Question

The partial pressure of O2 in mixture containing equal mass of O2,H2 and He at a total pressure of 7.5 atm is

Answer 1

By Dalton's law of Partial Pressures, the partial pressures of O2, N2 and CO2 will all add up to the total pressure. Thus, 46.5 kPa is equal to 5.6 plus 14.2 kPa plus the pressure of CO2.

Answer 2

Given:

A mixture of gas contains equal mass of $O_{2} ,H_{2},He$.

Total pressure of mixture is 7.5 atm.

To find:

the partial pressure of $O_{2}$.

Formula to be used:

$moles=\frac{given-mass}{molecular-mass}$

$mole-fraction=\frac{moles-of-oxygen}{total moles}$

$P_{O_{2} }=X_{O_{2}} P_{T}$

Step-by-step explanation:

Step 1 of 2

Let the mass of the gases be "x".

The moles of gases can be calculated as:

Oxygen gas:

$=\frac{given-mass}{molecular-mass} \\\\=\frac{x}{32}$

Hydrogen gas:

$=\frac{x}{2}$

Neon gas:

$=\frac{x}{20}$

Step 2 of 2

Mole fraction of oxygen gas is

$=\frac{moles-of-oxygen}{total moles}\\ \\=\frac{\frac{x}{32} }{\frac{x}{32}+\frac{x}{2}+\frac{x}{20}} \\\\=\frac{\frac{x}{32}}{\frac{5x+80x+8x}{160}} \\\\=\frac{x*160}{32*93x} \\\\=\frac{5}{93}$

The partial pressure of oxygen will be:

$P_{O_{2} }=X_{O_{2}} P_{T}\\\\P_{O_{2} }=\frac{5}{93} *7.5\\\\P_{O_{2} }=0.4032atm$

the partial pressure of O2 is 0.4032 atm.