TITLE.PM5

GASES AND VAPOUR MIXTURES 445

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\M-therm\Th9-3.pm5

3. In a mixture of gases, the partial pressure pi of any constituent gas can be found by
   (a)pi = niRT/V (b)pi = miRT/Vi
   (c)pi = niR 0 T/Vi (d)pi = niR 0 T/V
   where, R = Characteristic gas constant,
   R 0 = Universal gas constant,
   V = Volume of the mixture,
   Vi = Volume of the ith constituent, and
   T = Temperature of the mixture.


4. Mole fraction of a component of gas mixture is equal to
   (a)1/f (b) f^2
   (c)f (d) f/p
   where, f = Volume fraction, and
   p = Pressure of the mixture.


5. In a gaseous mixture the specific volume of each component is given by
   (a)V/m (b) Vi/mi
   (c)V/mi (d) none of the above.
   where, V = Volume of the mixture,
   Vi = Volume of the ith component,
   m = Mass of mixture, and
   mi = Mass of the ith component.

Answers

1. (b) 2. (d) 3. (d) 4. (c) 5. (c).

Theoretical Questions

1. Define the following terms :
   (i) Partial pressure (ii) Mole fraction
   (iii) Volume fraction of a gas constituent in a mixture.


2. Explain briefly Dalton’s law and Gibbs-Dalton law.


3. State and explain Amagat’s law or Leduc’s law.


4. Prove that the molar analysis is identical with the volumetric analysis, and both are equal to the ratio of
   the partial pressure to the total pressure.


5. Prove the following relation

M =

Σ

Σ

nM

n

ii

i

1

m

M

fi

∑ i

where, M = Molecular weight of the mixture,

ni = Number of moles of an any constituent,

mfi = Mass fraction of the constituent, and

Mi = Molecular weight of the constituent.

Unsolved Examples

1. 0.45 kg of carbon monoxide (28) and 1 kg of air at 15°C are contained in a vessel of volume 0.4 m^3.
   Calculate the partial pressure of each constituent and the total pressure in the vessel. The gravimetric
   analysis of air is to be taken as 23.3% oxygen (32) and 76.7% nitrogen (28).
   [Ans. pO 2 = 0.4359 bar ; pN 2 = 1.64 bar, pCO = 0.962 bar]