Transcript Ch.11 Molecular Composition of Gases

Avogadro’s Law: states that at the
same Temp & Pressure, equal
numbers of molecules (n) of
gases, occupy equal Volumes (V)
Avogadro also looked at the
relationship between the amount
of gas molecules and the Volume
they occupy.
For example, 2 moles of a gas will
occupy twice the space (Volume)
of 1 mole of gas.
V  n => V  1 => V = k
n
n
What counts for Volume is how
much of a gas you have, not
what kind of gas.
i.e.
A plot of a gas under
STP conditions will
produce a graph just like
the one to the right.
It turns out that any gas
will produce exactly the
same graph.
Therefore, the type of
gas will not change the
relationship.
And remember:1 mole = 6.02 x 1023 molecules
The Volume occupied by 1 mole of a gas is the
Molar Volume.
The Volume occupied by 1 mole of a gas @ STP
is known as the Standard Molar Volume of a
gas and is equal to 22.4 L. (~5.9 gallons)
 @ STP (= 0oC, 1atm. or 101.3kPa)
V = k = 22.4 L/mol. → V = n x 22.4 L/mol.
n
So 1 mole of any gas occupies 22.4L @ STP, but
their masses may be different.
What Volume does 0.75 moles of O2 occupy at STP?
What Volume does 75g of CO2 occupy at STP?
What is the mass of 75L of N2 gas at STP?
Density of Gases:
D=m
V
To calculate density, you need mass & volume.
Using standard volume (22.4L) for each mole of gas
and converting # moles (n) to mass (m) by
multiplying by molar mass (M), then:
D = m
= nxM →
D=
V
V
What is the density of O2 gas at STP?
@ STP
Volume-Volume Calculations
As discussed before, the Volumes of gases in a
chemical reaction are in direct proportion/ratio
to their coefficients in the balanced equation.
e.g.
N2 +
3H2 -> 2NH3
1
3
2
Volume: 1L
3L
2L
1.5L
4L
22.4L
…etc…
Volume-Mass or
Mass-Volume Calculations
Given
Find
Given
Volume-Mass problem
Find
Mass-Volume problem
“The Ideal Gas Law”
So far, we have formulas that tell us what
happens to a gas when you change certain
factors:- P, V & T (n is constant)
Now lets look at a gas in absolute terms where
no variables are changing.
Ideal Gas Law: the mathematical relationship
between __________________________ and
the number of moles of a gas.
So far, we know:
PV = k (Boyle’s)
V = k (Charles)
T
V = k (Avogadro’s)
n
Combine them to get:
=
=>
PV =
Let’s plot values for 1 mole of a gas (n=1):
P (kPa)
V (L)
T (K) k (L.kPa/mol.K)
from k=PV/nT
138.5
15
250
(STP)101.3
22.4
273
148.4
11.2
200
75
38.78
350
200
6.2325
150
Instead of using k, there is a special constant called the
_________ Constant (R)
So, Ideal Gas Law is:
=
where T is in K, V is in L
BUT there are several different values of R,
depending on what units of Pressure you use:
If P is in kPa,
atm,
mm Hg or torr
R=
R=
R=
L.kPa/mol.K
L.atm/mol.K
L.mm Hg/mol.K
H.M. moles of O2 gas are in a 15.0L balloon at
25oC and 0.975atm.
(5 step algebraic method)
H.M. moles of O2 gas are in a 15.0L balloon at
25oC and 0.975atm.
(Chain Method)
What is the Volume of 12g of CO2 at 27oC and 1.05atm?
What is the mass of 44.4L of H2S at 22oC and 750 Torr?
Let’s revisit an earlier problem and try it another way.
Finding Density of a Gas using Ideal Gas Law:
PV = nRT
and
D=m
V
Using m = n x M
=>
n=m
M
And sub into 1st eqn. =>
PV = mRT
M
Manipulate eqn to get m/V (=D) on one side:
m=
=D
Now we can calculate D at
V
at any Pressure & Temp.
What is the density of O2 gas in the room now?
What is the Volume of 12g of CO2 at 27oC and 1.05atm?
(Algebra – 5 step method)
Effusion & Diffusion
Diffusion: the gradual ______ of 2 gases due to their
spontaneous random motion.
Effusion: process where the molecules of a gas confined
in a container, randomly pass through a tiny opening
in the container.
Recall that KE = ½ mv2 and that 2 gases at the same
Temp. have the same Kinetic Energy.
So
Gas A
Gas B
@ same T.
KE(A)
=
KE(B)
½ mAvA2 =
½ mBvB2
If gas B is a lighter molecule (mB is lower) then vB must
be ________ for KE to remain the same.
The whole point:
Lighter, less dense gases travel ______ at the same
Temperatures.
Graham’s Law of Effusion
There is a direct relationship of the average Kinetic
Energy (KE) of a gas molecule to the Kelvin Temp.
KE = 3 RT kJ/mol. and remember KE = ½mv2(NA)
2
So ½mv2NA = 3 RT
→
vrms = 3RT
2
 mNA
Graham’s Law of Effusion “states that the rates of
effusion (r or v) of gases (at same T & P ) are
inversely proportional to the square root of their
Molar Masses”
What is the Kinetic Energy (per mole) of Oxygen gas
in a room that is at 20oC? And, what is the average
speed of an Oxygen molecule?
Let’s prove it:
We want to compare v2 to v1
i.e.
we want v2/v1
v2
v1
=
 3RT/m2 =
 3RT/m1
 m1
 m2 and m = n  M
v2
v1
=
 n M1
 n M2
 M1
 M2
So

=
rate of effusion of Gas 2 (v2 or r2)
rate of effusion of Gas 1 (v1 or r1)
=
M1
 M2
How many times faster is H2 gas compared to CO2
gas in a room at 20oC?
An unknown gas has a rate of effusion that is 2.5 times
faster than Chlorine gas. What is the Molar Mass of the
unknown gas?