A 10 litre flaskAt 298k(kelvin) contain a gaseous mixture of carbon monoxide and carbon dioxide .At a total pressure of 2.0 if 2.02 mole of carbon monoxide and find its partial pressure and also that of carbon dioxide?

Asked by hemanttkumarr | 1st Nov, 2017, 08:00: AM

Expert Answer:

Given:
Volume (V)= 10lit
Total pressure (P) = PCO + PCO2 = 2 bar let's convert it into atm
1 bar =0.986923 atm therefore, 2bar =1.973846 atm=1.97 atm
R= 0.0821 lit.atm/mole.kelvin
 
from the relation,
 
P V equals n R T
space n equals fraction numerator P V over denominator R T end fraction equals fraction numerator 1.97 cross times 10 over denominator 0.0821 cross times 298 end fraction equals fraction numerator 19.7 over denominator 24.47 end fraction equals space 0.81 space
G i v e n space n o. space o f space m o l e s space o f space C O space equals 0.02 space m o l e
n subscript t o t a l space equals n subscript C O space end subscript end subscript plus space n subscript C O end subscript subscript 2 space
n subscript C O end subscript subscript 2 space equals space space n subscript t o t a l space space minus space n subscript C O space end subscript end subscript
space space space space space space space space equals space 0.81 minus 0.20
space space space space space space space space equals space 0.61 space m o l e s

N o w comma space m o l e space f r a c t i o n space o f space C O space i n space t h e space m i x t u r e comma
X subscript C O end subscript space equals fraction numerator space n subscript C O end subscript space over denominator n subscript t o t a l end subscript end fraction space equals fraction numerator 0.20 over denominator 0.81 end fraction equals 0.24
N o w comma space m o l e space f r a c t i o n space o f space C O subscript 2 space i n space t h e space m i x t u r e comma
X subscript C O end subscript subscript 2 space equals fraction numerator space n subscript C O end subscript subscript 2 space over denominator n subscript t o t a l end subscript end fraction equals space fraction numerator 0.61 over denominator 0.81 end fraction equals 0.75

t h e r e f o r e space p a r t i a l space p r e s s u r e space o f space C O comma
P subscript C O end subscript space equals space m o l e space f r a c t i o n space o f space C O space cross times space P subscript t o t a l end subscript
P subscript C O end subscript space equals space 0.24 cross times 1.97
space P subscript C O end subscript equals space 0.47 space a t m

P subscript C O subscript 2 end subscript space equals space m o l e space f r a c t i o n space o f space C O space cross times space P subscript t o t a l end subscript
space P subscript C O subscript 2 end subscript equals space 0.75 space cross times space 1.97
space P subscript C O subscript 2 end subscript equals space 1.48 space a t m

Answered by Ramandeep | 1st Nov, 2017, 10:55: AM

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