Answer to Question #340730 in General Chemistry for Onie

Question

 A 3.0-L bulb containing He at 145 mm Hg is connected by a valve to a 2.0-L bulb containing Ar at 355 mm Hg (see figure below). Calculate the partial pressure of each gas and the total pressure after the valve between the flasks is opened. [PHe = 87 mm Hg; PAr = 140 mm Hg]

Answer 1

Initial volume of He, V = 3L

Initial pressure of He, P= 145 mmHg

After connecting the tube,

Final volume of He, V' = (total volume for the gas free to move) = (Total volume of the two tubes) = ( 3 + 2) L = 5 L

let, in the mixture pertial pressure of He = P' mmHg

We know, PV= P'V'

Or, P' = PV/V'

Or, P' = 3×145/5

Or, P' = 87 mmHg

Initial volume of Ar, V* = 3L

Initial pressure of Ar, P* = 145 mmHg

After connecting the tube,

Final volume of Ar, V" = (total volume of two tubes) = (Total volume of the two tubes) = ( 3 + 2 ) L

= 5 L

let, in the mixture pertial pressure of He = P" mmHg

We know, P*V* = P"V"

Or, P' = P*V*/V"

Or, P" = 2×355/5

Or, P'' = 142 mmHg

Hence, pertial pressure of He in the mixture = 87 mmHg

Hence, pertial pressure of Ar in the mixture = 142 mmHg

Total pressure = (P" + P') mmHg =( 87+142 ) mmHg = 229 mmHg