2a. Calculate the total pressure in the flask after all of the oxygen has been r

Question

2a. Calculate the total pressure in the flask after all of the oxygen has been removed from the air n the ask. Assume that the temperature is 25°C, the initial barometrie pressure is ssume that the temperature is 25°C, the initial barometric pressure is 760 torr and 2b. Actually, the air inside the flask was in contact with water. Water vapor accounted for 23.6 torr of the total 760 torr of "wet" air inside the flask. If the O2 was removed from this air, would the pressure be higher or lower than in 2a?

Explanation / Answer

Q2a) Given that :Initial pressure=P=760 torr and % of O2 gas in the flask=21%,so this gives mole fraction of O2 in the flask=X(O2)=21/100=0.21

Using Dalton's law of partial pressures,

p(O2)=partial pressure of O2=total pressure*mole fraction of O2

or,pO2=P*X(O2)=760 torr*0.21=159.6 torr

So, Pressure inside the flask=P-pO2 (Dalton's law of partial pressures,total pressure=sum of partial pressures)

Pressure inside the flask=760 torr-159.6 torr=600.4 torr

b)p(H2O)=partial pressure of water vapor=23.6 torr

Pair=p(water vapor)+p(O2)+p(rest air components)

It will be same as p(water vapor) was included in the 760 torr total air pressure of which only p(O2 ) was substracted.The rest air components included water vapor as well.

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