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Combined Gas Law

Combined Gas Law. The good news is that you don’t have to remember all three gas laws! Since they are all related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION! P 1 V 1 P 2 V 2 =

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Combined Gas Law

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  1. Combined Gas Law • The good news is that you don’t have to remember all three gas laws! Since they are all related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION! P1 V1 P2 V2 = T1 T2 No, it’s not related to R2D2

  2. Combined Gas Law If you should only need one of the other gas laws, you can cover up the item that is constant and you will get that gas law! = P1 V1 P2 Boyle’s Law Charles’ Law Gay-Lussac’s Law V2 T1 T2

  3. Combined Gas Law Problem A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm? Set up Data Table P1 = 0.800 atm V1 = 180 mL T1 = 302 K P2 = 3.20 atm V2= 90 mL T2 = ?? *Make volume units the same!

  4. Calculation P1 = 0.800 atm V1 = 180 mL T1 = 302 K P2 = 3.20 atm V2= 90 mL T2 = ?? P1 V1 P2 V2 = T1 T2 T2 = P2 V2 T1 P1 V1 T2 = 3.20 atm x 90.0 mL x 302 K 0.800 atm x 180.0 mL T2 = 604 K - 273 = 331 °C = 604 K

  5. Learning Check A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temperature in °C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg? Note: Volumes must be the same unit and pressures must be the same unit!

  6. 1. List variables: • V1 = 675 mL = 0.675 L • T1 = 35°C = 308 K • P1 = 0.850 atm • V2 = .315 L • P2 = 802 mmHg = 1.06 atm • Decide on the appropriate gas law: • Everything’s changing, so Combined! • 3. Rearrange to solve for unknown: • T2 = (P2) (V2) (T1) = 179.2 K = (178.4K) (P1) (V1)

  7. PV =nRT Ideal gas law • IF WE COMBINE ALL OF THE LAWS we’ve looked at TOGETHER - INCLUDING AVOGADRO’S LAW - WE GET:

  8. Ideal gas constant(R) • R IS A CONSTANT THAT CONNECTS THE 4 VARIABLES • R IS DEPENDENT ON THE UNITS OF THE VARIABLE FOR PRESSURE • TEMP IS ALWAYS IN KELVIN • VOLUME IS ALWAYS IN LITERS • PRESSURE IS IN EITHER atm OR mmHg OR kPa

  9. L•atm L•kPa R=.0821 R=8.314 L•mmHg R=62.4 mol•K mol•K mol•K • Because of the different pressure units we use there are 3 different values for “R”” • IF PRESSURE IS GIVEN IN atm • IF PRESSURE IS GIVEN IN mmHg • IF PRESSURE IS GIVEN IN kPa

  10. Learning Check Dinitrogen monoxide (N2O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23°C, what is the pressure (in atm) in the tank in the dentist office?

  11. Using Ideal gas law L•atm 0.0821 mol•K (2.86 mol)(.0821)(296K) PV = nRT 20.0 L 1. List variables: n = 2.86 moles • R • P • ? • T • V • 20.0 L • 296K 2. Rearrange to solve for unknown: P = nRT V 3, Plug & Chug: • =3.48 atm

  12. GAS DIFFUSION AND EFFUSION • Diffusion is the movement of molecules to fill a container • Effusion is the movement of molecules through a small hole into an empty container.

  13. Graham’s Law Rates of effusion of gases are inversely proportional to the square root of their molar masses, at constant temp. & pressure. Thomas Graham, 1805-1869. Professor in Glasgow and London. M = molar mass & Gas B is the heavier gas!

  14. Graham’s Law Molecules effuse thru holes in a rubber balloon – that’s the main reason they get ‘whimpy’ after awhile! They do this at a rate that is inversely proportional to molar mass. Therefore, He (4 g/mol) effuses more rapidly than O2 (32 g/mol) at same T. (It’s lighter!) He Lighter gases effuse faster than heavier ones

  15. Graham’s Law • We can use the entire equation to calculate the actual speed of gas particles, however… • We will just use the square root side to COMPARE rates of effusion (speeds) • Ex. Compare the rates of effusion of oxygen gas & hydrogen gas. • 1st – find their molar masses! • O2 = 32.0 g/molH2 = 2.0 g/mol 2nd -Put the heavier gas (Gas B) in the numerator! 4 32 g You’re not done yet! 2.0 g

  16. Graham’s Law • The number 4 is not much of a “comparison”! • You must put your answer in sentence form! • Try this: • “_______ gas travels (or effuses) at a rate ___ times faster than _________ gas.” So, the answer is… • “Hydrogen gas travels (or effuses) at a rate 4 times faster than oxygen gas.” (Lighter gas) (#) (heavier)

  17. Graham’s Law • You try it! • Compare the rates of effusion of Ar and nitrogen gas (N2) 39.9 g/mol 28.0 g/mol 39.9 28.0 1.19 “Nitrogen gas travels (or effuses) at a rate 1.19 times faster than argon gas.”

  18. All of these gas laws work just ducky assuming the gases are ‘ideal” (Points with “no volume” & “no mutual attraction” Most of the time, gases conform to ideal conditions. So, when are gases not “ideal”? Under conditions of low temperature & high pressure (force molecules close enough to affect each other!)

  19. Deviations from Ideal Gas Law • Real Molecules have volumes and attractive forces between them. • Gases are not “Ideal” under conditions of high pressure & low temperature which bring particles close enough together to affect each other!

  20. STOP HERE

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