# A flask contains three times as many moles as #H_2# as it does #O_2# gas. If hydrogen and oxygen are the only present gases, what is the total pressure flask if the partial pressure due to oxygen is #P#?

##### 1 Answer

#### Answer:

#### Explanation:

According to **Dalton's Law of Partial Pressures**, the **total pressure** of a gaseous mixture will be equal to the *individual* partial pressure of each component of the mixture.

Simply put, to get the total pressure of a gaseous mixture, you add up the partial pressure you'd get for each component of that mixture if it occupied the volume of the mixture **alone**.

Mathematically, this can written as

#color(blue)(|bar(ul(color(white)(a/a)P_"total" = sum_i P_icolor(white)(a/a)|)))" "# , where

**total pressure** of the gaseous mixture

**partial pressure** of component

In your case, the mixture is said to contain hydrogen gas, **mole ratio**.

This means that if you take *number of moles* of oxygen gas present in the mixture, you can say that

#n_(H_2) = 3 xx n_(O_2)" "color(purple)("(*)")#

Now, let's assume that the mixture is being held in a container that has a volume of *partial pressure* of oxygen gas, you can use the **ideal gas law** equation

#color(blue)(|bar(ul(color(white)(a/a)PV = nRT color(white)(a/a)|)))#

to express

#P * V = n_(O_2) * RT implies V = n_(O_2)/P * RT" "color(red)("(*)")#

Now, you can write the same thing for

#P^' * V = n_(H_2) * RT implies V = n_(H_2)/P^' * RT#

Use equation

#n_(O_2)/P * color(red)(cancel(color(black)(RT))) = n_(H_2)/P^' * color(red)(cancel(color(black)(RT)))#

Rearrange to find

#P^' = n_(H_2)/n_(O_2) * P#

Use equation

#P^' = (3 xx color(red)(cancel(color(black)(n_(O_2)))))/color(red)(cancel(color(black)(n_(O_2)))) * P#

#P^' = 3 xx P#

Finally, plug this into the equation that describes *Dalton's Law of Partial Pressures* to get the **total pressure** of the mixture,

#P_"total" = P^' + P#

#P_"total" = 3 xx P + P = color(green)(|bar(ul(color(white)(a/a)4 xx Pcolor(white)(a/a)|)))#