NOTES - H - Behavior of Gases -...
4
1 BEHAVIOR OF GASES •Gases have weight •Gases take up space •Gases exert pressure •Gases fill their containers Gases doing all of these things! Kinetic Theory of Gases The basic assumptions of the kinetic molecular theory are: Gases are mostly empty space The molecules in a gas are separate, very small and very far apart Kinetic Theory of Gases The basic assumptions of the kinetic molecular theory are: Gas molecules are in constant, chaotic motion Collisions between gas molecules are elastic (there is no energy gain or loss) Kinetic Theory of Gases The basic assumptions of the kinetic molecular theory are: The average kinetic energy of gas molecules is directly proportional to the absolute temperature Gas pressure is caused by collisions of molecules with the walls of the container Measurements of Gases To describe a gas, its volume, amount, temperature, and pressure are measured. • Volume: measured in L, mL, cm 3 (1 mL = 1 cm 3 ) • Amount: measured in moles (mol), grams (g) • Temperature: measured in KELVIN (K) • K = ºC + 273 • Pressure: measured in mm Hg, torr, atm, etc. • P = F / A (force per unit area) Moderate Force (about 100 lbs) Small Area (0.0625 in 2 ) Enormous Pressure (1600 psi) P = F /A
-
Upload
phungtuyen -
Category
Documents
-
view
242 -
download
1
Transcript of NOTES - H - Behavior of Gases -...
1
BEHAVIOR OF GASES
•Gases have weight
•Gases take up space
•Gases exert pressure
•Gases fill their containers
Gases doing all ofthese things!
Kinetic Theory of GasesThe basic assumptions of the kinetic molecular theory are:
Gases are mostlyempty space
The molecules in a gasare separate, verysmall and very farapart
Kinetic Theory of GasesThe basic assumptions of the kinetic molecular theory are:
Gas molecules are inconstant, chaoticmotion
Collisions between gasmolecules are elastic(there is no energygain or loss)
Kinetic Theory of GasesThe basic assumptions of the kinetic molecular theory are:
The average kineticenergy of gasmolecules is directlyproportional to theabsolute temperature
Gas pressure is causedby collisions ofmolecules with thewalls of the container
Measurements of Gases
To describe a gas, its volume, amount,temperature, and pressure are measured.• Volume: measured in L, mL, cm3 (1 mL = 1 cm3)• Amount: measured in moles (mol), grams (g)• Temperature: measured in KELVIN (K)
• K = ºC + 273• Pressure: measured in mm Hg, torr, atm, etc.
• P = F / A (force per unit area)
Moderate Force(about 100 lbs)
Small Area (0.0625in2)
Enormous Pressure(1600 psi)P = F /A
2
Bed of Nails
Large Surface Area(lots of nails)
ModerateForceSmall
Pressure
P = F / A
Units of Pressure
Units of Pressure: 1 atm = 760 mm Hg 1 atm = 760 torr 1 atm = 1.013 x 105 Pa 1 atm = 101.3 kPa 1 atm = 1.013 bar
Boyle’s Law
As P↑, V↓ (when T and n are constant) andvice versa…. INVERSE RELATIONSHIP
V α 1/P
P1V1 = P2V2
For a given number ofmolecules of gas at aconstant temperature, thevolume of the gas variesinversely with the pressure.
Example: A sample of gas occupies 12 L under a pressureof 1.2 atm. What would its volume be if the pressurewere increased to 3.6 atm? (assume temp is constant)
P1V1 = P2V2
(1.2 atm)(12 L) = (3.6 atm)V2
V2 = 4.0 L
Charles’ LawJacques Charles (1746-1828)The volume of a given number of molecules
is directly proportional to the Kelvin temperature.
As T↑, V ↑ (when P and n are constant) andvice versa…. DIRECT RELATIONSHIP
V α T2
2
1
1
T
V
T
V=
Example: A sample of nitrogen gas occupies 117 mL at100.°C. At what temperature would it occupy 234 mL if thepressure does not change? (express answer in K and °C)
V1 / T1= V2 / T2
(117 mL) / (373 K) = (234 mL) / T2
T2 = 746 K T2 = 473 ºC
3
Combined gas law
2
22
1
11
T
VP
T
VP=
This is for one gas undergoing changingconditions of temp, pressure, and volume.
Combining Boyle’s law (pressure-volume)with Charles’ Law (volume-temp):
122211TVPTVP =
Example 1: A sample of neon gas occupies 105 Lat 27°C under a pressure of 985 torr. Whatvolume would it occupy at standard conditions?
P1 = 985 torr V1 = 105 L T1 = 27 °C = 300. K
P2 = 1 atm = 760 torr V2 = ? T2 = 0 °C = 273 K
P1V1T2 = P2V2T1
(985 torr)(105 L)(273K) = (760torr)(V2)(300K)
V2= 124 L ~ 120 L
Example 2: A sample of gas occupies 10.0 L at 240.°C undera pressure of 80.0 kPa. At what temperature would the gasoccupy 20.0 L if we increased the pressure to 107 kPa?
P1 = 80.0 kPa V1 = 10.0 L T1 = 240 °C = 513 K
P2 = 107 kPa V2 = 20.0 L T2 = ?
P1V1T2 = P2V2T1
(80.0kPa)(10.0L)(T2) = (107kPa)(20.0L)(513K)
T2= 1372K≈ 1370K
Example 3: A sample of oxygen gas occupies 23.2 L at 22.2 °Cand 1.3 atm. At what pressure (in mm Hg) would the gas occupy11.6 L if the temperature were lowered to 12.5 °C?
P1 = 1.3 atm = 988 mmHg V1 = 23.2 L T1 = 22.2 °C = 295.2 K
P2 = ? V2 = 11.6 L T2 = 12.5 °C = 285.5 K
P1V1T2 = P2V2T1
(988mm Hg)(23.2L)(285.5K) = (P2)(11.6L)(295.2K)
P2= 1938 mm Hg ≈ 1900 mmHg
Gases: Standard Molar Volume &The Ideal Gas Law
Avogadro’s Law: at the same temperatureand pressure, equal volumes of all gasescontain the same # of molecules (& moles).
Standard molar volume = 22.4 L @STP
This is true of “ideal” gases at reasonabletemperatures and pressures ,the behavior ofmany “real” gases is nearly ideal.
Example: 1.00 mole of a gas occupies 36.5 L, andits density is 1.36 g/L at a given temperature &pressure.
a) What is its molecular weight (molar mass)?
!L
g 36.1=
mol 1.00
L 5.36g/mol 6.49
4
Example: 1.00 mole of a gas occupies 36.5 L, andits density is 1.36 g/L at a given temperature &pressure.
b) What is the density of the gas understandard conditions?
!mol
g 6.49=
L 22.4
mol 1g/L .212
