Question

A beaker of acetone (C3H60) is left in a 3.00 liter container at 298 Kelvin until the acetone vapor pressure in the box reaches the equilibrium value of 200.0 mmHg. The beaker of acetone is then removed (assume no gas particles escape) a. With the temperature held constant, what happens to the vapor pressure when the container is expanded to 9.00 L? b. Referring to part a., how many mols of acetone vapor are in the box before the expansion? How many mols of acetone vapor are present after the expansion? c. Now, the container is compressed to 1.50 Liters while temperature remains constant. What is the new pressure of the acetone vapor, immediately after the box is compressed? d. Referring to C, what is the percent humidity in the box immediately after it was compressed? e. What is likely to happen if the box remains at 1.50 Liters in volume? What will the final vapor pressure be? f. How many mols of gas are present in the gas phase now? g. How many grams of acetone are not in the gas phase, but still in the box?

Answer 1

a - By using the ideal gas equation-

PV=nRT

At equillibrium-

P₁V₁=P₂V₂

where P₁ = 200 mmHg

V₁ = 3 L

V₂ = 9 L

by putting all these formula, we get

P₂ = ( 200 * 3 )/ 9 = 66.67 mmHg
(o Again by using ideal gas - PV = nRT (T=const) Here at Equilibrium given - PV = P₂ V₂ P = 200 mm Hg V = 3L By V = lis putting P2 = all these values. mm Hg Pivi - 200XL - 400 15 for finding Relative humidity -RH = Partial pressure Vapour pressure & 100 Here , partial pressure = 200 mm Hg - Vapour pressure = 400 mm Hg RH = 600 mm Hg x 100 400 mm RH = 50 %

(6) For finding the motes we we ideal gas equation - PV = nRT = n = PV Before the expansion P = 200 mm Hg V = 3L n = R = Cras const - 8.312 J / mo T= 290K ? By putting in all these value nel can. find After the expansion - P= 66 67 rum Har h=2 R = 8.312 J/mol k T= 298 K putting these values we can find no a