Question

A sealed flask contains 0.50 g of water at 28 ∘C. The vapor pressure of water at this temperature is 28.36 mmHg.

What is the minimum volume of the flask in order that no liquid water be present in the flask?

Answer 1

This problem is providing us with the mass, 0.50 g, temperature, 28 °C and pressure, 28.36 mmHg, of steam and asks for the volume to not let the steam to condense. At the end, the result turns out to be 18.4 L.

Ideal gas

In chemistry, we can model the pressure-temperature-volume-mole behavior of gases via gas laws with different approaches. However, since this problem involves all these variables, we understand we need to use the ideal gas law:

$PV=nRT$

Where we should set the pressure in atmospheres, moles in mol and the temperature in kelvins, as well as solving for volume as it is the required:

$V=\frac{nRT}{P} \\\\V=\frac{(0.50g*\frac{1mol}{18.02g} )(0.08206\frac{atm*L}{mol*K})(28+273)K}{28.36mmHg*\frac{1atm}{760mmHg} } \\\\V=18.4L$

Learn more about the ideal gases: https://brainly.com/question/8711877

Answer 2

For this question, we need to use ideal gas law

$pV=nRT$

Given :

Mass of water = 0.50g

Temperature = 28°C

Converting to Kelvin Scale,

28 + 273.15 K

=> 298.15 K

Pressure = 28.36 mm of Hg

=> 0.03732 atm

We also know that R = 0.08205 L/atm/mol.K

We want to the Volume

Let's first calculate the number of moles (n) of H2O :

$n = \frac{m}{M}$, where m is the mass and M is the molar mass of water ( = 18,01528 g/mol )

n = 0.50g / 18,01528 g/mol

n = 0.02775 mol

$V= \frac{nRT}{p}$

$V= \frac{(0.02775mol \times 0.08205l/atm/mol.k \times 298.15k)}{0.03732atm}$

$= > V = 18.19L$

Therefore, the minimum volume = 18.19L

~Benjemin360