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Gas MixturesBecause the gas laws apply to ideal gases, they also apply to gas mixtures. Laws frequently used:
• Ideal gas law
• Dalton’s Law for partial pressures(including mole fraction)
Collection of Gases By Water Displacement
Example: 2KClO3(g) 2KCl(s) + 3O2(g)
Dalton’s Law of Partial Pressures
Dalton’s Law (based on Avogadro’s Law)
For two gases in a mixture:
Ptotal = total gas pressure
PA = partial pressure of gas ‘A’ (etc.)
Ptotal = PA + PB + ....
The total pressure of a gas mixture is the sum of the partial pressures of the components of the gas.
Dalton’s Law
PN2 = 0.78 atm
PO2 = 0.21 atm
Example:• air is 78% N2, 21% O2, and 1% other gases. At 1 atm:
Example - 1
A mixture of gases in scuba diving tank: (at 25oC, 1atm)He...... 46 L O2...... 12 L tank
volume = 5.0 L
¿
(a)partial pressure of each gas?
(b)total pressure inside tank?
Strategy – for (a)
(i) to use PV = nRT, we need the number of moles of each.
(ii) then we can determine the partial pressure for each
A mixture of gases in scuba diving tank: (at 25oC, 1atm)He...... 46 L O2...... 12 L tank volume = 5.0 L
¿
(i) number of moles of each gas?
(ii) partial pressure of each gas?
(a) partial pressure of each gas?
A mixture of gases in scuba diving tank: (at 25oC, 1atm)He...... 46 L O2...... 12 L tank volume = 5.0 L
¿
(b) total pressure inside tank?
Ptotal = PA + PB + ....
Ptotal = 9.3 atm + 2.4 atm = 11.7 atm
Example - 2Potassium chlorate (KClO3) was heated in a test tube and decomposed by the following reaction:
¿Question:
a. What is the partial pressure of O2 in the gas collected?
b. What was the mass of KClO3 in the original sample?
2KClO3(s) 2KCl(s) + 3O2(g)
The oxygen is collected by water displacement at 22oC, at a total pressure of 754 torr, for a total volume of 0.650 L. (PH2O = 21 torr.)
¿
2KClO3(s) 2KCl(s) + 3O2(g)
(a) What is the PO2?
Ptotal = PO2 + PH2O
PO2 = 754 torr – 21 torr = 733 torr
¿
2KClO3(s) 2KCl(s) + 3O2(g)
(b) What was the mass of KClO3 in the original sample?
iii. Number of moles of O2
ii. Number of moles of KClO3 / i. Grams KClO3
Example - 3Two bulbs are separated by a valve.
¿
Bulb Gas Pinit Vinit
A Ne 1.09 atm 1.12 LB CO 0.773 atm 2.18 L
When the valves are opened, and the gases are allowed to reach equilibrium, what is the final pressure inside the bulbs?
(Assume constant temperature.)
¿
Strategy
(iv) Final pressure
(iii) pressure from PV = nRT (Ttotal = VA + VB + VC)
(ii) total number of moles (ntotal = nA + nB + nC)
(i) mole of each gas
¿
(i) mole of each gas:
(ii) total moles:
(We don’t know Temp, but RT will cancel when calculating Ptotal)
¿
(iii) PTotal:
Mole Fraction
The ratio of the number of moles of given component in a mixture to the total number of moles in the mixture.
for component gas, A:
Combining ideal gas laws for both, and cancelling out R, V and T, and rearrange:
The fraction is called the mole fraction of A (= XA)
PA = Xa
Ptotal