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Gas Laws Q4U1 (13d). Gas Pressure Gas Pressure is related to the mass of the gas and to the motion of the gas particles Gas molecules move and bounce.

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Presentation on theme: "Gas Laws Q4U1 (13d). Gas Pressure Gas Pressure is related to the mass of the gas and to the motion of the gas particles Gas molecules move and bounce."— Presentation transcript:

1 Gas Laws Q4U1 (13d)

2 Gas Pressure Gas Pressure is related to the mass of the gas and to the motion of the gas particles Gas molecules move and bounce off the walls of their container – These collisions cause gas pressure Pressure is a force per unit area – Standard Units of pressure = Pascal(Pa) – 1 Pa is the pressure of 1 Newton per square meter (N/m 2 ) 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 psi 1.00 atm = 760 mm Hg = kPa =14.7 psi

3 Atmospheric Pressure The pressure exerted on the Earth by the gasses in the atmosphere Absolute 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 psi Convert 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 end P gas =P atmosphere + P h3 Solve for gas pressure if: P h3 = 185 mmHg Atmospheric pressure is kPa 1. Convert mmHg to kPa # kPa = (185mm/1)(1kPa/7.50mm)=24.7 kPa 2.Add kPa to atmospheric pressure Pressure 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 mm 7.50 mm = 1 kPa Pressure of gas = (165.0 mm/1) ( 1kPa/7.5 mm) = 22.0 kPa

7 Try some yourself 1.Closed manometer containing sulfur dioxide gas (SO2). Height difference in mmHg is 560, what is the pressure of SO2 in kPa? 2. 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? 3.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 1. Pressure of SO2 = (560.0 mm/1) ( 1kPa/7.5 mm) = 74.6 kPa 2.# kPa = (78.0mm/1)(1kPa/7.50mm)=10.4 kPa Pressure of H 2 = 100.7kPa kPa= kPa 3.# kPa= (26.0mm/1)(1kPa/7.50mm)= 3.47kPa Pressure of N2 = -3.47kPa =96.1kPa notice the negative number, less than air pressure

9 Homework Complete 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 gas Gas volumes change significantly with small changes in temperature and pressure These changes can be defined by equations called the gas laws. Gas laws are only valid for ideal gasses Ideal gases: do not exist, but are a model, they have no attractive force and no volume

11 Boyles Law; Relates Pressure and Volume If the pressure on an ideal gas is increased, the volume decreases When the pressure is doubled ( ), the volume of the gas is half ( ) what it was If the pressure is cut( ) by half, the volume is doubled ( ). Boyles 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 pressure V 2 = V 1 X P 1 /P 2 – To find new pressure, you need the original pressure AND the change in volume --- New pressure = old pressure X volume Ratio P 2 = P 1 X V 1 /V 2 - 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 Boyles Law Example Problems 1.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) V 2 = V 1 X P 1 /P 2 (425mL/1) ( 9.8 kPa/9.4 kPa) = mL 2.Calculate 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 = P 1 X V 1 /V 2 (95.0 kPa/1) (219.0 mL/125.0mL) = kPa

13 Charless Law; relates Temperature and Volume In an ideal gas, with constant pressure, if the temperature increases the volume will increase. Charless 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 change V 2 = V 1 X T 2 /T 1 New temperature = old temp (K) X volume change T 2 = T 1 X V 2 /V 1 - 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 Charless Law Example Problem 1.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 K K= degrees C K = degrees C = 301 K and K = C = 273 K (Temp increase means Volume increase: Ratio > than 1) New Volume = (457mL/1) (301 k/273 k) = 504 mL

15 Charless Law Example Problem #2 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) 10.0 °C = 283 °K T 2 = T 1 X V 2 /V 1 = (283 °K/1) (1225 mL/733 mL) = 473 °K Convert °K to °C °C = 473 K – 273 = 200 °C

16 Practice use of Boyles Law Complete the problem solving packet, Pg ; problems 1-5 all parts Show all work Label all units Due tomorrow (complete for homework if necessary)

17 Practice use of Charless Law Complete the problem solving packet, Pg problems 1-3 all parts Show all work Label all units Due tomorrow (complete for homework if necessary)

18 Combined Gas Law: when both temperature and pressure change occur Boyles and Charless 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 ratio V 2 =V 1 (P 1 /P 2 )(T 2 /T 1 ) *STP=273K and 101.3kPa First: Construct a press-volume-temp data table ValueOld ConditionsNew ConditionsWhat happens to gas Volume? PVTPVT

19 Practice: Combined Gas Law ValueOld ConditionsNew ConditionsWhat happens to gas Volume? PVTPVT 96.0 kPa 502 mL K kPa ? 273 K Decrease ? Decrease Calculate the volume of a gas at STP if 502 mL of the gas are collected at 29.7 C and 96.0 kPa Convert Celsius to Kelvin temps. ( = 302.7) Organize the data in the table. Calculate: V 2 =V 1 (P 1 /P 2 )(T 2 /T 1 ) V 2 = 502 mL ( 96.0kPa/ 101.3kPa) (273K/ 302.7k) V 2 = 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? V 2 =V 1 (P 1 /P 2 )(T 2 /T 1 ) = 400 mL (94.7 kPa/101.3 kPa) (273 k/293 K) = 348 mL ValueOld ConditionsNew ConditionsWhat happens to gas Volume? PVTPVT 94.7 kPa 400. mL 293 K kPa ? 273 K Decrease ? Decrease

21 Homework: Complete Handout, Pg #s 10-14, all parts Show all of your work label all of your units DUE TOMORROW!!!

22 Gay- Lussacs Law: relates Temperature and pressure The Pressure of a gas is DIRECTLY proportional to the absolute temperature when the volume is unchanged. P 1 T 2 =P 2 T 1 or P 1 = P 2 T 1 T 2 When temp increases, pressure increases Must use Kelvin temperatures!

23 Gay- Lussacs 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? P 1 = 4.40atm P 1 T 2 =P 2 T 1 T 2 = ? K and ?C 298 X 6.50 atm = 440 K P 2 = 6.50 atm 4.40 atm T 1 = =298 K 440.K = (440. – 273) C =167°C

24 Practice Problems; Gay- Lussacs Law Complete the problem solving packet, Pg20- 24, problems 1-3 all parts Show all work Label all units

25 Avogadros 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: V 1 n 2 = V 2 n 1 When any three of the four quantities in the equation are known, the fourth can be calculated. – For example, if n 1, V 1 and V 2 are known, the n 2 can be solved by the following equation: – n 2 = V 2 x (n 1 /V 1 )

26 Avogadros 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 V 1 n 2 = V 2 n 1 (5.00 L) (1.80 mol) = (x) (0.965 mol)

27 Avogadros 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 = mol 2) Use Avogadro's Law: V 1 /n 1 = V 2 /n L / mol = 2.70 L / x x = mol 3) Compute grams of He added: mol mol = mol mol x 4.00 g/mol = 0.7 grams of He added

28 Progression of Laws Boyles', 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 Law The 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 Equation: PV = nRT P is pressure in kPa V is volume in cubic decimeters T is temperature in K n represents the number of moles of the gas R 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 gas – Moles (n) = mass(m)/ molecular mass(M) PV = m RT M

31 Amount of Gas Present Amount of gas present is related to its pressure, temperature and volume using the Ideal Gas Law Applying the gas law allows for the calculation of the amount of gaseous reactants and products in reactions

32 Application Of Gas Laws Example 1. 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 = PV RT Convert degree C to Kelvin, 35.0C = = 308K n = PV RT n= 95.4 kPa 1250 mL 1L 8.31 L kPa/mol K 308 K 1000 mL n= mol

33 Gas Laws Example 2 2.A 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 x10 4 Pa. Calculate the molecular mass of the gas using the ideal gas equation. n= mass/molecular mass(M) M = mRT PV M= g 8.31 L kPa K 1000 mL 1000 Pa 9.86 x10 4 Pa mol K 258 mL 1L 1kPa M= g/mol

34 Ideal Gas Law Problems 1.How many moles of He are contained in a 5.00L canister at 101 kPa and 30.0 dC? 2.What is the volume of mol Ne at kPa and 27.0 dC? 3.How much Zn must react in order to form 15.5 L of H 2 gas at 32.0 dC and 115 kPa? Zn + H 2 SO 4 ZnSO 4 + H 2

35 Ideal Gas Law Problem Solutions 1.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 mol 8.31 X 303 = mol

36 Ideal Gas Law Problem Solutions 2. V= nRT P V = ( mol Ne) (8.31 kPa L) (300 K) ( 1 mol K) kPa V= 99 L Ne

37 Ideal Gas Law Problem Solutions 3.First determine moles PV=nRT n=PV n= (115 kPa) ( 15.5 L) RT ( 8.31 kPa L) (305) ( 1 mol K ) n= mol H2 Now 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 Gas LawRelatesEquationUnit BoylesPressure to VolumeP 1 V 1 =P 2 V 2 L or kPa CharlessTemperature to Volume T 1 V 2 =T 2 V 1 K or L Combined Gas law Temperature, pressure and volume V 2 =V 1 (P 1 /P 2 )(T 2 /T 1 )K, L and kPa Gay- Lussacs Law Temperature and pressure P 1 T 2 =P 2 T 1 K or kPa Avogadros Law volume to molesV 1 / n 1 = V 2 / n 2 kPa or mol Ideal gas LawPressure, volume, temp and moles PV=nRT and PV = m RT M Mol, L, K, or kPa


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