# A 60.0 mL sample of CO_2 gas is collected over water at 70.0° and 101.3 kPa. What is the volume of the dry gas at STP?

Dec 23, 2015

$\text{33.5 mL}$

#### Explanation:

The important thing to remember about gases that are being collected over water is that the total pressure measured includes the vapor pressure of water at that given temperature.

In this case, the collection tube will hold a gaseous mixture that contains molecules of carbon dioxide, ${\text{CO}}_{2}$, and molecules of water, $\text{H"_2"O}$.

This means that you can use Dalton's Law of Partial Pressures to determine the partial pressure of carbon dioxide in this mixture.

color(blue)(P_"total" = sum_i P_i)" ", where

${P}_{\text{total}}$ - the total pressure of the mixture
${P}_{i}$ - the partial pressure of gas $i$

In this case, the vapor pressure of water at ${70.0}^{\circ} \text{C}$ is equal to $\text{31.09 kPa}$

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

This means that you have

${P}_{\text{total}} = {P}_{C {O}_{2}} + {P}_{{H}_{2} O}$

${P}_{C {O}_{2}} = \text{101.3 kPa" - "31.09 kPa" = "70.21 kPa}$

Now, STP conditions imply a temperature of ${0}^{\circ} \text{C}$ and a pressure of $\text{100 kPa}$. Since you're looking to find the volume of the gas at STP, and keeping in mind that the number of moles of gas remains constant, you can use the combined gas law equation

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

${P}_{1}$, ${V}_{1}$, ${T}_{1}$ - the pressure, volume, and temperature of the gas at an initial state
${P}_{2}$, ${V}_{2}$, ${T}_{2}$ - the pressure, volume, and temperature of the gas at a final state

Plug in your values and solve for ${V}_{2}$ - do not forget to convert the temperature from degrees Celsius to Kelvin!

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

V_2 = (70.21 color(red)(cancel(color(black)("kPa"))))/(100.0color(red)(cancel(color(black)("kPa")))) * ( (273.15 + 0)color(red)(cancel(color(black)("K"))))/((273.15 + 70.0)color(red)(cancel(color(black)("K")))) * "60.0 mL"

${V}_{2} = \textcolor{g r e e n}{\text{33.5 mL}}$

The answer is rounded to three sig figs.