Boyle’s Law P1V1 2 - · PDF fileBoyle’s Law At constant temperature (in Kelvin)...
-
Upload
truongduong -
Category
Documents
-
view
217 -
download
3
Transcript of Boyle’s Law P1V1 2 - · PDF fileBoyle’s Law At constant temperature (in Kelvin)...
Boyle’s Law
At constant temperature (in Kelvin) and constant
moles of gas, the pressure of a gas is inversely
proportional to the gas volume.
1
Final condition:
P2 at V2
P1V1 = P2V2 = k
Initial condition:
P1 at V1
Boyle’s Law
At constant temperature (in Kelvin) and constant
moles of gas, the pressure of a gas is inversely
proportional to the gas volume.
2
P1V1 = P2V2 = k
Practice Problem
A sample of chlorine gas occupies a volume
of 946 mL at a pressure of 726 mmHg.
What is the pressure of the gas (in mmHg)
if the volume is reduced at constant
temperature to 154 mL?
3
P1 x V1 = P2 x V2
P1 = 726 mmHg
V1 = 946 mL
P2 = ?
V2 = 154 mL
P2 = P1 x V1
V2
726 mmHg x 946 mL
154 mL = = 4460 mmHg
Charles’s Law
At constant pressure and moles of gas, the volume
of a gas is directly proportional to the absolute
temperature in kelvin.
4
Hg
Hg
gas
gas
𝑽𝟏
𝑻𝟏=
𝑽𝟐
𝑻𝟐= 𝒌
Practice Problem
A sample of carbon monoxide gas occupies
3.20 L at 125 0C. At what temperature (in
Kelvin) will the gas occupy a volume of
1.54 L if the pressure remains constant?
5
V1 = 3.20 L
T1 = 398.15 K
V2 = 1.54 L
T2 = ?
T2 = V2 x T1
V1
1.54 L x 398.15 K
3.20 L = = 192 K
V1/T1 = V2/T2
T1 = 125 (0C) + 273.15 (K) = 398.15 K
Avogadro’s Law
At a specified pressure
and temperature, the
volume of a gas is
proportional to its
number of moles.
6
𝑽𝟏
𝒏𝟏=
𝑽𝟐
𝒏𝟐= 𝒌
Breathe in, breathe out
7
Ears Popping
8
http://abetterchemtext.com/gases/ear_pop.htm
Under normal conditions:
pressure inside the Eustachian
tube is equal to the pressure
on the ear canal.
At higher altitudes: pressure
inside the Eustachian tube is
greater than the pressure on
the ear canal.
Up, up, and away
9
What is STP?
Scientists defined a standard laboratory
temperature and pressure called STP.
What is the volume occupied by 1 mole of
any gas at STP?
10
Temperature = 273.15 K (0 °C)
Pressure = 1 atm = 760 torr
𝑉 =𝑛𝑅𝑇
𝑃=
1 𝑚𝑜𝑙 0.0821 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙−1 𝐾−1 273.15 𝐾
1 𝑎𝑡𝑚
𝑉 = 22.4 𝐿 standard molar volume
Standard Molar Volume
11
12
Ideal Gas Law
13
PV = nRT
R = PV
nT =
1 atm x 22.414 L
1 mol x 273.15 K =
0.0821 atm·L
mol·K
R is the universal gas constant;
the numerical value of R depends on the units used.
Ideal Gas Law and Gas Density
14
PV = nRT and d = m
V
PV = m
MW
RT
MW x P
RT V
m = d =
, n = m
MW
Learning Check
Rate the gases according to density from
lowest to highest assuming all of the gases
are maintained at the same pressure and
temperature.
15
MW x P
RT d =
16
Use gas laws to determine a balanced
equation
The piston-cylinder is depicted before and after
a gaseous reaction that is carried out at
constant pressure. The temperature is 150 K
before the reaction and 300 K after the
reaction. (Assume the cylinder is insulated.)
17
Which of the following balanced equations describes the reaction?
(1)A2(g) + B2(g) → 2AB(g) (2) 2AB(g) + B2(g) → 2AB2(g)
(3) A(g) + B2(g) → AB2(g) (4) 2AB2(g) + A2(g) + 2B2(g)
Dalton’s Law of Partial Pressure
The total pressure of a mixture of gases is the
sum of the pressure of the individual gases.
V and T
are
constant
P1 P2 Ptotal = P1 + P2 18
Type equation here.
20
We can express the partial pressure Pi in a mixture
of gases in terms of the total gas pressure, PT, and
the mole fraction of each gas in the mixture.
𝑃𝑇 = 𝑃𝐴 + 𝑃𝐵 =𝑛𝐴𝑅𝑇
𝑉𝑇+
𝑛𝐵𝑅𝑇
𝑉𝑇
𝑃𝑇 = 𝑛𝐴 + 𝑛𝐵
𝑅𝑇
𝑉𝑇
𝑃𝐴
𝑃𝑇=
𝑛𝐴𝑅𝑇𝑉𝑇
𝑛𝑇𝑅𝑇𝑉𝑇
=𝑛𝐴
𝑛𝑇=
𝑛𝐴
𝑛𝐴 + 𝑛𝐵
𝑃𝐴 =𝑛𝐴
𝑛𝑇𝑃𝑇
𝑃𝐵 =𝑛𝐵
𝑛𝑇𝑃𝑇
𝑷𝒊 = 𝒙𝒊𝑷𝑻
Dalton’s Law
Factor out RT/V
Get ratio of PA/PT
Ptotal = P1 + P2