Solving gas law problems

1
Solving gas law problems

• Figuring out which equation to use
• Relative cues
• Look for two volumes, or two pressures or two
  temperatures or two number of moles
• Look for change or new or will be
• Absolute cues
• Only one of each variable given
• R often given
• Often if g of gas given (need to get to moles)
• Reducing/rearranging equation to get unknown
• Changing all variables to appropriate units that
  can be used in equations
• Always have to use K
• Absolute always have to use L, atm, moles
• Relative can use different volume and pressure
  units as long as consistent

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Example 1 of using strategy

• Problem What is the volume, in milliliters,
  occupied by 89.2 g CO2 (g) at 37 ºC and 737 mmHg?
• Solution
• Cues
• g of CO2 ?absolute
• Only one of each variable ? absolute
• Equation need volume ? V nRT/P
• Conversions
• 89.2 g of CO2 x 1 mole CO2/44.0 g CO2 2.03
  moles
• 37 273 310 K
• 737/760 0.970 atm
• R 0.082 L-atm/mole-K
• Substitute and solve V 2.03x0.082x310/.970
  53.1 L

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Example 2 of using strategy

• Problem A sample of gas has a volume of 4.25 L
  at 25.6 ºC and 748 mmHg. What will be the volume
  of this gas at 26.8 ºC and 742 mmHg? Solution
• Cues
• Will be ? relative
• Two temperature two pressures
• Reduce equation (moles constant) ? P1V1/n1T1
  P2V2/n2T2
• Conversions
• T1 25.6 273 298.6 K T2 26.8 273
  299.8 K
• Can use pressure values as given and L value
• Substitute and solve P1V1/T1 P2V2/T2
• (748)(4.25)/298.6 (742)V2/299.8
• ALGEBRA!!!!
• V2 4.30 L

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Some more problems 1

• A light bulb with an internal pressure of 720.
  torr at 20ºC is thrown into an incinerator
  operating at 750ºC. What internal pressure must
  the light bulb be able to withstand if it does
  not break?
• Relative P and T only make sure convert T to K
• A 12.8-L cylinder contains 35.8 g O2 at 46 ºC.
  What is the pressure of this gas, in atmospheres?
• Absolute convert g O2 to moles convert T to K

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Some more problems 2

• A 72.8-L constant-volume cylinder containing 1.85
  mol He is heated until the pressure reaches 3.50
  atm. What is the final temperature in degrees
  Celsius?
• Absolute only one of each OK to go ? get T
• BUT ? convert back to oC
• If a 1500.-mL sample of air at 22 ºC is cooled
  enough to cause the volume to decrease to 750. mL
  at constant pressure, what is the final Celsius
  temperature required? What was the temperature
  change?
• Relative Two volumes V (use mL) and T only
  convert T to K
• Get K answer ? convert back to oC also get
  temperature change

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Gas Stoichiometry

• Stoichiometry with gases obeys the same rules as
  stoichiometry with g moles molarity
• Usually relationships in equations in which gas
  is consumed or produced
• Two general problems
• Information about gases known and information
  about other components in equation sought
• Information about other components known and
  information about gases in equation sought
• Can expand switching yard!!

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Gas Stoichiometry
PV nRT
8
How many mg of magnesium metal must react with
excess HCl(aq) to produce 28.50 mL of H2(g),
measured at 26 ºC and 758 Torr? Mg (s) 2 HCl
(aq) ? H2 (g) MgCl2 (aq)
PV/RT? moles of H2 ? moles of Mg ? g of Mg
n 1.16 x10-3 moles H2
1.16 x10-3 moles H2
1 mole Mg 1 mole H2
1.16 x10-3 moles Mg
24.3 g Mg 1 mole Mg
0.0282 g Mg
28.2 mg Mg
0.0282 g Mg
9
What volume of CO2 (g) is produced at a
temperature of 20. ºC and a pressure of 1.00 atm
by the fermentation of 500. g of glucose?
C6H12O6 ? 2 C2H5OH 2 CO2 (g)
g of glucose ?moles of glucose ? moles of CO2 ?V
of CO2 ( PV/RT)
V 134 L
5.56 mole CO2
2 mole CO2 1 mole glucose
2.78 mole glucose
1 mole glucose 180 g glucose
500 g glucose
0.0282 g Mg
10
A 3.57-g sample of a KCl-KClO3 mixture is
decomposed by heating and produces 119 mL O2 (g),
measured at 22.4 ºC and 738 mmHg. What is the
mass percent of KClO3 in the mixture? 2 KClO3
(s) ? 2 KCl(s) 3 O2 (g)
Moles of O2 ? moles of KClO3 ? g of KClO3 ?
KClO3 of 3.57g
n 4.77 x10-3 moles O2
4.77 x 10-3 mole O2
2 mole KClO3 3 mole O2
3.18 x 10-3 mole KClO3
122.6 g KClO3 1 mole KClO3
0.390 g/3.57 g x 100
10.9
0.390 g KClO3
0.0282 g Mg
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Complex gas stoichiometry

• How many liters of SO3(g) can be produced by the
  reaction of 1.15 L SO2(g) and 0.65 L O2(g) if all
  three gases are measured at the same temperature
  and pressure?
• 2 SO2(g) O2(g) ? 2 SO3(g)
• Since R is a constant the term RT/P is a constant
  since we are not changing temperature and
  pressure
• We can therefore compare volumes in balanced
  equations just like we can compare moles
• This only applies if T and P are constant
• This is a limiting reagent problem
• SO2 O2
• HAVE 1.15 L 0.65 L
• NEED 2(0.65)L ½ (1.15)L
• 1.30 L 0.575 L
• SO2 is the limiting reagent since we need 1.30 L
  to react with O2 but only have 1.15 L. Therefore
  we will form the same amount of SO3 gas
• 1.15 L