Question

how do u solve for d) on the percent humidity ?

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? El. Referring to C, what is the percent humidity! n the box immediately after it was Compressed? E. WHAT 15 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

This can be answered simply by taking acetone vapour as an ideal gas.

By ideal gas law

PV = nRT

Where,

n is no of mole

R is Gas constant

P is pressure of ideal gas taken

and V is volume of gas or volume of container in which it is filled.

now for the part D only we have to solve C part solve.

on solving C part by using Boyel's law

P1V1 = P2V2

new pressure of acetone vapour is found to be 400 mm Hg.

but in the he ask about percent Humidity which refer to water vapour present in container as compare to when it is saturated with water vapour.

since at this temperature V.P. of water not given so it is impossible to calculate.  

by using Data for water it is 23.8 mm Hg at 298 K temperature.

But According to question there is no water vapour present in the close container therefore percentage Humidity is 0% .

Answer: 0%