2 Gas PressureGas Pressure is related to the mass of the gas and to the motion of the gas particlesGas molecules move and bounce off the walls of their containerThese collisions cause gas pressurePressure is a force per unit areaStandard Units of pressure = Pascal(Pa)1 Pa is the pressure of 1 Newton per square meter (N/m2)Normal air pressure at sea level is kilopascals (kPa)kPa = 760mm Hg, 1kPa = 7.5 mm Hg (1 kPa=1000Pa)Other units of pressure, Atmosphere or psi1.00 atm = 760 mm Hg = kPa =14.7 psi
3 Atmospheric PressureThe pressure exerted on the Earth by the gasses in the atmosphereAbsolute pressure will include the pressure on a closed system, as indicated by a gauge PLUS the pressure exerted by the atmosphere.Ex.The pressure gauge on a bicycle tire reads 44 psi, what is the absolute pressure?44psi+14.7psi =59 psiConvert to kPa;59 psi X (101.3kPa/14.7 psi) =410kPa
4 Barometer vs. Manometer: Both measure pressure Barometer: is always closed end
5 Calculations for Open Ended System If mmHg is higher in open endPgas =Patmosphere + Ph3Solve for gas pressure if:Ph3 = 185 mmHgAtmospheric pressure is kPaConvert mmHg to kPa# kPa = (185mm/1)(1kPa/7.50mm)=24.7 kPaAdd kPa to atmospheric pressurePressure of gas = = 124kPa
6 Closed system: Used to measure pressure of a gas The closed arm is filled with gas, what is the pressure of the gas, in kPa?The difference in the Hg level is 165 mm7.50 mm = 1 kPaPressure of gas = (165.0 mm/1) ( 1kPa/7.5 mm) = 22.0 kPa
7 Try some yourselfClosed manometer containing sulfur dioxide gas (SO2). Height difference in mmHg is 560, what is the pressure of SO2 in kPa?An open monometer containing Hydrogen. Hg level is 78.0 mmHg in the arm connected to the air (open end), air pressure is 100.7kPa. What is the pressure of Hydrogen in kPa?An open monometer containing Nitrogen. Hg level is 26.0 mmHg in the arm connected to the gas, air pressure is 99.6kPa. What is the pressure of nitrogen in kPa?
8 Answers Pressure of SO2 = (560.0 mm/1) ( 1kPa/7.5 mm) = 74.6 kPa # kPa = (78.0mm/1)(1kPa/7.50mm)=10.4 kPaPressure of H2= 100.7kPa kPa= kPa# kPa= (26.0mm/1)(1kPa/7.50mm)= 3.47kPa Pressure of N2 = -3.47kPa =96.1kPanotice the negative number, less than air pressure
9 HomeworkComplete section review #1-8 handout p 73
10 Mathematic Relationships There is a Mathematical Relationship between Pressure, Temperature and Volume of a constant amount of gasGas volumes change significantly with small changes in temperature and pressureThese changes can be defined by equations called the gas laws.Gas laws are only valid for ideal gassesIdeal gases: do not exist, but are a model, they have no attractive force and no volume
11 Boyle’s Law; Relates Pressure and Volume If the pressure on an ideal gas is increased, the volume decreasesWhen the pressure is doubled ( ), the volume of the gas is half ( ) what it wasIf the pressure is cut( ) by half, the volume is doubled ( ).Boyle’s Law: GAS VOLUME AND PRESSURE, at constant temperature, ARE INVERSELY PROPORTIONAL! ( )(when one goes up the other goes down)To find the new volume, you need the original volume AND the change in pressureV2 = V1 X P1/P2To find new pressure, you need the original pressure AND the change in volume --- New pressure = old pressure X volume RatioP2 = P1 X V1/V2- Mentally decide if the new volume is larger or smaller than the old, arrange the pressure ratio accordingly (If larger use fraction greater than 1, If smaller use fraction less than 1)
12 Boyle’s Law Example Problems If 425 mL of O2 are collected at a pressure of 9.80 kPa what volume will the gas occupy if the pressure is changed to 9.40 kPa?the pressure decreases from 9.8 to 9.4 kPa. (V will increase)V2 = V1 X P1/P2(425mL/1) ( 9.8 kPa/9.4 kPa) = mLCalculate the pressure of a gas that occupies a volume of mL, if at a pressure of 95.0 kPa, it occupies a volume of mL.the volume decreases from mL to 125.0mL.(P will increase)P2 = P1 X V1/V2(95.0 kPa/1) (219.0 mL/125.0mL) = kPa
13 Charles’s Law; relates Temperature and Volume In an ideal gas, with constant pressure, if the temperature increases the volume will increase.Charles’s Law: at a constant pressure, the volume of a gas is directly proportional to its Kelvin temperature.(you must express temp in degrees K)New volume =old volume X degree K temp changeV2 = V1 X T2/T1New temperature = old temp (K) X volume changeT2 = T1 X V2/V1- Mentally decide if the new volume is larger or smaller than the old, arrange the pressure ratio accordingly (If larger use fraction greater than 1, If smaller use fraction less than 1)
14 Charles’s Law Example Problem What volume will a sample of nitrogen occupy at 28.0 degrees C if the gas occupies a volume of 457 mL at a temperature of 0.0 degrees C? Assume the pressure remains constant.convert temp to KK= degrees CK = degrees C = 301 Kand K = C = 273 K(Temp increase means Volume increase: Ratio > than 1)New Volume = (457mL/1) (301 k/273 k) = 504 mL
15 Charles’s Law Example Problem #2 If a gas occupies a volume of 733 mL at 10.0 dC, at what temperature in C degrees, will it occupy a volume of 1225 mL if the pressure remains constant?(increased volume means increased temperature)Convert °C to °K (°K = °C + 273) °C = 283 °KT2 = T1 X V2/V1= (283 °K/1) (1225 mL/733 mL) = 473 °KConvert °K to °C°C = 473 K – 273 = 200 °C
16 Practice use of Boyle’s Law Complete the problem solving packet, Pg 12-16; problems 1-5 all partsShow all workLabel all unitsDue tomorrow (complete for homework if necessary)
17 Practice use of Charles’s Law Complete the problem solving packet, Pg problems 1-3 all partsShow all workLabel all unitsDue tomorrow (complete for homework if necessary)
18 Combined Gas Law: when both temperature and pressure change occur Boyle’s and Charles’s laws used together make the combined gas law.A pressure ratio and a Kelvin temperature ratio are needed to calculate the new volume.New Volume = old volume X pressure ratio X Kelvin temperature ratioV2 =V1(P1/P2)(T2/T1) *STP=273K and 101.3kPaFirst: Construct a press-volume-temp data tableValueOld ConditionsNew ConditionsWhat happens to gas Volume?PVT
19 Practice: Combined Gas Law Calculate the volume of a gas at STP if 502 mL of the gas are collected at 29.7 C and 96.0 kPaConvert Celsius to Kelvin temps. ( = 302.7)Organize the data in the table.ValueOld ConditionsNew ConditionsWhat happens to gas Volume?PVT96.0 kPa502 mL302.7 K101.3 kPa?273 KDecreaseCalculate:V2 =V1(P1/P2)(T2/T1)V2 = 502 mL ( 96.0kPa/ 101.3kPa) (273K/ 302.7k)V2 = 429 mL
20 Try this on your own!If 400 ml of oxygen are collected at 20.0C, and the atmospheric pressure is 94.7 kPa, what is the volume of the oxygen at STP?V2 =V1(P1/P2)(T2/T1)= 400 mL (94.7 kPa/101.3 kPa) (273 k/293 K)= 348 mLValueOld ConditionsNew ConditionsWhat happens to gas Volume?PVT94.7 kPa400. mL293 K101.3 kPa?273 KDecrease
21 Homework: Complete Handout, Pg 78-79 #’s 10-14, all partsShow all of your worklabel all of your unitsDUE TOMORROW!!!
22 Gay- Lussac’s Law: relates Temperature and pressure The Pressure of a gas is DIRECTLY proportional to the absolute temperature when the volume is unchanged.P1T2=P2T1 or P1 = P2T T2When temp increases, pressure increasesMust use Kelvin temperatures!
23 Gay- Lussac’s Law Example A cylinder of gas has a pressure of 4.40 atm at 25 C. At what temperature, in Celsius, will it reach a pressure of 6.50 atm? P1 = 4.40atm P1T2=P2T1 T2= ? K and ?C 298 X 6.50 atm = 440 K P2 = 6.50 atm 4.40 atm T1 = =298 K 440.K = (440. – 273) C =167°C
24 Practice Problems; Gay- Lussac’s Law Complete the problem solving packet, Pg20-24, problems 1-3 all partsShow all workLabel all units
25 Avogadro’s Law; relates volume to moles Avogadro's Law states: for a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas.Avogadro's law relates the quantity of a gas and its volume.According to Avogadro's Law: V1n2 = V2n1When any three of the four quantities in the equation are known, the fourth can be calculated.For example, if n1, V1 and V2 are known, the n2 can be solved by the following equation:n2 = V2 x (n1/V1)
26 Avogadro’s Law Practice Example #1: 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)?Answer: this time I'll use V1n2 = V2n1(5.00 L) (1.80 mol) = (x) (0.965 mol)
27 Avogadro’s Law Practice Example #2: A cylinder with a movable piston contains 2.00 g of helium, He, at room temperature. More helium was added to the cylinder and the volume was adjusted so that the gas pressure remained the same. How many grams of helium were added to the cylinder if the volume was changed from 2.00 L to 2.70 L? (The temperature was held constant.)Solution:1) Convert grams of He to moles:2.00 g / 4.00 g/mol = mol2) Use Avogadro's Law:V1/n1 = V2/n22.00 L / mol = 2.70 L / xx = mol3) Compute grams of He added:0.675 mol mol = mol mol x 4.00 g/mol = 0.7 grams of He added
28 Progression of LawsBoyles', Charles', and Avogadro's laws combine to form the ideal gas law, which is the uber law of gases.The ideal gas law can be manipulated to explain Dalton's law, partial pressure, gas density, and the mole fraction. It can also be used to derive the other gas laws.
29 Ideal Gas LawThe ideal gas law is an ideal law. It operates under a number of assumptions.The two most important assumptions are that the molecules of an ideal gas do not occupy space and do not attract each other.These assumptions work well at the relatively low pressures and high temperatures, but there are circumstances in the real world for which the ideal gas law holds little value.
30 Ideal Gas Law PV = m RT Equation: PV = nRT P is pressure in kPa V is volume in cubic decimetersT is temperature in Kn represents the number of moles of the gasR is a constant , using these units, is 8.31 L kPa/mol K(R can have different units when needed)This equation can be used to determine the molecular mass of a gasMoles (n) = mass(m)/ molecular mass(M)PV = m RTM
31 Amount of Gas PresentAmount of gas present is related to its pressure, temperature and volume using the Ideal Gas LawApplying the gas law allows for the calculation of the amount of gaseous reactants and products in reactions
32 Application Of Gas Laws Example How many moles of gas will a 1250 mL flask hold at 35.0 degrees C and a pressure of 95.4 kPa?PV=nRT or n = PVRTConvert degree C to Kelvin, 35.0C = = 308Kn = PVn= kPa mL L8.31 L kPa/mol K K mLn= mol
33 Gas Laws Example 2A flask has a volume of 258 mL. A gas with mass g is introduced into the flask at a temperature of K and a pressure of 9.86 x104 Pa. Calculate the molecular mass of the gas using the ideal gas equation.n= mass/molecular mass(M)M = mRTPVM= g L kPa K mL Pa9.86 x104 Pa mol K mL L kPaM= g/mol
34 Ideal Gas Law ProblemsHow many moles of He are contained in a 5.00L canister at 101 kPa and 30.0 dC?What is the volume of mol Ne at kPa and 27.0 dC?How much Zn must react in order to form 15.5 L of H2 gas at 32.0 dC and 115 kPa?Zn + H2SO4 ZnSO4 + H2
35 Ideal Gas Law Problem Solutions How many moles of He are contained in a 5.00L canister at 101 kPa and 30.0 dC?n= VP/RT(5.00 L) (101 kPa)( 8.31 kPa L) 303 K(1 mol K)= 5.0 X 101 X 1 mol8.31 X 303= mol
36 Ideal Gas Law Problem Solutions V= nRTPV = ( mol Ne) (8.31 kPa L) (300 K)( 1 mol K)0.505 kPaV= 99 L Ne
37 Ideal Gas Law Problem Solutions First determine molesPV=nRT n=PV n= (115 kPa) ( 15.5 L)RT ( 8.31 kPa L) (305)( 1 mol K )n= mol H2Now determine mass of Zn (molar ratio and molar mass)(0.703 mol H2) (1 mol Zn) ( g Zn)(1) (1 mol H2) ( 1 mol Zn)= 46.0 g Zn
38 Compare and Contrast gas Laws RelatesEquationUnitBoyle’sPressure to VolumeP1V1=P2V2L or kPaCharles’sTemperature to VolumeT1V2=T2V1K or LCombined Gas lawTemperature, pressure and volumeV2 =V1(P1/P2)(T2/T1)K, L and kPaGay- Lussac’s LawTemperature and pressureP1T2=P2T1K or kPaAvogadro’s Lawvolume to molesV1 / n1 = V2 / n2kPa or molIdeal gas LawPressure, volume, temp and molesPV=nRT andPV = m RTMMol, L, K, or kPa