# Question e6a5f

Jan 21, 2016

Here's what I got.

#### Explanation:

A wet gas is simply a term used to denote gases that are collected over water and, as a result, are mixed with water vapor.

In this regard, wet oxygen is a mixture of dry oxygen and water vapor.

So, you need to collect a volume of wet oxygen gas that would be equivalent to a $\text{337 mL}$ of dry oxygen gas at $\text{1.00 atm}$.

The trick here is to realize that the total pressure of the oxygen gas - water vapor mixture will be equal to the atmospheric pressure measured in the lab.

P_"lab" = overbrace(P_"dry oxygen" + P_"water vapor")^(color(purple)("the pressure of the wet oxygen gas"))

In order to be able to find the volume of dry oxygen gas, you need to know the vapor pressure oif water at ${15}^{\circ} \text{C}$. The listed value is equal to

${P}_{\text{water" = "12.73 torr}}$

http://www.endmemo.com/chem/vaporpressurewater.php

This means that at $\text{744 torr}$, the pressure of the dry oxygen gas would be

${P}_{\text{dry oxygen" = "744 torr" - "12.73 torr" = "731.27 torr}}$

Assuming that the temperature remains constant, you can use Boyles' Law to find the volume of wet oxygen gas that must be collected at $\text{744 torr}$ in order to produce that volume of dry oxygen

$\textcolor{b l u e}{{P}_{1} {V}_{1} = {P}_{2} {V}_{2}} \text{ }$, where

${P}_{1}$ - the pressure of the dry oxygen that's part of the oxygen - water vapor mixture
${V}_{1}$ - the volume of the wet oxygen gas
${P}_{2}$, ${V}_{2}$ - the pressure and volume of the dry oxygen gas

Rearrange to get

${V}_{1} = {P}_{2} / {P}_{1} \cdot {V}_{2}$

Plug in your values to get - remember that you have

$\text{1 atm " = " 760 torr}$

V_1 = (760color(red)(cancel(color(black)("torr"))))/(731.27color(red)(cancel(color(black)("torr")))) * "337 mL" = color(green)("350. mL")

Alternatively, you can actually use the ideal gas law equation to get the same result.

$\textcolor{b l u e}{P V = n R T}$

The trick this time is to realize that you must have equal numbers of moles of oxygen gas in both cases. So, you need

${n}_{\text{dry oxygen}} = \frac{{P}_{2} \cdot {V}_{2}}{R T}$

n_"dry oxygen" = (1 color(red)(cancel(color(black)("atm"))) * 337 * 10^(-3)color(red)(cancel(color(black)("L"))))/(0.0821 (color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 15.0)color(red)(cancel(color(black)("K"))))#

${n}_{\text{dry oxygen" = "0.014245 moles}}$

When collected over water, the pressure of the gas will be equal to $\text{731.27 torr}$. This means that the same amount of gas would be present in a volume of

${V}_{1} = \frac{{n}_{\text{dry oxygen}} \cdot R T}{P} _ 1$

${V}_{1} = \left(0.014245 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))) * 0.0821 (color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 15.0)color(red)(cancel(color(black)("K"))))/(731.27/760color(red)(cancel(color(black)("atm}}}}\right)$

${V}_{1} = \textcolor{g r e e n}{\text{350. mL}}$

The answer is rounded to three sig figs.