Download Document

Applying the ideal gas equation
Copyright © 2011 Pearson
Canada Inc.
General Chemistry: Chapter 6
Slide 1 of 19
Using the Gas Laws
Copyright © 2011 Pearson
Canada Inc.
General Chemistry: Chapter 6
Slide 2 of 19
Avogadro’ Hypothesis:
• Equal volumes of different gases at the same T
and P contain equal numbers of molecules (or,
equal numbers of moles of gas). Neglecting
history (Avogadro’s elegant experiments!), we
can apply the Ideal Gas Law Equation to two
gases (Gas 1 and Gas 2).
• n1 = P1V1/RT1 and n2 = P2V2/RT2
• If P1=P2 and V1=V2 and T1=T2 then n1=n2
Class Example – Avogadro’s
Hypothesis:
• At a given T and P, 8.00 g of oxygen gas (O2(g))
has a volume of 8.00 L. At the same T and P 10.0
L of a gas having the molecular formula XO2 has a
mass of 20.0 g. Identify element X.
Partial Solution: Apply Avogadro’s Hypothesis
Here: Number of moles of = Number of moles of
O2(g) per liter
XO2(g) per liter
# Moles O2(g) per liter = 0.250 mol/8.00L
= 0.0313 mol∙L-1
“Aside”: Moles O2 = 8.00g/(32.0 g.mol-1) = 0.250mol
Completion of “XO2 example” in class
Ideal Gas Law and Molecular
Formulas:
• In high school you used % composition data
for compounds to derive corresponding
empirical formulas. The Ideal Gas Law Eqtn
can be used to determine molar masses.
Combining an empirical formula with a molar
mass allows a molecular formula to be
determined. Empirical formulas specify
relative numbers of atoms of each element.
Knowing “too much chemistry” Can lead you
astray. How?
Molecular Formulas using PV=nRT
• Elemental analysis shows that a compound
containing carbon, hydrogen and fluorine ,
CxHyFz, is 63.17 % carbon, 3.53 % hydrogen
and 33.30 % fluorine by mass. At 44.2 oC a
2.400 g sample of this substance is completely
evaporated in a previously empty 2.50 L
container and a gas pressure of 22.2 kPa is
observed. Determine (a) the empirical formula
of the compound and (b) the molecular
formula of the compound.
A Step at a Time?
• One possible strategy is:
• Step 1: Use mass % composition data to
determine the empirical formula for CxHyFz.
• Step 2: Use the ideal gas law equation to get
(a) the number of moles of CxHyFz in 2.400g of
compound and (b) the molar mass of CxHyFz.
• Step 3: Combine the results to Step 1 and Step
2 to find the molecular formula of CxHyFz.
Empirical Formula of CxHyFz
•
Mass of C atoms
Mass of H atoms
Mass of F atoms
Moles of C atoms
Moles of H atoms
Moles of F atoms
• “Knowing too much chemistry” one might be
tempted to use 2.016 g.mol-1 for the molar
mass of H. Why is this absolutely wrong?
We’ll complete this problem in class.
6-5 Gases in Chemical Reactions
•Stoichiometric factors relate gas quantities to
quantities of other reactants or products.
•Ideal gas equation relates the amount of a gas to
volume, temperature and pressure.
•Law of Combining Volumes can be developed using
the gas law.
Copyright © 2011 Pearson
Canada Inc.
General Chemistry: Chapter 6
Slide 10 of 41
6-6 Mixtures of Gases
•Partial pressure
–Each component of a gas mixture exerts a
pressure that it would exert if it were in the
container alone.
• Gas laws apply to mixtures of gases.
• Simplest approach is to use ntotal, but....
Copyright © 2011 Pearson
Canada Inc.
General Chemistry: Chapter 6
Slide 11 of 41
The total pressure of a mixture of gases is the sum of
the partial pressures of the components of the mixture.
Figure 6-12
Dalton’s law of partial pressures illustrated
Copyright © 2011 Pearson
Canada Inc.
General Chemistry: Chapter 6
Slide 12 of 41
Partial Pressure
Ptot = Pa + Pb +…
Va = naRT/Ptot
and
Vtot = Va + Vb+…
na
naRT/Ptot
Va
=
=
ntotRT/Ptot
ntot
Vtot
Pa
Ptot
Recall
na
=
ntot
a
na
naRT/Vtot
=
=
ntotRT/Vtot
ntot
Copyright © 2011 Pearson
Canada Inc.
General Chemistry: Chapter 6
Slide 13 of 41