Question

Question in the attachment!

#ASAP

@Chem experts ,kindly answer.

Note:- Don't even think of answering if you don't know the answer, just for the sake of points .

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Answer 1

Answer:

question:- A flask of volume V containing few drops of a liquid 'X' is filled with air so that the pressure in flask is 700 mm. Then the contents are shifted to another flask of volume 2 at the same temperature. What would be the pressure (in mm) in the new flask if the vapour pressure of liquid X' is 100 mm that temperature?

Sol.

Pr = 200 mm Hg PHO = 93 mm Hg

PH20 = 93 mm Hg

P1 = 200 mmHg 93 mmHg = 107 mmHg

We know, P V = K at constant T.

P1V1 = P1V2

107 x1= P2 x 2

P2 = 53.5 mm Hg

Pmix = (53.5+ 93) mmHg 146.5 mm Hg

hope it helps you!!

Answer 2

Heya mate,

Let V1 be the volume of liquid X in the original flask.

Given pressure of flask P1 = 700 mm Hg.

The another volume is = V/2 ( volume of the new flask which is shifted from the previous flask ).

Given, vapour pressure of liquid X in new flask = 100 mm Hg.

So, by applying Boyle's Law, we get:

• P1 × V1 = P2 × V2
• 700 × V/V = P2 × 1/2
• 700 × 1 = P2 × 1/2
• Therefore, P2 = 700 × 2 = 1400 mm Hg.
• Now, the vapor pressure exerts on the liquid, so the pressure of the liquid X in the new flask is = 1400 + 100 = 1500 mm Hg.

So, The required answer is = 1500 mm Hg.