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Gas Mixtures Dalton’s Law

Gas Mixtures Dalton’s Law. Gas Mixtures--Partial Pressure. So far: pure gases Many gases are actually mixtures of two or more gases: air: O 2 , N 2 , H 2 O, etc How do mixtures of gases behave?. Gas Mixtures--Partial Pressure. P= 6 psi. P= 8 psi. P= 9 psi. O 2 (g). N 2 (g).

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Gas Mixtures Dalton’s Law

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  1. Gas MixturesDalton’s Law

  2. Gas Mixtures--Partial Pressure • So far: pure gases • Many gases are actually mixtures of two or more gases: • air: O2, N2 , H2O, etc • How do mixtures of gases behave?

  3. Gas Mixtures--Partial Pressure P= 6 psi P= 8 psi P= 9 psi O2 (g) N2 (g) CO2(g)

  4. P Gas Mixtures--Partial Pressure What happens when you put all three samples of gas together into one container (the same size container as each was in alone)? • The gases form a homogeneous mixture. • The pressure in the container increases, V and T stay the same • How do you know what the new pressure will be?

  5. Gas Mixtures--Partial Pressure • Each gas in a mixture behaves independently of the other gases present. • Each gas exerts its own pressure on the container. • PO = pressure exerted by O2 • PN = pressure exerted by N2 • PCO = pressure exerted by CO2 2 2 2

  6. Gas Mixtures--Partial Pressure • Partialpressure: the pressure exerted by a particular gas present in a mixture • Dalton's Law of Partial Pressure:The total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone. • Ptotal = P1 + P2 + P3 + ………

  7. Gas Mixtures--Partial Pressure • Ptotal = PO + PN + PCO • So for this example: Ptotal = 6 psi + 8 psi + 9 psi = 23 psi 2 2 2

  8. Partial Pressure • In other words, at constant T and V, • Ptotal depends only on the total number of moles of gas present • Ptotal is independent of the type (or types) of gases present.

  9. Partial Pressure-Mole Fraction • When describing a mixture of gases, it is useful to know the relative amount of each type of gas. • Mole fraction (X):a dimensionless number that expresses the ratio of the number of moles of one component compared to the total number of moles in a mixture.

  10. Mole Fraction • If a gas mixture contains 5.0 mol O2 (g), 3.0 mol H2O (g), and 12.0 mol N2 (g), XO= • On the exam, you must be able to calculate the mole fraction of each component of a gas mixture. nO2 5.0 mol = 0.25 = nt 20.0 mol

  11. Partial Pressure • The partial pressure of a gas in a mixture can be found: PA = XA Ptotal where PA = partial pressure of gas A XA = mole fraction of gas A Ptotal = total pressure of mixture

  12. Partial Pressure Calculation A mixture of gases contains 0.51 mol N2, 0.28 mol H2, and 0.52 mol NH3. If the total pressure of the mixture is 2.35 atm, what is the partial pressure of H2? PH2 = XH2 Ptotal 0.28 mol XH2= = 0.21 0.28 mol + 0.51 mol + 0.52 mol PH2 = 0.21 x 2.35 = 0.50 atm

  13. Dalton’s law of Partial Pressure The simple gas laws and idea gas law apply to individual gases as well as to a mixture of nonreactive gases. What is responsible for the pressure in a gas mixture? Since the pressure of pure gas at constant V and T is proportional to its amount (P=nRT/V), the pressure contribute from each individual gas in a mixture is also its amount in the mixture. In other words,the total pressure exerted by a mixture of gases in a container at constant V and T is equal to the sum of the partial pressure of each individual gas in the container, a statement known as Dalton’s law of partial pressure Ptot = PA + PB+ PC….at constant V and T Where PA, PB, PC …. refer to the pressure each individual gas would have if it were alone.

  14. Dalton’s law of partial pressures The pressure exerted by a particular gas in a mixture, PA, PB, PCis called partial pressure and refer to the pressure each individual gas would exert if it were alone in the container. That is, Here V = VA =VB = V tot The Concentration of any individual component in a gas mixture is usually expressed as mole fraction (X). The mole fraction of a component in a mixture is the fraction of moles of that component in the total moles of gas mixture.

  15. Which can be rearrange to solve PA, the partial pressure of component A: PA = XA·Ptot

  16. At constant P and T, where P = PA =PB = PC = Ptot, the volumeof each gas would individually occupy at a pressure equal to Ptot is VA = nART/Ptot ; VB =nBRT/Ptot … and so on.

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