1 / 22

Ideal Gas Law

Ideal Gas Law. Boyle’s law : V a (at constant n and T ). V a. nT. nT. nT. P. P. P. V = constant x = R. 1. P. Ideal Gas Equation. This equation is a combination of 3 simpler gas relationships:. Charles’ law : V a T (at constant n and P ).

kalyca
Télécharger la présentation

Ideal Gas Law

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Ideal Gas Law

  2. Boyle’s law: V a (at constant n and T) Va nT nT nT P P P V = constant x = R 1 P Ideal Gas Equation This equation is a combination of 3 simpler gas relationships: Charles’ law: VaT(at constant n and P) Avogadro’s law: V a n(at constant P and T) R is the gas constant PV = nRT

  3. What does R=? R = (1 atm)(22.414L) PV = nT (1 mol)(273.15 K) The conditions 0 0C and 1 atm are called standard temperature and pressure (STP). Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L. PV = nRT R = 0.082057

  4. Ideal Gas Problem #1 1 mol HCl V = n = 49.8 g x = 1.37 mol 36.45 g HCl 1.37 mol x 0.0821 x 298.15 K V = 0.950 atm nRT L•atm P mol•K What is the volume (in liters) occupied by 49.8 g of HCl at a temperature of 25.0°C and a pressure of 0.950 atm? T = 25.0 + 273.15 = 298.15 K PV = nRT P = 0.950 atm At STP this same mass of HCl occupied a volume of 30.6 L V = 35.3 L

  5. Ideal Gas Problem #2 1 mol HCl V = n = 49.8 g x = 1.37 mol 36.45 g HCl 1.37 mol x 0.0821 x 273.15 K V = 1 atm nRT L•atm P mol•K What is the volume (in liters) occupied by 49.8 g of HCl at STP? T = 0 0C = 273.15 K PV = nRT P = 1 atm In the last question, this same mass of HCl at 25°C and 0.95 atm had a volume of 35.3 L V = 30.6 L

  6. Ideal Gas Problem #3 1 mol CO2 P = n = 25.0 g x = 0.568 mol 44.01 g CO2 0.568 mol x 0.0821 x 296.65 K P = 6.00 L nRT L•atm V mol•K What is the pressure (in atmospheres) of a gas in a 6.00 L container filled with 25.0 g of carbon dioxide at 23.5°C? T = 23.5°C+273.15 = 296.65 K PV = nRT V = 6.00 L P = 2.31 atm

  7. Ideal Gas Problem #4 1 mol N2 T = n = 15.0 g x = 0.535 mol 28.01 g N2 6.75 atm x 2.00 L T = 0.535 mol x 0.0821 PV L•atm nR mol•K What is the temperature (in °C) of a gas in a 2.00 L container filled with 15.0 g of nitrogen with a pressure of 6.75 atm? P = 6.75 atm PV = nRT V = 2.00 L T = 307.35 K – 273.15 = 34.2°C

  8. Ideal Gas Problem #5 n = 3.25 atm x 1.50 L n = 0.0821 x 288.15 K PV L•atm RT 32.00 g O2 mol•K 1 mol O2 What mass of oxygen gas occupies a volume of 1.50 L at 15.0 °C and 3.25 atm of pressure? T = 15.0°C+273.15 = 288.15 K PV = nRT V = 1.50 L P = 3.25 atm 0.200 mol = 6.41 g O2 n = 0.200 mol

  9. Ideal Gas Problem #6 (Making additional conversions) V = 765 mL = 0.765 L n = P = 333 kPa = 3.29 atm 3.29 atm x 0.765 L n = 0.0821 x 285.45 K PV L•atm RT 32.00 g O2 mol•K 1 mol O2 What mass of oxygen gas occupies a volume of 765 mL at 12.3 °C and 333 kPa of pressure? T = 12.3°C+273.15 = 285.45 K PV = nRT 0.107 mol = 3.44 g O2 n = 0.107 mol

  10. Ideal Gas Problem #7 (Making additional conversions) V = 439 mL = 4.39x10-4 L n = 1.25 atm x 4.39x10-4 L n = 0.0821 x 375.05 K PV L•atm RT 222 g Rn mol•K 1 mol Rn What mass of radioactive radon gas occupies a volume of 439 mL at 101.9 °C and 950 mmHg of pressure? T = 101.9°C+273.15 = 375.05 K PV = nRT P = 950 mmHg = 1.25 atm n = 1.782x10-5 mol 1.782x10-5 mol = 4.00x10-3 g Rn

  11. Since Before & After Calculations In some situations, we know the amounts of all 4 variables and our task is to determine one of them under new conditions where one or more of the others are changing. Here is how we use PV=nRT for this situation: We can use it to represent a “before” and “after” set of conditions like this: 1 = before or initial and 2 = after or final

  12. Ideal Gas Problem #8 (Before & After) P1 = 1.20 atm P2 = ? T1 = 291 K T2 = 358 K = P2 P1 T2 358 K T1 T2 T1 291 K = 1.20 atm x P2 = P1 x Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18°C is heated to 85°C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? n, and V are constant so they can be left out of the equation. Therefore: = 1.48 atm

  13. Dalton’s Lawof Partial Pressures

  14. Mixtures of gases require special handling when doing calculations. Collected gas Evaporated water

  15. Ptotal = P + P Collected gas Evaporated water

  16. PT = PO + PH O 2 2 Bottle being filled with oxygen gas NaClO3 Bottle full of oxygen gas and water vapor 2 NaClO3 (s)  2 NaCl (s) + 3 O2 (g)

  17. Dalton’s Law of Partial Pressures V and T are constant P1 P2 Ptotal= P1 + P2

  18. Pressure of Collected Gas Alone Pcollected gas = Ptotal - PH2O Vapor pressure of water at this temperature Total pressure of gas in container

  19. PA = nART nBRT V V PB = nB nA cB = cA = nA + nB nA + nB mole fraction (ci) = ni nT Consider a case in which two gases, A and B, are in a container of volume V. nA is the number of moles of A nB is the number of moles of B PT = PA + PB PA = cAPT PB = cBPT Pi = ciPT

  20. Gas Problem #9 0.116 8.24 + 0.421 + 0.116 A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116 moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)? Pi = ciPT PT = 1.37 atm cpropane = = 0.0132 Ppropane = 0.0132 x 1.37 atm = 0.0181 atm

  21. P V Chemistry in Action: Scuba Diving and the Gas Laws

More Related