question on finding the volume of vapor and the volume of liquid

rinchan4

1. The problem statement, all variables and given/known data

On a hot 35 degrees celsius day you perspire 1.0 kg of water during your workout.
A) what volume is occupied by the evaporated water?
B) By what factor is this larger than the volume occupied by the liquid water?

2. Relevant equations

PV=nRT
PV=NKbT
density of water is 1000 kg/m3= M/V
n=Mass/ Molar mass


[b]3. The attempt at a solution[/b

I really have no idea on how to start this problem. I did solve for n=M/Mmol=1000g/18.02g/mol= 55.49 moles but i don't even know if I need it. Any help would be greatly appreciated.

rinchan4

I just need a hint to get going in the right direction. I'm thinking that this problem may have something to do with the phase diagram of water?

Redbelly98

Welcome to PF.

What's missing here is the pressure of water. Has your class or book discussed the concept of vapor pressure ? There might be a table of water's vapor pressure in your book, for different temperatures. That information is needed to solve this.

rinchan4

Yeah we didn't talk at all about vapor pressure and the only thing that it says in my book is that we can ignore vapor pressure if we keep the tempature low. I did find that the vapor pressure of water is 2.4 kPa and 20 degrees celcius.

rinchan4

I tried using the density of vapor 0.0022(since i knew the mass of 1kg) and from that find the volume of 454.5m3 but that wasn't right

ideasrule

You had pV=nRT and you had both n and T, so all you need is p. Imagine trapping all of the water vapor in a box that's free to expand or contract whenever it needs to. What would the pressure of the water vapor be?

rinchan4

would be the same as the atmospheric pressure of 101300 Pa?

ideasrule

Yeah.

rinchan4

so then...
P= 101300 Pa
n= 1000/18.02=55.49moles
T= 308 k
R=8.31
Solve for V and I get 1.402 m3

rinchan4

awesome thank you so much. I was able to figure out part b. Thank you!!! :)