an introduction to gases chapter 13. kinetic molecular theory postulate #1 –gases consist of tiny...
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An Introduction to Gases
Chapter 13
Kinetic Molecular Theory
• Postulate #1– Gases consist of tiny particles (atoms or
molecules)
• Postulate #2– These particles are so small, compared
with the distances between them, that the volume (size) of the individual particles can be assumed to be negligible (zero).
– Gases are COMPRESSIBLE
Kinetic Molecular Theory
• Postulate #3– The particles are in constant random
motion, colliding with the walls of the container. These collisions with the walls cause the pressure exerted by the gas.
Kinetic Molecular Theory
• Postulate #4– The particles are assumed not to
attract or to repel each other.
• Postulate #5– The average kinetic energy of the gas
particles is directly proportional to the Kelvin temperature of the gas.
Avogadro’s Hypothesis
• At the same temp & pressure, equal volumes of gas hold same number of molecules.
• V and n are directly related.V and n are directly related.
twice as many molecules
Pressure
Pressure of air is Pressure of air is measured with a measured with a BAROMETERBAROMETER(developed by Torricelli in (developed by Torricelli in 1643)1643)
Pressure
• What is Pressure?What is Pressure?
• What tool do we use to measure it?What tool do we use to measure it?
Pressure
1.1. mmHg (or Torr)mmHg (or Torr)2.2. Atmospheres (atm)Atmospheres (atm)3.3. Pascals (used in physics: Pascals (used in physics:
1 pascal = 1 newton per square meter)1 pascal = 1 newton per square meter)4. 4. psi psi
Equivalences:Equivalences:1 atm = 760 mmHg1 atm = 760 mmHg1 atm = 101,325 Pa = 101.325 kPa1 atm = 101,325 Pa = 101.325 kPa1 atm = 14.7 psi
Pressure Calculation
What is 475 mm Hg expressed in atm?
475 mm Hg=
1 atm
760 mm Hg0.625 atm
The pressure of a tire is measured as 294,000 pascals. What is this pressure in atm?
Pressure Calculation
294, 000 Pa
101, 325 Pa
1 atm= 2.90 atm
Dalton’s Law
“The Law of Partial Pressure”
• The total pressure of a mixture of gases is the sum of the partial pressures of the gases in the mixture.
Ptotal = PA + PB + PC
Gas Laws Calculations
Get out a calculator!!!
The Gas Law
PV=nRT
P = pressure ( atm or kPa )V= volume ( L )
n= number of moles (mol)T= temperature (K)
R – The Proportionality Constant
Value depends on units
8.314L (kPa)mol (K)
0.0821L (atm)mol (K)
Or
The Gas Law – Problem If 7.0 moles of an ideal gas has a volume
of 12.0 L with a temperature of 300. K, what is the pressure in kPa?
P (12.0 L) =(7.0 mol)(300
K)8.31
4
L (kPa)
mol (K)
PV = nRT
P = 1454.95 kPa P = 1500 kPa
Combined Gas Law
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ii
TRn
TRn
VP
VP
Let’s say we have some O2 gas AND we change some conditions. Would there be anything similar between the two gases?
Combined Gas Law – Problem
You have 3 moles of a solution at 300. K and 15 atm in a 2 L container. If the container is heated to 350. K and the volume decreased to 1 L, what will the new pressure be?
P1 15 atm P2 want
V1 2 L V2 1 L
n1 3 moles n2 3 moles
R1 constant R2 constant
T1 300. K T2 350. K
Combined Gas Law – Problemc
P1V1=
n1R1T1
P2V2 n2R2T2
P1V1=
T1
P2V2 T2
If we know that R1 = R2 and the mass is constant then
(15 atm)(2 L) = (300. K)
P2(1L) (350. K)
Replace with numbers
Combined Gas Law – Problem
(15 atm)(2 L) = (300. K)
P2(1L) (350. K)
(15 atm)(2 L)(350. K) =
P2(1L)(300. K)
P2 = 35 atm
Pressure & Volume
• At constant Temperature• Pressure and Volume vary
inversely.– Why? – More collisions More pressure
P1V1 = P2V2
P & V – Example Problem If you start with 0.500 L of a gas at 7.0 atm and you move the gas to a container with 3.5 L available, how much pressure will the gas exert?
7 atm (0.500 L)= P2
3.5 L
P1 (V1) = P2 (V2)
7.0 atm (0.500 L) = P2 (3.5 L)
1.0 atm = P2
Put a few drops of water in a can. Heat the can until the water boils. What is happening to the gas inside? Now flip the can over
into cold water. Predict what do you predict will happen?
Demo
On a Larger Scale
On a Larger Scale
Temperature & Volume
At constant PressureVolume & Temperature vary directly.– Why?– More collisions More Volume
V1=
V2
T1 T2
T & V – Example Problem
If a gas is in a balloon with a volume of 12.0 L and at a temperature of 300. K, what will the volume be if you place the balloon in a freezer at 250. K?
V1=
V2
T1 T2
12.0 L=
V2
300. K250.
K
12.0 L (250. K) = V2
300. K
10.0 L = V2
S.T.P.
• Standard Temperature and PressureStandard Temperature and Pressure
These are conditions that are universalThese are conditions that are universal
Standard Temperature: Standard Temperature:
0ºC or 273K0ºC or 273K
Standard Pressure: Standard Pressure:
1atm or 101.325kPa1atm or 101.325kPa
S.T.P. – Example Problem
What is the volume What is the volume of 1 mole of a gas of 1 mole of a gas at STP?at STP?
P 1 atm
V want
n 1 mole
R 0.0821 (L)(atm)/(K)(mole)
T 273 KPV = nRT
(1atm)V = (1 mole)(0.0821 [Latm/Kmole])(273K)
V= 22.4 L