# Question #e6a5f

##### 1 Answer

#### Answer:

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 **dry oxygen gas** at

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

#P_"water" = "12.73 torr"#

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

This means that at

#P_"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 **dry oxygen**

#color(blue)(P_1V_1 = P_2V_2)" "# , where

**dry oxygen** that's **part of the oxygen - water vapor mixture**

**wet oxygen gas**

Rearrange to get

#V_1 = P_2/P_1 * V_2#

Plug in your values to get - remember that you have

#"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.

#color(blue)(PV = nRT)#

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_"dry oxygen" = (P_2 * V_2)/(RT)#

#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_"dry oxygen" = "0.014245 moles"#

When collected **over water**, the pressure of the gas will be equal to

#V_1 = (n_"dry oxygen" * RT)/P_1#

#V_1 = (0.014245 color(red)(cancel(color(black)("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"))))#

#V_1 = color(green)("350. mL")#

The answer is rounded to three sig figs.