# 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?

Nov 15, 2017

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.

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} p V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

or

$p = \frac{n R T}{V}$

The number of moles $n$ is given by

$n = \frac{m}{M}$

So,

$p = \frac{m R T}{M V}$

Calculate p_text(O₂)

$m \textcolor{w h i t e}{l} = \text{7.67 g}$
$R \textcolor{w h i t e}{l} = \text{0.082 06 L·atm·K"^"-1""mol"^"-1}$
$T \textcolor{w h i t e}{l} = \text{65 °C" = "338.15 K}$
$M = \text{32.00 g·mol"^"-1}$
$V \textcolor{w h i t e}{l} = \text{9.77 L}$

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 ${p}_{\textrm{N e}}$

$m \textcolor{w h i t e}{l} = \text{2.84 g}$
$R \textcolor{w h i t e}{l} = \text{0.082 06 L·atm·K"^"-1""mol"^"-1}$
$T \textcolor{w h i t e}{l} = \text{65 °C" = "338.15 K}$
$M = \text{20.18 g·mol"^"-1}$
$V \textcolor{w h i t e}{l} = \text{9.77 L}$

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"