Question

A gas mixture containing only helium and neon is 27.1 %% neon (by volume) and has a total pressure of 759 mmHgmmHg. What is the partial pressure of neon?

Answer 1

Step 1 - Understanding the relation between volume percent and parcial pressure

The parcial pressure of a gas in a mixture can be obtained by the following expression:

$p_p=x\times p_t$

In the equation above, pp represents the parcial pressure and pt the total pressure. x is the molar fraction, which can be defined as:

$x=\frac{n_{gas}}{n_{\text{total}}}$

I.e., is the quociente between the number of moles of the gas you want to calculate the parcial pressure of and the total number of moles.

Since, for a gas, the number of mols is proportional to the volume, the molar fraction is equal the volume percent:

$n\propto V\rightarrow x=\frac{kV_{gas}}{kV_{\text{total}}}$

K is a constant of proportionallity. It doens't matter its value, since it will be cancelled. We obtain thus:

$x=\frac{V_{gas}}{V_{\text{total}}}$

I.e., the molar fraction is exactly equal the volume percentage.

Step 2 - Using the relation between parcial pressure and volume percentage to solve the exercise

Since, as we saw, the molar fraction x is exactly equal to the volume percentage, we can substitute it by the volume percentage in the formula for the parcial pressure:

$p_p=V_{\text{percentage}}\times p_t$

From the exercise, we know that:

$\begin{gathered} V_{\text{percentage}}=27.1\text{ \% = 0.271} \\ \\ p_t=759\operatorname{mm}Hg_{} \end{gathered}$

Substituting these values on the equation we obtain the parcial pressure of Neon:

$p_p=0.271\times759=205.6\text{ mmHg}$

The parcial pressure of Neon is thus 205.6 mmHg.