Mole concept applied to gases

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Mole concept applied to gases

• 1.4.4 Apply Avogadros law to calculate reacting
  volumes of gas
• 1.4.5 Apply the concept of molar volume at
  standard temperature and pressure in calculations
• 1.4.6 Solve problems between temperature,
  pressure and volume for a fixed mass of an ideal
  gas.
• 1.4.7 Solve problems relating to the ideal gas
  equation, PVnRT
• 1.4.8 Analyse graphs relating to the ideal gas
  equation.

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Avogadros Hypothesis

• At a constant temperature and pressure, a given
  volume of gas always has the same number of
  particles.
• The coefficients of a balanced reaction is the
  same ratio as the volumes of reactants and
  products

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2CO (g) O2 (g)? 2CO2(g)

• For the above example, it is understood that half
  the volume of oxygen is needed to react with a
  given volume of carbon monoxide.
• This can be used to carry out calculations about
  volume of gaseous product and the volume of any
  excess reagents.

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Example

• 10cm3 of ethyne (C2H2) is reacted with 50cm3 of
  hydrogen to produce ethane (C2H6), calculate the
  total volume and composition of the remaining gas
  mixture, assuming constant T and P.

1st get balanced equation C2H2(g) 2H2(g) ?
C2H6(g)
2nd look at the volume ratios 1 mol ethyne to 2
mol of hydrogen, therefore 1 vol to 2 vol
3rd analyse If all 10cm3 of ethyne is used, it
needs only 20cm3 of hydrogen, therefore hydrogen
is in excess by 50cm3-20cm3 30 cm3. In the
end youll have 10 cm3 Ethane and the leftover 30
cm3 hydrogen
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Molar volume

• The temperature and pressure are specified and
  used to calculate the volume of one mole of gas.
• Standard temperature and pressure (STP) is at sea
  level 1 atm 101.3 kPa and 0oC 273 K this
  volume is 22.4 dm3 (or 22.4 L)

Molar gas volume, Vm. It contains 6.02 x 1023
molecules of gas
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Example

• Calculate how many moles of oxygen molecules are
  there in 5.00 dm3 at STP
• n VSTP 5.00 0.223 mol
• 22.4 dm3 22.4 dm3

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Boyles Law (1659)

• Boyle noticed that the product of the volume of
  air times the pressure exerted on it was very
  nearly a constant, or PVconstant.
• If V increases, P decreases proportionately and
  vice versa. (Inverse proportions)
• Temperature must be constant.
• Example A balloon under normal pressure is blown
  up (1 atm), if we put it under water and exert
  more pressure on it (2 atm), the volume of the
  balloon will be smaller (1/2 its original size)
• P1V1P2V2

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Charles Law (1787)

• Gas expands (volume increases) when heated and
  contracts (volume decreases) when cooled.
• The volume of a fixed mass of gas varies directly
  with the Kelvin temperature provided the pressure
  is constant. V constant x T
• V1 V2
• T1 T2

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Gay-Lussacs Law

• The pressure of a gas increases as its
  temperature increases.
• As a gas is heated, its molecules move more
  quickly, hitting up against the walls of the
  container more often, causing increased pressure.
• P1 P2
• T1 T2

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Laws combined

• P1V1 P2V2
• T1 T2
• T must be in Kelvins, but P and V can be any
  proper unit provided they are consistently used
  throughout the calculation

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Practice

• If a given mass of gas occupies a volume of 8.50
  L at a pressure of 95.0 kPa and 35 oC, what
  volume will it occupy at a pressure of 75.0 kPa
  and a temperature of 150 oC?

1st convert oC to K 35 273 308 K 150
273 423 K
2nd rearrange equation and solve problem V2
V1 x P1 x T2 8.50 x 95.0 x 423 14.8 L P2
x T1 75.0 x 308
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Temperature

• Kelvin temperature is proportional to the average
  kinetic energy of the gas molecules.
• It is a measure of random motion of the gas
  molecules
• More motion higher temperature

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Ideal gas behaviour

• Ideal behaviour is when a gas obeys Boyles,
  Charles and Gay-Lussacs laws well
• At ordinary temperature and pressures, but there
  is deviation at low temperature and high pressures

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Ideal gas

• where all collisions between molecules are
  perfectly elastic and in which there are no
  intermolecular attractive forces.
• Its like hard spheres bouncing around, but NO
  interaction.

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Ideal gas law

• PV nRT
• P pressure (kPa)
• Volume (dm3)
• n number of moles
• Runiversal gas constant 8.3145 J mol-1 K-1
• T temperature (K)

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Example

• 3.376 g of a gas occupies 2.368 dm3 at 17.6 oC
  and a pressure of 96.73 kPa, determine its molar
  mass.
• PV nRT rearrange equation for n

n PV/RT (96.73 x 2.368) / (8.314 x 290.6)
0.09481 mol Molar mass mass/ mole
3.376 g / 0.09481 mol 35.61
g/mol