Presentation on theme: "An Introduction to Gases Chapter 13. Kinetic Molecular Theory Postulate #1 –Gases consist of tiny particles (atoms or molecules) Postulate #2 –These particles."— Presentation transcript:
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.
Avogadros 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 measured with a BAROMETER (developed by Torricelli in 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.mmHg (or Torr) 2.Atmospheres (atm) 3.Pascals (used in physics: 1 pascal = 1 newton per square meter) 4. psi Equivalences: 1 atm = 760 mmHg 1 atm = 101,325 Pa = kPa 1 atm = 14.7 psi
Pressure Calculation What is 475 mm Hg expressed in atm? 475 mm Hg = 1 atm 760 mm Hg 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
Daltons 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. P total = P A + P B + P C
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 L (kPa) mol (K) L (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)8.314 L (kPa) mol (K) PV = nRT P = kPa P = 1500 kPa
Combined Gas Law Lets say we have some O 2 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? P1P1 15 atmP2P2 want V1V1 2 LV2V2 1 L n1n1 3 molesn2n2 R1R1 constantR2R2 T1T KT2T K
Combined Gas Law – Problemc P1V1P1V1 = n1R1T1n1R1T1 P2V2P2V2 n2R2T2n2R2T2 P1V1P1V1 = T1T1 P2V2P2V2 T2T2 If we know that R 1 = R 2 and the mass is constant then (15 atm)(2 L) = (300. K) P 2 (1L)(350. K) Replace with numbers
Combined Gas Law – Problem (15 atm)(2 L) = (300. K) P 2 (1L)(350. K) (15 atm)(2 L)(350. K) = P2P2 (1L)(300. K) P 2 = 35 atm
Pressure & Volume At constant Temperature Pressure and Volume vary inversely. –Why? –More collisions More pressure P 1 V 1 = P 2 V 2
P & V – Example Problem If you start with 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) =P2P2 3.5 L P 1 (V 1 ) = P 2 (V 2 ) 7.0 atm (0.500 L) = P 2 (3.5 L) 1.0 atm = P 2
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
Temperature & Volume At constant Pressure Volume & Temperature vary directly. –Why? –More collisions More Volume V1V1 = V2V2 T1T1 T2T2
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? V1V1 = V2V2 T1T1 T2T L = V2V K250. K 12.0 L (250. K) = V2V K 10.0 L= V2V2
S.T.P. Standard Temperature and PressureStandard Temperature and Pressure These are conditions that are universal Standard Temperature: 0ºC or 273K Standard Pressure: 1atm or kPa
S.T.P. – Example Problem What is the volume of 1 mole of a gas at STP? P1 atm Vwant n1 mole R (L)(atm)/(K)(mole) T273 K PV = nRT (1atm)V = (1 mole)( [Latm/Kmole])(273K) V= 22.4 L