# Question #956b2

##### 1 Answer

#### Explanation:

The idea here is that the **volume** of a gas, **proportional** to the number of moles of gas, **Avogadro's Law** here.

#V prop n#

The balloon is said to contain **parts** helium gas to **part** air, which implies that the volume of helium gas is **twice as large** as the volume of air.

Consequently, you can say that the balloon contains **twice as many** moles of helium gas than moles of air. In other words, if you take **moles** of helium gas.

Mathematically, you can show that this is the case because

#V_"He" prop n_"He" " " -># for helium

#V_"air" prop n_"air" " " -># for air

gets you

#V_"He"/V_"air" = n_"He"/n_"air"#

And since you know that

#V_"He" = 2 * V_"air"#

you will end up with

#n_"He" = (2 * color(red)(cancel(color(black)(V_"air"))))/color(red)(cancel(color(black)(V_"air"))) * n_"air"#

#n_"He" = 2 * n_"air"#

Now, the **partial pressure** of a gas that's part of a gaseous mixture depends on the **mole fraction** of the gas in the mixture **Dalton's Law of Partial Pressures** here.

The **mole fraction** of helium, **total number of moles** of gas present in the balloon.

Since we've said that for

#n + 2n = 3n#

This means that you have

#chi_"He" = (color(blue)(cancel(color(black)(n))) color(red)(cancel(color(black)("moles"))))/(3color(blue)(cancel(color(black)(n))) color(red)(cancel(color(black)("moles")))) = 1/3#

The partial pressure of helium in the balloon will be

#P_"He" = chi_ "he" * P_"total"#

Here **total pressure** in the balloon, is equal to

Plug in your values to find

#color(darkgreen)(ul(color(black)(chi_"He"))) = 1/3 * "100.1 kPa" = color(radkgreen)(ul(color(black)("33.37 kPa")))#

The answer is rounded to four **sig figs**, the number of sig figs you have for the total pressure in the balloon.