Dalton's Law and Eudiometry

Daltonsinglequotes Law of partial pressure

Partial pressure: Partial pressure is the pressure exerted by one gas in a mixture of gases if it was the only gas occupying the whole volume at that temperature.Therefore, each gas has its own partial pressure and the total pressure is sum of partial pressures of all the gases present in the mixture.This is true for non-reacting, ideal gases though real gases do not deviate much from the law.

Daltonsinglequotes Law state that " the total pressure of a mixture of non-reacting, ideal gases is the sum total of their individual partial pressures." 

That is, if there n gases in a mixture and partial pressure of each gas i is denoted by P_i then, the total pressure P is defined as :

                                                                ? P_total = P₁ + P₂ + ....... P_n

 The individual pressures can be calcuated if the amount of each gas is known.

That is, if the moles of gas P₁ is n₁ and gas P₂ is n₂ and so on ..

P1 = ( n₁/ n_total ) P  = X₁P , where n_total is the total number of moles in the system and X₁ is mole fraction of gas 1.

? Partial pressure of a gas in a mixture = Mole fraction of the gas in the mixture X Total pressure of the system.

Hence, %age of a gas in the mixture = (Its partial pressure / total pressure of the system) X 100.

Application of the daltonsinglequotes law of the partial pressure:

Daltonsinglequotes law of partial pressure is used in calculation the vapour pressure of water or the aqueous tension.A lot of gases in laboratory are collected by the downward displacement of water.The gases collected by this method also has water molecules mixed with the gas  molecules.Therefore, the total pressure of the mixture is the pressure of the gas plus the aqueous tension which is equal to vaopur pressure of water at that temperature.

?P_total = P_drygas + Aq. tension

Partial Pressure in a binary Mixture:

Lets suppose there is a binary mixture having componets singlequoteasinglequote and singlequotebsinglequote with partial pressures P_a and P_b and mole fraction X_a and X_b respectively.

It can be seen from the graph as:

The line joining Pa and Pb denotes total pressure at any given time. The graph shows that X_a = 0 then , total pressure = P_b and when X_b = 0 then total pressure = P_a the partial pressure of component singlequoteasinglequote .At any given time P_total = P_a + P_b

  [Solved Example]

Example6:7.96 grams of NH₃ is collected by the downward displacement of water at a total pressure of 2.68 atm, at a total volume of 4.55L at a temperature of 27⁰C.Find the vapour pressure of water vapour.

Solution: Using the ideal gas equation,  P = nRT / V,

?n = 7.96/17 = 0.47 moles.

P = .47 X 0.0821 X (273+27) / 4.55 = 2.53 atm.

, Aqueous tension = (2.68 - 2.53)atm = 0.15 atm.

[/Solved Example]                                

[Solved Example]

Example7:A 900 ml of a mixture of oxygen and ozone weigh 1.4 gram at STP. Calculate the volume of oxygen present in the mixture.

Solution:Let the volume of oxygen present be V.

, Volume of ozone present = (900 - V).

At STP , 1 mole of any gas occupies 22400 mL , so we can find out the number of moles of both gases.

Moles of oxygen = V / 22400

Mass of oxygen = (V / 22400) X 32  ..................... (i)

Moles of ozone = (900 - V). / 22400 

Mass of ozone = (900 - V). / 22400  X 48 ..................... (ii)

Now, given mass of mixture =  2 grams.

using (i) and (ii) we get,

(V / 22400) X 32  + (900 - V). / 22400  X 48 = 1.4

? V = 740 mL.

[/Solved Example]

Eudiometry: (Application of gay lussacsinglequotes law of combining volumes)

Eudiometry is the science of determining the constituents of a gaseous mixture by means of air or determining the amount of oxgen present in the air or the purity of it. It also used in determining the %age composition of hydrocarbons and hence determining their molecular formulae.It is done as follows:

1. The hydrocarbon is burnt in a closed tube with an excess quantity of oxygen,so that C and H are oxidized to CO₂ and H₂O respectively.

2. The mixture is then cooled, due to which H₂O condenses into liquid and the volume of the mixture is decreased.

3.NaOH is used to seperate CO₂ from the mixture as NaOH reacts with CO₂ to form Na₂CO3 and H₂O and the volume of the solution decreases furthur.

4. The excess O₂ is absorbed with the pyrogallol solution resulting a furthur decrease in volume.

Using the volume contracted at every stage , we can find out the amount of CO₂ , H₂O formed , hence the molecular formula can be found out.

 [SolvedExample]

Example8: 10 ml of a gaseous hydrocarbon was burnt with 50 ml of O₂ in an eudiometer tube.The volume of products after being cooled to room temperature was 35 ml, when treated with NaOH the volume contracted by 30 ml.Find the molecular formula of the hydrocarbon.

Solution: Let the hydrocarbon be C_xH_y .
The balanced reaction for its combustion will be:

C_xH_y(g) + (χ + y/4)O₂ ? χCO2 + (y/2)H₂O(g).

From gay lussacsinglequotes law of combining volume we get:

1 volume of C_xH_y  = (χ + y/4)volume of O2 = χ volume of CO2 (y/2) volume of H₂O.

Contraction in volume = 1 + (χ + y/4) - χ ( volume of water is not considered as it becomes liquid at room temperature).

                                    = 1 + y/4 . ( for one volume of C_xH_y ) 

For, 10 volume of C_xH_{y ,} contraction = 10₍1 + y/4)

Contraction in volume here ,. 50 + 10 - 35 = 25.

10 (1 + y/4) = 25

?y = 6.

Now, CO₂ formed = 10χ = 30 ml (volume contracted after treatment by NaOH)

?χ = 3.

We can cross check our answer by using  the data regarding unused O₂.

Unused O₂ = volume left after treatment with NaOH = 5 ml .

Volume of O₂ used = 45 ml.

1 volume of C_xHy reacts with (χ + y/4) volume of  O₂,

10 volumes of C_xHy will react with 10 ( χ + y/4) volume of  O₂ which is 45 here.

Putting , χ = 3 and y = 6 it can be verified that 10 ( χ + y/4) is 45 .

 [/SolvedExample]