III. Ideal Gas Law - Sykes Chemistry Classes
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Transcript III. Ideal Gas Law - Sykes Chemistry Classes
Learning Log
Why
are you advised to open
windows slightly if a tornado
approaches?
Ch. 10 & 11 - Gases
Dalton’s Law
Ideal Gas Law
(p. 322-325, 340-346)
A. Dalton’s Law of Partial Pressures
Total
pressure of a
mixture of gases is equal
to the sum of the partial
pressures of the
component gases.
• In the absence of a
chemical reaction
PT = P1+ P2+ P3+…
Practice Problem
A mixture of four gases exerts a total
pressure of 1200 mm Hg. Gases A and
B each exert 420 mm Hg. Gas C exerts
280 mm Hg. What pressure is exerted
by gas D?
PT = P1 + P2 + P3 + P4
1200 = 420 mm Hg + 420 mm Hg
+ 280 mm Hg + P4
1200 = 1120 mm Hg + P4
P4 = 80 mm Hg
B. Vapor pressure of water
Gases
are often collected in lab by
water displacement and are mixed
with water vapor
Patm = Pgas + PH2O
To determine the pressure of the
gas collected – subtract the vapor
pressure of the water at that
temperature from the current
atmospheric pressure
C. Ideal Gas Law
The
mathematical relationship
among pressure, volume,
temperature and the number of
moles of a gas.
Derived by combining the gas laws.
PV=nRT
D. Ideal Gas Constant
Merge the Combined Gas Law with Avogadro’s Principle:
PV
V
k
=R
nT
T
n
IDEAL GAS CONSTANT
R=0.0821 Latm/molK
R=8.315 dm3kPa/molK
E. Ideal Gas Law Problems
Calculate
the pressure in atmospheres
of 0.412 mol of He at 16°C & occupying
3.25 L.
GIVEN:
WORK:
P = ? atm
PV = nRT
n = 0.412 mol
P(3.25)=(0.412)(0.0821)(289)
L
mol Latm/molK K
T = 16°C = 289 K
V = 3.25 L
P = 3.01 atm
R = 0.0821Latm/molK
E. Ideal Gas Law Problems
Find
the volume of 85 g of O2 at 25°C
and 104.5 kPa.
GIVEN:
WORK:
V=?
85 g 1 mol = 2.7 mol
n = 85 g = 2.7 mol
32.00 g
T = 25°C = 298 K PV = nRT
P = 104.5 kPa
(104.5)V=(2.7) (8.315) (298)
kPa
mol
dm3kPa/molK K
R = 8.315 dm3kPa/molK
V = 64 dm3
F. Finding Molar Mass
from the Ideal Gas Law
n =
mass
n =m
Molar mass
M
PV=nRT
PV = mRT
M
M = mRT
PV
OR
G. Finding Density
from the Ideal Gas Law
D = mass
volume
or
M = mRT
PV
M = DRT
P
D = MP
RT
D=m
V
PRACTICE PROBLEMS
At
28°C and 0.974 at, 1.00 L of gas has
a mass of 5.16 g. What is the molar
mass of this gas?
P
= 0.974 atm
T = 28°C = 273 = 301 K
M
V = 1.00 L
m = 5.16 g
= mRT
PV
= (5.16 g) (0.0821 L∙atm/mol∙K) (301K)
(0.974 atm)(1.00 L)
= 131g/mol
PRACTICE PROBLEMS
What
is the density of a sample of
ammonia gas, NH3, if the pressure is
0.928 atm and the temperature is
63.0°C?
P = 0.928 atm
M = 17.034 g/mol
D = MP
RT
=
T = 63.0°C + 273 = 336 K
R = 0.0821 L∙atm/mol∙K
(17.034 g/mol)(0.928 atm)
(0.0821L∙atm/mol∙K)((336 K)
= 0.572 g/L NH3
Homework Assignment
Workbook.
Complete
problems #1 and 2
on pp. 173-174, 1-2 p. 178,
1-2 p. 180
Practice Test Part 2
P.
181 – 182 #1-7, 12, 18 - 21
Practice test Part 2
Workbook
p. 182 #1-7