Ideal Gas Law Calculations
Missing Variable
and
Before & After Problems
(and Stoichiometry)
Algorithm for Gas Law Problems
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Create a PVnT list
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Decide which form of the IGL to use
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For Missing Variable problem: PV = nRT [or (PV)/(nT) = R]
For B & A problem:
(P1V1)/(n1T1) = (P2V2)/(n2T2)
 Any variables that don’t change will cancel, leaving the
Combined Gas Law or a minor Gas Law (forget them)
Attend to units
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One list means you have a “Missing Variable” problem
Two lists means you have a “Before & After” problem
For Missing Variable, all units must match R (especially P units)
 R = .0821 atm L/mol K OR 62.4 mmHg L/mol K
For B & A, units must simply match each other (there is no R)
T is always in Kelvins
Solve for the missing variable (before or after substitution)
Gas Law - sample problem #1
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A sample of gas has a volume of 22,400 mL at a
temperature of 273 K and a pressure of
760. mmHg. How many moles of gas is in the
sample?
V = 22,400 mL
T = 273 K
P = 760. mmHg
n = ___?___
Gas Law - sample problem #1
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P = 760. mmHg
V = 22,400 mL
n = ___?___
T = 273 K
One list: use PV = nRT, solved for n = PV/RT
Match units to R = 62.4 mmHg L/mol K
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Change 22,400 mL to 22.4 L
n = (760.)(22.4)/(62.4)(273)
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= 1.00
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= 1.00 mol
Gas Law - sample problem #2
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What pressure is the exerted by 100. grams of H2 gas
in a rigid 1.00 liter container at 100.˚C?
V = 1.00 L
T = 100.˚C + 273 = 373 K
P = ___?___
n=
?
(two unknowns - yikes!)
n = 100.g ÷ 2.02 g/mol = 49.5 mol
One list ---> Use PV = nRT solved for P = nRT/V
P = nRT/V = (49.5)(.0821)(373)/(1.00)
= 1520
= 1520 atm
Gas Law - Sample Problem #3
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Bacteria in sewage produce 37.6 mL of CH4 gas at
31˚C and 753 mmHg. What is the volume of this gas
at STP?
V = 37.6 mL
T = 31 ˚C + 273 = 304 K
P = 753 mmHg
n = ___?___ mol
But wait, there’s more …
T2 = 0˚C + 273 = 273 K
P2 = 1 atm
V2 = ___?___ L
n2 = ___?___ mol
Gas Law - Sample Problem #3
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V = 37.6 mL
T = 31 ˚C + 273 = 304 K
P = 753 mmHg
n = ___?___ mol
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V2 = ___?___
T2 = 0˚C + 273 = 273 K
P2 = 1 atm
n2 = ___?___ mol
Two lists ---> use PV/nT = PV/nT
Since n isn’t given, it isn’t changing,
Canceling n leaves P1V1/T1 = P2V2/T2
Match units: change P2 to 760. mmHg, V2 will be in mL
Plug ‘n’ Chug: (753)(37.6)/(304) = (760)V2/(273)
V2 = (753 x 37.6 x 273)/(304 x 760) = 33.5
V2 = 33.5 mL
Gas Law - Sample Problem #3
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Alternate Solution
V = 37.6 mL
T = 31 ˚C + 273 = 304 K
P = 753 mmHg
n = ___?___ mol
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V2 = ___?___ mL
T2 = 0˚C + 273 = 273 K
P2 = 1 atm = 760. mmHg
n2 = ___?___ mol
Multiply original Volume (37.6 mL) by P and T factors.
Since P is increasing, that will cause V to decrease. The P factor
must be less than 1.
Since T is decreasing, that will cause V to decrease. The T
factor must be less than 1.
V2 = (37.6 mL) (753/760) (273/304) = 33.5 mL
Stoichiometry with Gases
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You’ve memorized the molar volume of a gas at STP
= 22.4 L
From Sample Problem #1, you notice that the molar
volume factoid is simply a particular solution to the
IGL. Forget it now.
We are no longer restricted to STP. We can calculate
the volume of a gas at any T & P by using PV=nRT
Old: V=22.4 L for n=1 mol, T=273K and P=1atm
New: V = nRT/P
with no restrictions at all.
Gas Stoich. - Sample Problem
(#5.69 on Page 226)
LiOH is used in spacecraft to absorb the CO2 exhaled by
astronauts. The reaction produces lithium carbonate
and water. What volume of CO2 gas at 21˚C and 781
mmHg could be absorbed by 327 g of LiOH?
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Eq’n: 2 LiOH + 1 CO2 ---> 1 Li2CO3+ 1 H2O
 Change given to moles: 327 g LiOH ÷ 23.9 g/mol
=13.7 mol LiOH
 Change moles LiOH to moles CO2:
13.7 mol LiOH x 1CO2/2 LiOH = 6.84 mol CO2
 Change moles CO2 to liters CO2 (not at STP)
V = nRT/P = (6.84 x 62.4 x 294)/781 = 161 L
Dalton’s Law of Partial Pressures
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Deals with mixtures of gases
Assigns a “partial” pressure to each component of the
mixture based on the mole fraction
The simple concept: If one third of the moles in a
mixture is H2 gas (i.e., mole fraction H2=1/3), then
one third of the total pressure is H2’s partial pressure
The sum of all partial P’s = total P
The Key: In dealing with a mixture of gas, know
whether you’re using Ptotal or Ppartial
npartial = PpartialV/(RT)
ntotal = PtotalV/(RT)
2nd Key: There is no Vpartial. Every gas occupies all of
the Volume
An Illustration of Dalton’s Law of Partial Pressures
Copyright
©
Houghton Mifflin Company. All rights reserved
5-16AB
Partial Pressure - Sample Problem
(#5.79 on Page 226)
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The gas from a certain volcano has the following composition
in mole percent: 65.0% CO2, 25.0% H2, 5.4% HCl, 2.8% HF,
1.7% SO2, and 0.1% H2S. What is the partial pressure of each
gas if the atmospheric pressure is 760. mmHg?
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PCO2 = .650 x 760 mmHg = 494 mmHg
PH2 = .250 x 760 mmHg = 190 mmHg
PHCl = .054 x 760 mmHg = 41 mmHg
PHF = .028 x 760 mmHg = 21 mmHg
PSO2 = .017 x 760 mmHg = 12 mmHg
PH2S = .001 x 760 mmHg = .8 mmHg
Ptot = (494+190+41.0+21.3+12.9+.8) = 760.mmHg
Partial Pressure - Sample Problem
(#5.75 on Page 226)
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A mixture of gas contains 0.0200 mol He and 0.0100 mol H2 in a 5.00 L
container at 10.0˚C. Calculate PHe, PH2, and Ptotal.
PHe = nHeRT/V = (.0200 x 62.4 x 283)/5.00 = 70.7 mmHg
PH2 = nH2RT/V = (.0100 x 62.4 x 283)/5.00 = 35.3 mmHg
Ptotal = ntotalRT/V = (.0300 x 62.4 x 283)/5.00 = 106.0 mmHg
OR Ptotal = PHe + PH2 = (70.7 + 35.3) = 106.0 mmHg
Also: Mol Fraction He x Ptotal = (.0200/.0300)106.0 mmHg
= 70.7 mmHg
 Mol fraction H2 x Ptotal = (.0100/.0300)106.0 mmHg
= 35.3 mmHg
Collection of a Gas over Water
Copyright
©
Houghton Mifflin Company. All rights reserved
5-17
Partial Pressure - Sample Problem
Collecting Gas over Water (#5.81 on Page 226)
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Formic acid, HCOOH, is a convenient source of carbon monoxide. When
warmed with sulfuric acid, formic acid decomposes to give H2O and CO. If
3.85 L of CO was collected over water at 25˚C and 689 mmHg, how many
grams of formic acid was consumed.
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Eq’n: 1 HCOOH ----> 1 H2O + 1 CO
Change given to moles: An IGL calc ---> Make list, etc…
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Change mol CO to mol HCOOH:
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Remember to subtract VP of H2O since gas is collected over water
n = PV/(RT) = (689 - 24)mmHg(3.85 L)/(62.4 x 298 K)
n = .138 mol CO
.138 mol CO x 1 HCOOH/1 CO = .138 mol CO
Change mol HCOOH to g HCOOH:
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.138 mol HCOOH x 46.0 g/mol = 6.348 g HCOOH