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Ideal-Gas Equation

Combining these, we get. nT P. V . Ideal-Gas Equation. V  1/ P (Boyle’s law) V  T (Charles’s law) V  n (Avogadro’s law). So far we’ve seen that. nT P. nT P. V . V = R. Ideal-Gas Equation. The relationship. then becomes. or. PV = nRT. Ideal-Gas Equation.

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Ideal-Gas Equation

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  1. Combining these, we get nT P V Ideal-Gas Equation V 1/P (Boyle’s law) VT (Charles’s law) Vn (Avogadro’s law) • So far we’ve seen that

  2. nT P nT P V V= R Ideal-Gas Equation The relationship then becomes or PV = nRT

  3. Ideal-Gas Equation PV = nRT • P = pressure • V = volume • n = amount • T = temperature • R = gas constant

  4. Gas Laws • The value of the gas constant (R) depends on the units of P, V, n, and T. • T must always be in Kelvin • n is usually in moles • If P (atm) and V (L), • then R = 0.08206 atm.L mol.K • If P (torr) and V (L), • then R = 62.36 L.torr mol.K I will give you these on the test.

  5. Gas Laws • The ideal gas law is used to describe the behavior of an idealgas. • Idealgas:hypothetical gas that obeys kinetic molecular theory and the ideal gas law

  6. Gas Laws • The ideal gas law is used in calculations for a specific sample of gas that has a constant T, P, V, and n. • i.e. no changes are being made to the sample of gas • If you know 3 of the 4 variables, you can calculate the other using the ideal gas law.

  7. Gas Laws Calculate the volume of 1.00 mol of an ideal gas at 1.00 atm and 0.00oC. Given: P = 1.00 atm n = 1.00 mol T = 0.00oC = 273 K Find: V PV = nRT

  8. Gas Laws PV = nRT Solve for VV = nRT P Use R = 0.08206 atm.L/molK • V = 22.4 L

  9. Gas Laws • The temperature and pressure used in the previous problem are commonly used to report the properties of gases. • Standard Temperature and Pressure (STP): 0oC and 1 atm

  10. Gas Laws • Molar volume: • the volume one mol of a gas occupies (L/mole) • At STP, one mole of an ideal gas has a molar volume of 22.4 L: 22.4 L 1 mol

  11. Think How many molecules are in 22.4 L of an ideal gas at STP? Hint: How many moles are there under these conditions?

  12. Gas Laws • The ideal gas law applies only to ideal gases. • Does not always accurately describe real gases • The molar volumes for many real gases at STP differ slightly from 22.4 L/mol. • In most cases, the differences between ideal gas behavior and real gas behavior is so small that we can ignore it.

  13. Gas Laws A weather balloon contains 4.75 moles of He gas. What volume does the gas occupy at an altitude of 4300 m if the temperature is 0.oC and the pressure is 0.595 atm? Given: P = 0.595 atm T = 0.oC = 273K n = 4.75 mol Find: V PV = nRT

  14. Gas Laws PV = nRT solve for V V = nRT P 0.08206 atm L 273 K 4.75 mol = 179 L V= 0.595 atm mol K

  15. Gas Laws A used aerosol can contains 0.0173 mol of gas and has a volume of 425 mL. Calculate the pressure in the can if it is accidentally heated to 395oC. (Warning: Don’t do this!!) Given:V = 425 mL n = 0.0173 mol T = 395oC + 273 = 668K Find:P PV = nRT

  16. PV = nRT nRT solve for P P = V 0.0173 mol 0.08206 atm L 668 K P = = 2.23 atm 425 mL mol K 1 L/1000 mL

  17. Gas Laws A tire with an interior volume of 3.50 L contains 0.357 mol of air at a pressure of 2.49 atm. What is the temperature of the air in the tire in K? In oC? Given: P = 2.49 atm V = 3.50 L n = 0.357 mol Find: T (K) PV = nRT

  18. Gas Laws PV = nRT solve for T PV 2.49 atm 3.50 L mol K = T = 0.357 mol 0.08206 atm L nR = 297 K = 297 K -273 = 24C

  19. Gas Laws • The ideal gas law was useful in determining the properties of a specific sample of gas at constant T, P, V, and n. • We often need to know how a change in one (or more) properties impacts the other properties for a sample of a gas.

  20. Combined Gas Laws • Fixed number of moles of gas (n= constant) and R is a constant: • PV/T = nR = constant P1V1 = P2V2 T1 T2 where subscripts mean initial (1) and final (2) conditions of P, V, T

  21. Gas Laws A helium-filled balloon occupies 6.00 L at 19.5oC and 0.989 atm. What volume will the balloon occupy on top of Pike’s Peak if the pressure is 0.605 atm and the temperature is constant? Given: V1 = 6.00L P1 = 0.989 atm T1 = T2 = 19.5oC P2 = 0.605 atm Find: V2 Since T is constant, P1V1 = P2V2

  22. Gas Laws P1 V1 = P2 V2 (0.989 atm) x (6.00 L) = (0.605 atm) x V2 V2 = 9.81 L

  23. P1V1 = P2V2 T1 T2 Gas Laws Suppose a used aerosol can contains a gas at 0.989 atm at 23oC. If this can is heated to 425oC, what is the pressure inside the can? Given: P1 = 0.989 atm T1 = 23oC + 273 = 296K T2 = 425 + 273 = 698K V1 = V2 Find: P2 = 2.33 atm

  24. m V PM RT = Densities of Gases • We know that moles  molecular mass (g/mol) = mass n M = m • So multiplying both sides (of PV = nRT when rearranged to n/V=P/RT) by the molecular mass (M) gives

  25. m V PM RT d = = Densities of Gases • Mass  volume = density • So, Note: One only needs to know the molecular mass (M), the pressure, and the temperature to calculate the density of a gas.

  26. dRT P PM RT d = M = Molecular Mass We can manipulate the density equation to enable us to find the molecular mass of a gas: Becomes

  27. Examples What is the density of helium gas at 1.00 atm and 25oC? Given: P = 1.00 atm T = 25 + 273 = 298K He gas: 4.00 g/mol Find: d • d = PM • RT 1.00 atm 4.00 g mol K = = 0.164 g/L 0.08206 atm L 298 K mol Note: Density of gases units are: g/L

  28. Gas Laws--More Applications Example: What is the average molar mass of dry air if it has a density of 1.17 g/L at 21oC and 740.0 torr? Given: P = 740.0 torr T = 21 + 273 = 294K d = 1.17 g/L Find:M (molar mass) • d = PM • RT

  29. M = dRT • P Solution: M= • d = PM • RT 62.36 L torr 1.17 g 294K L 740.0 torr mol K = 29.0 g/mol

  30. Gas Laws--More Applications • Understanding the properties of gases is important because gases are often the reactants or products in a chemical reaction. • Often need to calculate the volume of gas produced or consumed during a reaction

  31. grams A grams A Moles A Moles A Molar mass Molar mass PV = nRT Molar ratio Molar ratio grams B grams B Moles B Moles B Molar mass Molar mass PV = nRT Gas Laws--More Applications Gas Data A PA, VA, TA Gas Data B PB, VB, TB

  32. Gas Laws--More Applications The air bag in a car is inflated by nitrogen gas formed by the decomposition of NaN3: 2 NaN3(s)  2 Na (s) + 3 N2 (g) If an inflated air bag has a volume of 36 L and is to be filled with N2 gas at a pressure of 1.15 atm at a temperature of 26C, how many grams of NaN3 must be decomposed?

  33. PV = nRT Gas Laws--More Applications Moles A Moles N2 Gas Data N2 Molar ratio Molar ratio grams B grams NaN3 Moles B Moles NaN3 Molar mass Molar mass

  34. PV = nRT Gas Laws--More Applications 36L N2, 1.15 atm 26oC Moles A Moles N2 Find Molar ratio Molar ratio Given grams B grams NaN3 Moles B Moles NaN3 Molar mass Molar mass

  35. Gas Laws--More Applications n = PV = (1.15 atm) ( 36 L) mol.K RT (299 K) 0.08206 atm.L n = 1.7 mol N2 g NaN3 = 1.7 mol N2 x 2 mol NaN3 x 65.0 g NaN3 3 mol N2 1 mol NaN3 g NaN3 = 74 g NaN3

  36. Gas Laws--More Applications Example: How many mL of oxygen gas can be collected at STP when 1.0 g of KClO3 decomposes: 2 KClO3 (s)  2 KCl (s) + 3 O2 (g)

  37. PV = nRT Gas Laws--More Applications mL O2 at STP (1.0atm, 0oC) Moles A Moles O2 Given Molar ratio Molar ratio Find grams B grams KClO3 Moles B Moles KClO3 Molar mass Molar mass

  38. Gas Laws--More Applications Mol O2 = 1.0 g KClO3 x 1 mole KClO3x 3 mol O2 122.5 g 2 mol KClO3 mol O2 = 0.012 mol O2 nRT (0.012 mol)0.0821 atm.L (273K) x 103 mL 1.0 atm mol.K 1 L V = 270 mL V= = P

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