# Suppose 0.50 g of pure water is sealed in an evacuated 5.0-L flask and the whole assembly is heated to 60 degrees C. Will any liquid water be left in the flask or does all of the water evaporate?

Jul 11, 2016

There will be no liquid water left in the flask.

#### Explanation:

One way to solve this problem is to use the Ideal Gas Law to calculate the moles of water vapour that would be needed to fill the flask.

We could compare this value with the moles of liquid water, to see which is greater.

The Ideal Gas Law is

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

We can rearrange this to get

$n = \frac{P V}{R T}$

The vapour pressure of water at 60 °C is 149.4 torr.

P = 149.4 color(red)(cancel(color(black)("torr"))) ×("1 atm")/(760 color(red)(cancel(color(black)("torr")))) = "0.1966 atm"
$V = \text{5.0 L}$
$R = \text{0.082 06 L·atm·K"^"-1""mol"^"-1}$
$T = \text{(60 + 273.15) K" = "333.15 K}$

n = (0.1966 color(red)(cancel(color(black)("atm"))) × 5.0 color(red)(cancel(color(black)("L"))))/("0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 333.15 color(red)(cancel(color(black)("K")))) = "0.0360 mol"

Thus, it takes 0.0360 mol of water vapour to fill the flask at 60 °C.

The mass of liquid water that we have is 0.50 g.

$\text{Moles of H"_2"O" = 0.50 color(red)(cancel(color(black)("g H"_2"O"))) × (1 "mol H"_2"O")/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) = "0.0277 mol H"_2"O}$

This is less than 0.0360 mol, so all the water will have evaporated in the flask.