Question

A mixture of neon and oxygen gases, in a 9.77 L flask at 65 °C, contains 2.84 grams of neon and 7.67 grams of oxygen. The partial pressure of oxygen in the flask is ? atm and the total pressure in the flask is atm?

Answer 1

`p_text(O₂) = "0.681 atm"; p_text(tot) = "1.081 atm"`

Explanation:

We can use the Ideal Gas Law to calculate the partial pressures.

`color(blue)(bar(ul(|color(white)(a/a)pV = nRTcolor(white)(a/a)|)))" "`

or

`p = (nRT)/V`

The number of moles `n` is given by

`n = m/M`

So,

`p = (mRT)/(MV)`

Calculate `p_text(O₂)`

`m color(white)(l)= "7.67 g"`
`Rcolor(white)(l) = "0.082 06 L·atm·K"^"-1""mol"^"-1"`
`Tcolor(white)(l) = "65 °C" = "338.15 K"`
`M = "32.00 g·mol"^"-1""`
`V color(white)(l)= "9.77 L"`

∴ `p = (7.67 color(red)(cancel(color(black)("g"))) × "0.082 06" color(red)(cancel(color(black)("L")))·"atm"color(red)(cancel(color(black)("·K"^"-1""mol"^"-1"))) × 338.15 color(red)(cancel(color(black)("K"))))/(32.00 color(red)(cancel(color(black)("g·mol"^"-1"))) × 9.77 color(red)(cancel(color(black)("L")))) = "0.681 atm"`

Calculate `p_text(Ne)`

`m color(white)(l)= "2.84 g"`
`Rcolor(white)(l) = "0.082 06 L·atm·K"^"-1""mol"^"-1"`
`Tcolor(white)(l) = "65 °C" = "338.15 K"`
`M = "20.18 g·mol"^"-1""`
`V color(white)(l)= "9.77 L"`

∴ `p = (2.84 color(red)(cancel(color(black)("g"))) × "0.082 06" color(red)(cancel(color(black)("L")))·"atm"color(red)(cancel(color(black)("·K"^"-1""mol"^"-1"))) × 338.15 color(red)(cancel(color(black)("K"))))/(20.18 color(red)(cancel(color(black)("g·mol"^"-1"))) × 9.77 color(red)(cancel(color(black)("L")))) = "0.3997 atm"`

Calculate the total pressure

`p_text(tot) = p_text(O₂) + p_text(Ne) = "(0.681 + 0.3997) atm" = "1.081 atm"`