Equal masses of methane and hydrogen are mixed in an empty container. At 25 degree Celsius. The fraction of total pressure exerted by hydrogen will be?

Jun 9, 2017

$\frac{8}{9}$

Explanation:

The idea here is that the partial pressure of hydrogen gas will depend on

• the mole fraction of hydrogen gas in the mixture
• the total pressure of the mixture

Now, let's say that we are mixing $m$ $\text{g}$ of hydrogen gas and $m$ $\text{g}$ of methane.

Use the molar masses of the two gases to express the number of moles of each component of the mixture in terms of their mass $m$

m color(red)(cancel(color(black)("g"))) * "1 mole H"_2/(2.0color(red)(cancel(color(black)("g")))) = (m/2) ${\text{moles H}}_{2}$

m color(red)(cancel(color(black)("g"))) * "1 mole CH"_4/(16.0color(red)(cancel(color(black)("g")))) = (m/16) ${\text{moles O}}_{2}$

The total number of moles of gas present in the mixture will be equal to

$\left(\frac{m}{2}\right) \textcolor{w h i t e}{.} \text{moles" + (m/16)color(white)(.)"moles" = (9/16 * m)color(white)(.)"moles}$

Now, the mole fraction of hydrogen gas is calculated by dividing the number of moles of hydrogen gas by the total number of moles of gas present in the mixture.

chi_ ("H"_ 2) = ((color(red)(cancel(color(black)(m)))/2)color(red)(cancel(color(black)("moles"))))/((9/16 * color(red)(cancel(color(black)(m))))color(red)(cancel(color(black)("moles")))) = 1/2 * 16/9 = 8/9

By definition, the partial pressure of hydrogen gas in the mixture is equal to

P_ ("H"_ 2) = chi_ ("H"_ 2) * P_"total"

In your case, this will be equal to

P_ ("H"_ 2) = 8/9 * P_"total"