# Question effc6

May 26, 2017

P_("H"_2"O") = 0.54 "atm"

P_("He") = 0.66 "atm"#

#### Explanation:

We can use the equation

${P}_{a} = \left({\chi}_{a}\right) \left({P}_{\text{total}}\right)$

where ${P}_{a}$ is the partial pressure of a component $a$,
${\chi}_{a}$ is the mole fraction of gas $a$, and
${P}_{\text{total}}$ is the total pressure of the system, in this case $1.200 \text{atm}$.

Essentially, the partial pressure of a gas is simply the mole fraction of the gas multiplied by the total pressure.

We're given the percentage of each gas in the mixture, and the mole fraction is just that percentage divided by $100$ ($0.45$ for water vapor, and $0.55$ for helium).

Let's plug in the mole fraction of each gaseous component into the first equation to find the partial pressure of each gas:

${P}_{\text{H"_2"O") = (0.45)(1.200 "atm") = color(red)(0.54 "atm}}$

${P}_{\text{He" = (0.55)(1.200 "atm" = color(blue)(0.66 "atm}}$

We can check our answers by summing the partial pressure of each gas, and seeing if it adds to the total pressure:

$0.54 \text{atm" + 0.66 "atm" = 1.20 "atm}$