A cylinder is found to have a total pressure of 1519.5kPa. If there were 2.0 moles of oxygen, 5.0 moles of He, and 7.0 moles of nitrogen, what would the partial pressure of each be?

Answer 1

Answer 2

First of all, know that the total pressure of any gaseous mixture is the sum of all the partial pressure contributed by each component of the gas. In the other words, if you add all the partial pressures you should get the total pressure of the system. So, how do we calculate the partial pressure? Here is the formula to remember.

p_i = xi.P_t

Where pi is the partial pressure of one of the component

xi is the mole fraction of that component [But what is mole fraction? ==> Well, see below]

Pt is the total pressure

Mole fraction of a component is = [moles of the component/Total number of moles of all the component]

So in the given problem, mole fraction of oxygen = 2/(2+5+7) = 2/14 = 0.14₃

mole fraction of He = 5/14 = 0.35₇

mole fraction of nitrogen = 7/14 = 0.50

(Notice that I usually carry one extra degit in my intermediate answers so that it won't lead to a rounding error in the final answer). Some professors/teachers do assign partial credit for significant figures, so watch carefully for those easy points!

Now we can calculate partial pressure of each of these gases by multiplying each of these mole fractions with the total pressure according to the formula.

partial pressure of oxygen = (0.14₃) * (1519.5 kpa) = 217.29 kpa = 220 kpa (2 Sig Fig)

partial pressure of He = (0.35₇) * (1519.5 kpa) = 542.46 kpa = 540 kpa (2 sig Fig)

partial pressure of nitrogen = (0.50) * (1519.5 kpa = 759.75 kpa = 760 kpa (2 sig fig)

[Please review the rules of sig figs. If you have questions, let me know]