2.
Equal volumes of gases at the same
temperature and pressure contain equal
numbers of molecules.
Amadeo Avogadro, the ugliest
of the dead chemists
3.
If we can fix, for example, the temperature
and pressure at STP, we can then compare
how many molecules are present by simply
comparing volumes. We find that one mole
of any gas – ANY gas – has a volume of
22.41410 liters.
4. What does this mean?
For starters, if we
convert any amount of
gas at any temperature
and pressure to STP
using the combined gas
law, we can convert the
resulting volume to
moles using this
conversion factor.
Image from TutorVista
5. How many moles of
gas are present in 46.5
liters of gas at 28 ° C
and 2.75 atm of
pressure?
Strategy? Use the 4-Step
method to solve for
V2, then use dimensional
analysis to convert from
liters to moles
4-Step method
Write the given
Rearrange the
equation
Plug in numbers and
units
Solve
6.
P1 = 2.75 atm
V1 = 46.5 L
T1 = 28 ° C + 273 = 301 K
P2 = 1.00 atm
V2 = ?
T2 = 273 K
7. PLUG IN VALUES WITH LABELS
DOES THIS MAKE SENSE?
Notice, BTW, that the
pressure goes down, which
will make the volume go up.
Also, the temperature is
lower at STP; that makes
the gas shrink. This is
consistent with what we see
as we disaggregate the
parts of this equation.
8. SOLVE
V2 = 116 Liters at STP
(3 sig figs)
USE DIMENSIONAL ANALYSIS TO CONVERT
FROM LITERS AT STP TO MOLES.
9. Shouldn’t this be easier?
If we know that 1 mole =
22.41 L at 273.15K and
1.000 atm, can’t we
organize our work so
that we don’t have to do
this step by step? Can’t
we come up with some
kind of
multiplier, something to
include the 22.41, the
1.000 atm, the 273.15 K?
Let’s fool around with
the math…
Random mole from internet, lost url
13. Gosh, if I can calculate moles, I can find mass, or number of molecules…
wait a sec, isn’t this how the stoichiometry chapter got started?
14.
15.
End of chapter 11, page 392
#40-52
Do NOT ignore sig figs. We always use one
more sig fig in “book values” than in the
givens, so that when we round back, we get
the right answer. There is actually one
correct answer for each question.
image fromwww.bowlinggreen.kctcs.edu
