2. Gases ~ An Overview and Review of Concepts and Laws J. Baumwirt, Chemistry Granada Hills Charter High School From a compilation of different online and textbook resources for instructional purposes ONLY. Reproduction of this PowerPoint is prohibited due to copyright laws.

3. States of Matter Gases are only one form of matter Note the relative distance in particles of a gas as compared to other states of matter This will be an important factor as we study the properties of gases Solid Liquid Gas Plasma

4. Properties of Gases Gases are composed of atoms or molecules Gas particles are far apart and therefore are the most compressible state of matter Gases have lower densities that solids or liquids Gases will mix evenly and completely when confined to the same container

5. Properties of Gases Gases expand to fill any container Because atoms and molecules are so small a container of gas is mostly empty space Gaseous particles have relatively few attractions or repulsions between particles under normal conditions

6. Variables that Affect the Behavior of a Gas Pressure Volume Amount of Gas Temperature

7. Gases exert pressure on any surface they contact We can look at pressure simplistically as the number of times the particles hit the walls of the container Pressure Pressure = Force per unit Area P = Force Area (Force = mass x acceleration) ( “Particle in a Box” model )

8. Units of Pressure Pressure can be designated in a variety of different units: –mmHg or torr –atmospheres, atm –Pascals, Pa –psi (pounds per square inch) –Bars Their equivalencies are as follows: 1 atm = 760 mmHg = 760 torr = 101,325 Pa = 14.7 psi In Chemistry these are the units we will use most commonly mmHg or torr atmospheres, atm 1 atm = 760 mmHg = 760 torr

9. How Pressure is Measured The Barometer was invented by EvangelistaTorricelli Origin of the pressure unit in mmHg is due to the use of the metric system to measure of the height of the mercury column. This unit is also known as a torr It is the atmospheric pressure that pushes the mercury up the inverted glass tube The height (h) above the level of the mercury in the dish is then read in mmHg Why was mercury used?

10. a) P gas = P atm - h 2 b) P gas = P atm + h 2 Manometers (mă  năh΄  mә  tŭr) There are two types of manometers Open ended manometers with one end open to the atmosphere Closed end manometers where one end is a vacuum Pressure readings require finding the difference between the “legs” of the manometer Open ended devices require access to a barometer to find the pressure of the atmosphere. (illustrated here ash1,h1, h3)h3) h2,h2, c) P gas = h 1 The two illustrations of an open ended manometer show: a) a gas with a pressure less than atmospheric pressure b) a gas with a pressure greater than atmospheric pressure

11. The Ideal Gas Law Most gases behave “ideally” under normal conditions The Ideal Gas Law equates the variables of pressure, volume, amount of gas and temperature that affect gas behavior together in one equation: P V = n R T

12. Pressure is in the units of atmospheres, torr or mmHg Units of the Ideal Gas Law PV = nRT Temperature in Kelvin R = gas constant 0.0821 Latm = 62.37 Ltorr molK molK Amount of gas is in the unit of moles = n Volume in Liters (L) Recall that  C + 273.15 = K The R constant is chosen to match the unit of Pressure used. Recall that 1 torr = 1 mmHg 62.37 L mmHg 0.0821 Latm atmospheres 62.37 Ltorr torrmmHg

13. 760 atm15667.7 mmHg A 12.25 L cylinder contains 75.5 g of neon at 24.5 o C. Determine the pressure of the cylinder of gas. PV = nRT P = V = n = R = T = ? 12.25 L 75.5 g mol = 374 mol 20.18 g 62.4 L mmHg mol K 24.5 o C + 273 = 297.5 K P = nRT V = (3.74 mol)(62.4 L mmHg )(297.5K) (12.25 L) molK = 5667.7 mmHg = 5670 mmHg What is this in atm? atm7.46 =

14. Alterations of the Ideal Gas Law The Ideal Gas Law is used to find one aspect about a gas: –Pressure, Volume, number of moles or Temperature Through mathematical substitutions, the variables can be extended to –Molecular mass and –Density

15. Using Subsitutions with PV=nRT n in the equation = moles But how do we find the number of moles of a substance? –moles = –Substituting this back into the equation: grams of substance molar mass PV = RT grams Mwt Molar mass (or molecular weight) = Mwt The variables have now been extended to include mass and molecular weight.

16. More substitutions into PV=nRT Taking the previous substituted equation: The equation can be rearranged to solve for density. Density = PV = RT grams Mwt Mass Volume PV = RT grams Mwt mass = grams P  Mwt RT grams V = The consideration here is to remember that density units are somewhat altered now as they are in grams per Liter.

17. What is the density of carbon dioxide gas at 25 o C and 725 mmHg pressure? What do we do now? P = V = n = R = T = 725mmHg 62.4 L mmHg molK 25  C + 273 = 298 K grams ? Mwt CO 2 = 44.0 g/mol grams V = Density 725 mmHg  44.0 g/mol 62.4 L mmHg /mol K  298 K grams V = P  Mwt RT grams V = ? 1.72 g/L grams V = PV = RT grams Mwt

18. Combined Gas Law If there is a change in conditions such as a change in pressure or volume of a gas, the ideal gas law can be converted to an equality called the Combined Gas Law: Setting the equation equal to R: Then since R is constant for any gas, the following can be used to calculate a change for any of the variables. If a value is constant (does not change) it can be cancelled out and eliminated from the equation. Condition 1Condition 2 P1V1n1T1P1V1n1T1 P2V2n2T2P2V2n2T2 = PV nT = R

19. A balloon contains helium gas with a volume of 2.60 L at 25 o C and 768 mmHg. If the balloon ascends to an altitude where the helium pressure is 590 mmHg and the temperature is 15 o C, what is the volume of the balloon? What type of problem is this? There are 2 sets of conditions. Yikes!

20. P1=V1=T1=P1=V1=T1= P2=V2=T2=P2=V2=T2= 768 mmHg 2.60 L 25 o C + 273 = 298 K 590 mmHg 15  C + 273 = 288 K ? = (768 torr)(2.60 L)(288 K) (590 torr) (298 K) = 3.27 L A balloon contains helium gas with a volume of 2.60 L at 25 o C and 768 mmHg. If the balloon ascends to an altitude where the helium pressure is 590 mmHg and the temperature is 15 o C, what is the volume of the balloon? Condition 1: Condition 2: 12 211 2 TP TVP V = P 1 V 1 = P 2 V 2 n 1 T 1 n 1 T 2 P 1 V 1 = P 2 V 2 T 1 T 2 n is constant

21. Avogadro’s Law and Standard Temperature and Pressure Avogadro’s Law states that equal volumes of any two gases (Ideal) at the same temperature and pressure contain the same number of molecules. STP: Pressure 1 atm (760 mm Hg) Temperature 0 o C (273 K) Standard At STP one mole of ideal gas occupies 22.4 L (Looks like another conversion factor to me!) 1 mol gas/22.4L STANDARD TEMPERATURE & PRESSURE