Question

The vapor pressure of liquid propanol, C3H7OH, is 100 mm Hg at 326 K. A sample of C3H7OH is placed in a closed, evacuated 522 mL container at a temperature of 326 K. It is found that all of the C3H7OH is in the vapor phase and that the pressure is 59.0 mm Hg. If the volume of the container is reduced to 377 mL at constant temperature, which of the following statements are correct?

A. Some of the vapor initially present will condense.

B. Only propanol vapor will be present.

C. No condensation will occur.

D. The pressure in the container will be 100 mm Hg.

E. Liquid propanol will be present.

Answer 1

Ans :- option B and C are correct .

Explanation :-

From Mathematical expression of Boyle's law i.e

P₁V₁ = P₂ V₂ at constant Temperature ................(1)

here P₁ = 59.0 mm Hg

V₁ = 522 mL

P₂ = ? and

V₂ = 377 mL

From equation (1) , we have

P₂ = P₁V₁ / V₂

P₂ = 522 mL x 59.0 mm Hg / 377 mL

P₂ = 81.69 mm Hg

Since with decrease in volume , pressure increases it means C₃H₇OH remains in vapour phase .

Hence statements B and C are the correct .