2.0  L)   P=0.40  atm     For  O : 2  5.0  L(P)=2.0  atm(3.0  L)   P=1.2  atm     P total    PN2  P = O2    +  1.2  =  1.6  atm   Collecting  Gas  over  Water     -­‐Gases  are  often  collected  by  having  them  displace  water  from  a  container.     •   The  problem  is  that  because  water  evaporates,  water  vapor  is  also   collected.   -­‐Partial  pressure  of  water  vapor  (vapor  pressure)  depends  only  on  the  temperature   (Table  11.3)   -­‐Ex.  A  gas  sample  is  collected  over  water.  The  gas  sample  has  a  total  pressure  of  758.2     mmHg  at  25°C.  From  a  vapor  pressure  of  water  table,  the  partial  pressure  of  the     water  vapor  is  23.78  mmHg  at  25°C.     Partial  Pressure=  758.2  -­‐  23.78  =  734.4  mmHg   Gases  in  Chemical  Reactions:  Stoichiometry  Revisited     -­‐mass  A  →  amount  A  (in  moles)  →  amount  B  (in  moles)  →  mass  B   Chem  130  Lecture  Notes  1.19.16  and  1.21.16     OR  use  P,V,T  of  gas  A  to  find  the  amount  in  moles  and  continue  from  there.     -­‐When  gases  are  at  STP,  use  1  mole=22.4  L     -­‐Pressure  here  could  also  be  partial  pressure.     *MOST  IMPORTANT  CONCEPT  IN  CHAPTER  11*   Ex.  Methanol  can  be  synthesized  by  the  reaction:     CO(g)+2H (g)  →CH O2  (g)   3            What  volume  (in  L)  of  2  gas  at  a  T  of  355  K  and  a  P  of  738  mmHg  do  we  need  to  synthesize                 35.7  g  CH OH?   3   Plan  to  Solve:       1.  g  CH OH3  →  mole  CH OH   3     2.  mol  CH OH  3  mol  H   2     3.  mol  H , 2P,  T  →  V H2   Ex.  How  many  grams  of  water  form  when  1.24  L  of  H  gas  at  STP  completely  reacts  with  O ?   2 2   2H (g)2  +  O (g) 2→  2H O  (g)   2   T=273  K;  P=1  atm;  1.24  L  H     2   *STP  Special  Condition:       *Use  Ideal  Gas  Law  if  you  forget  STP        1.24  L  2    1  mole  *  2  mol  *  18.02  g       special  conditions                            22.4  L          2  mol   1  mol                 =0.998  g  H O   2 Kinetic  Molecular  Theory     -­‐The  particles  of  the  gas  (either  atoms  or  molecules)  are  constantly  moving.     -­‐The  attraction  between  particles  is  negligible.     -­‐There  is  a  lot  of  empty  space  between  the  gas  particles  compared  to  the  size  of     particles.     -­‐Average  kinetic  energy  is  directly  proportional  to  Kelvin  T   •   As  temperature  increases,  average  speed  increases   -­‐Collision  of  one  particle  with  another  is  completely  elastic   •   may  exchange  energy,  but  no  overall  loss  of  energy   Temperature  and  Molecular  Velocities     -­‐Average  kinetic  energy  of  one  mole 2 of  gas  particles:       KE =(1avgN mu   a N =Avogadro's  Number       KE=1/2mv   2     m=  mass  of  particle   2           u =  average  square  of  velocity     -­‐Square  Root  of  u =u 2 rms   Root  mean  Square  Velocity  of  a  Particle     -­‐From  Kinetic  Molecular  Theory:  u =  square  root  of  (3RT)/M   rms   -­‐Lighter  particles=faster  average  velocity     -­‐Temperature  increases=more  spread  out   Diffusion  and  Effusion     -­‐Diffusion:gas  molecules  spreading  from  high  to  low  concentration     -­‐Effusion:  gas  molecules  escape  via  small  hole  to  vacuum.     -­‐Graham's  Law  of  Effusion:  rate  =  square  root A of  M /M B A                    B                   -­‐Lighter  particles  effuse  more  quickly  than  heavier  particles.     Chem  130  Lecture  Notes  1.19.16  and  1.21.16      


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