Presentation on theme: "The Kinetic Molecular Theory explains the forces between molecules and the energy that they possess. This theory has 3 basic assumptions. Kinetic Molecular."— Presentation transcript:
The Kinetic Molecular Theory explains the forces between molecules and the energy that they possess. This theory has 3 basic assumptions. Kinetic Molecular Theory
1. Matter is composed of small particles (molecules). Volume is the measure of space in between the molecules and not the space the molecules contain themselves.
2. The molecules are in constant motion. Solid - Molecules are held close to each other by their attractions of charge. They will bend and/or vibrate, but will stay in close proximity.
Liquid - Molecules will flow or glide over one another, but stay toward the bottom of the container. Motion is a bit more random than that of a solid.
Gas - Molecules are in continual straight line motion. The kinetic energy of the molecule is greater than the attractive force between them, thus they are much farther apart and move freely of each other.
3. When the molecules collide with each other, or with the walls of a container, there is no loss of energy.
Properties of Gases Gases can be compressed into smaller volumes ↑ P causes ↓ V so: ↑ P causes ↑density
Properties of Gases Gases exert pressure on surroundings. So to confine a gas, pressure must be exerted on it. (Newton’s third law: for every action, there is an equal and opposite reaction)
Properties of Gases Gases expand without limits, therefore a gas will completely fill the container it occupies
Gas Properties Collisions between gas particles and between particles and the container walls are elastic collisions
Properties of Gases Gas particles are in constant, rapid, random motion. Have kinetic energy
Gas Properties Fluidity Gas particles glide easily past one another since the attractive forces are minimal.
Properties of Gases Gases diffuse into one another. Gases mix completely, ie gases are miscible. The amount of gas can be known if the pressure, temperature and volume are known.
TEMPERATURE AND KINETIC ENERGY TEMPERATURE AND KINETIC ENERGY Different substances have the same average KE when they are at the same temperature. The velocity of the particles depends on their atomic or molecular mass (formula weight). The greater their mass, the slower their velocity.
Pressure Pressure – force / unit area Units: Newtons / m 2 (Pascals) Atmospheres (atm) mm Hg Torr 1 atm = 14.7 psi = 1.01 x 10 5 Pa = 760 mm Hg = 760 Torr =101.3kPa
GAS LAWS THREE VARIABLES THAT DETERMINE GAS PRESSURE: Temperature Amount (# of moles) Volume
GAS LAWS THREE VARIABLES THAT DETERMINE GAS PRESSURE: Temperature – the average Kinetic Energy of the gas. KE = ½ mv² If the Temperature ↑ then the Pressure ↑
GAS LAWS THREE VARIABLES THAT DETERMINE GAS PRESSURE: Amount (# of moles) – the number of particles of a gas in a closed container If the Amount ↑ then the Pressure ↑
GAS LAWS THREE VARIABLES THAT DETERMINE GAS PRESSURE: Volume of the sealed, but flexible container of the gas If the Volume ↑ then the Pressure ↓ If the Volume ↓ then the Pressure ↑
Gas Laws Boyle’s Law ↑ P causes ↓ V P 1 V 1 = P 2 V 2 PV = constant
The PEEP in the Vacuum
STP Standard Temperature and Pressure (STP) is defined as 0 o C and 1 atm of pressure.
A balloon initially occupies 12.4 L at 1.00 atm. What will be the volume at atm? P1=P1=1.00 atm V1=V1=12.4 L P2=P2=0.800 atm V2=V2= ? P 1 V 1 = P 2 V 2 P 1 V 1 = V 2 P atm * 12.4L = V atm 1.00 atm * 12.4L = V 2 *0.800 atm V2=V2= 15.5L
A 5.00 L container is filled with gas at 1.00 atm. A piston then compresses the gas to L. What is the new pressure? 1.00 atm 5.00 L 0.250L P 1 = V 1 = P 2 = V 2 = P 1 V 1 = P 2 V 2 P 1 V 1 = V 2 P 2 P 2 ? 1.00 atm* 5.00 L = 0.250L *X X = 20.0atm
Example #3: 2.50 L of a gas was at an unknown pressure. However, at standard pressure, its volume was measured to be 8.00 L. What was the unknown pressure? End E 4/7
Charles Law ↑ T causes ↑ V As the temperature increases, the volume increases. (in K not o C) V/T = constant V 1 /T 1 = V 2 /T 2
Charles’s Law If the amount of a gas and its pressure remain constant, then As the temperature of a gas increases, the volume of the gas increases. ↑ T = ↑ V As the temperature of a gas decreases, the volume of the gas decreases ↓ T = ↓ V
(T 2 *V 1 )/T 1 = V 2 A 450 mL piston assembly containing gas initially at 293 K is suddenly heated to 450 K. What will be the new volume of the piston assembly? V1=V1=450 mL T1=T1=293 K V2=V2=? T2=T2=450K V 1 /T 1 = V 2 /T mL = V 2 (450K *450 mL)=V 2 /293 K
Example #2: 4.40 L of a gas is collected at 50.0°C. What will be its volume upon cooling to 25.0°C?
Example #3: 5.00 L of a gas is collected at 100 K and then allowed to expand to 20.0 L. What must the new temperature be in order to maintain the same pressure?
KELVINS EVERY TEMPERATURE USED IN A CALCULATION MUST BE IN KELVINS, NOT DEGREES CELSIUS.
Gay –Lussac’s Law The pressure of a fixed mass of gas at a constant volume varies directly with the Kelvin Temperature Or Increased Temperature = Increased Pressure
Gay –Lussac’s Law If the temperature of a container is increased, the pressure increases. ↑ T = ↑ P If the temperature of a container is decreased, the pressure decreases. ↓ T = ↓ P
P 1 P 2 T 1 T 2 Don’t forget temperature is in Kelvin Gay –Lussac’s Law =
If the pressure of a chamber was 100 atm at 600 K, what will be the temperature of the chamber if the pressure was raised to 600 atm? P1=P1= T1=T1= P2=P2= T2=T2= 100 atm 600K 600 atm ? K P 1 = P 2 T 1 T atm 600K = 600 atm ? K
100 atm 600K = 600 atm ? K 100 atm600K = 600 atm *? K *
100 atm 600K = 600 atm ? K 100 atm 600K = 600 atm *? K ? K = atm* K 100 K ? K = 3600K
Example #2: 5.00 L of a gas is collected at 22.0°C and mmHg. When the temperature is changed to standard, what is the new pressure?
Avogadro's Law Gives the relationship between volume and amount when pressure and temperature are held constant. V1 / n1 = V2 /n2 or V1 = V2 n1 n2
Avogadro's Law If the amount of gas in a container is increased, the volume increases. ↑ n = ↑ V If the amount of gas in a container is decreased, the volume decreases. ↓ n = ↓ v
5.00 L of a gas is known to contain mol. If the amount of gas is increased to 1.80 mol, what new volume will result (at an unchanged temperature and pressure)?
THE GAS LAWS Boyle’s P 1 V 1 = P 2 V 2 Charles’s V 1 / T 1 = V 2 / T 2 Gay-Lussac's P 1 / T 1 = P 2 / T 2 Avogadro's V 1 / n 1 = V 2 / n 2
Combined Gas Law Write Boyle's Law: P 1 V 1 = P 2 V 2 Multiply by Charles Law: P 1 V 1 2 / T 1 = P 2 V 2 2 / T 2 Multiply by Gay-Lussac's Law: P 1 2 V 1 2 / T 1 2 = P 2 2 V 2 2 / T 2 2 Take the square root to get the combined gas law: P 1 V 1 / T 1 = P 2 V 2 / T 2 or P 1 V 1 T 2 = P 2 V 2 T 1
2.00 L of a gas is collected at 25.0°C and mmHg. What is the volume at STP? P 1 =P 2 = V 1 =V 2 = T 1 =T 2 = mm Hg 2.00 L ( ) 760 mm Hg (0+273) X P 1 V 1 / T 1 = P 2 V 2 / T mm Hg 2.00L / 298 K = 760 mm Hg *X / 273 K V 2 = 1.79 L
Ideal Gas Law To derive the Ideal Gas Law, we combine all the gas laws Each of the laws has a constant, k. Each k is different. The product of all the constants (k) is written as R. R depends on the units used. R
Ideal Gas Law PV= nRT P = Pressure V = Volume (must be in Liters ) n = number of moles T = Temperature(must be in Kelvin ) R =Universal Gas Constant (Write this down) R = (L*atm)/(mol*K) or R = (L*kPa)/(mol*K) or R = (L*mm Hg)/(mol*K)
How many moles of gases are contained in a can with a volume of 555 mL and a pressure of kPa at 20 o C? (R = L*kPa)/(mol*K) N = mole
Calculate the pressure exerted by 43 moles of nitrogen in a 65 L cylinder at 5 o C? (Since there are no pressure units given you can use any R value you want) P= 15 atm P= 1529 kPa P= mmHg
If I have 4 moles of a gas at a pressure of 5.6 atm and a volume of 12 liters, what is the temperature? PV=nRT P= 5.6 atm V= 12 L n = 4 mole R= (L*atm)/(mol*K) T= x 5.6 atm *12 L = 4 mol * L* atm *X mol K 67.2 atm L = x L atm/K 67.2atm L / L atm /K = x K
A steel tank having a volume of 5.82 L contains 8.78 g of O 2 under a pressure of 1.35 atm. What is the temperature -- in degrees Celsius -- of the O 2 ? PV=nRT R = (L*atm)/mol*K) 1.35 atm *5.82L = (8.78g/32g/mol)(0.0821L atm/mol K) T atm L = T atm L/K T= K = = 76.5 o C
A sample of an unknown weighing grams is found to occupy L at 22.0°C and.9737atm How many moles of the gas are present? (0.97 atm) (2.85 L) = (n) ( L atm / mol K) (295.0 K) atm L = n *24.20 L atm/mol atm L = n L atm mole mole = n Mole buck Question What is the molar mass of the unknown gas?