A mixture of two gases has a total pressure of 5.7 atm. If one gas has a partial pressure of 4.1 atm, what is the partial pressure of the other gas?

Jun 6, 2016

I got $\text{1.6 atm}$.

DALTON'S LAW OF PARTIAL PRESSURES

The equation that sums up what we have to do in this problem (pun intended) is Dalton's law of partial pressures:

$\setminus m a t h b f \left({P}_{\text{tot}} = {\sum}_{i = 1}^{N} {P}_{i}\right)$

$= {P}_{1} + {P}_{2} + . . . + {P}_{N}$

which really says that the total pressure is the sum of all the individual pressures, known as partial pressures.

(Note that this equation is constructed for ideal gases only, so it doesn't work perfectly for real gases.)

USING DALTON'S LAW OF PARTIAL PRESSURES

By the language of the question, we already know that there are only two gases in the closed container; let's call them gas $1$ and gas $2$.

So, from the question, what we have at our disposal, and what we are solving for, are:

${P}_{\text{tot" = "5.7 atm}}$
${P}_{1} = \text{4.1 atm}$
P_2 = ???

Therefore, the unknown partial pressure is:

$\textcolor{b l u e}{{P}_{2}} = {P}_{\text{tot}} - {P}_{1}$

$= 5.7 - 4.1$

$=$ $\textcolor{b l u e}{\text{1.6 atm}}$

UNSTATED ASSUMPTIONS IN THE PROBLEM

When we are looking at ideal gases, we can say that their partial pressure contributions are similar at any total pressure for any identity gas. (This is not true for real gases.)

Another unstated assumption is that this is a closed rigid container, so we have a constant volume. That allows us to assume that the pressures are additive (but that still doesn't entirely work unless the gases are truly ideal).