Question

The vapor pressure of liquid ethyl acetate, CH_3 COOC_2 H_5, is 40.0 mm Hg at 282 K. A sample of CH_3 COOC_2 H_5 is placed in a closed, evacuated container of constant volume at a temperature of 411 K. It is found that all of the CH_3 COOC_2 H_5 is in the vapor phase and that the pressure is 60.0 mm Hg. If the temperature in the container is reduced to 282 K, which of the following statements are correct? Choose all that apply. No condensation will occur. Some of the vapor initially present will condense. The pressure in the container will be 41.2 mm Hg. Only ethyl acetate vapor will be present. Liquid ethyl acetate will be present.

Answer 1

Vapor pressure of CH3COOC2H5 is given as 40 mmHg at 282 K, that is, if the temperature falls to 282 K, the liquid will boil only if the pressure is greater than or equal to 40 mmHg. Now we have gas at 411 K and 60 mmHg. If the temperature is reduced to 282 K, its reduced pressure can be calculated according to ideal gas law at constant volume given as P1/T1 = P2/T2. Therefore 60/411 = P2/282. Therefore P2 = 41.2 mmHg.

Now since this pressure is greater than its vapor pressure at that temperature, there will be no liquid and all the CH3COOC2H5 will be in vaporized phase. No condensation will occur.