# A 1.00 L flask contains nitrogen gas. what is the final partial pressure?

## A 1.00 L container contains nitrogen gas at 300k and a pressure of 2.3 atm. 0.30 mol of O2(g) is added to the flask and allowed to mix. Then a lid is opened to allow 0.040 moles of molecules to leave. What is the partial pressure of oxygen in the final mixture?

Jun 16, 2018

$\text{6 atm}$.

#### Explanation:

To answer this question, we'll need to use two formulas:

• The ideal gas law, or $p V = n R T$.
• An expression of Dalton's law of partial pressures, or
${P}_{a} = {x}_{a} \times {P}_{\text{total}}$,
where $a$ is a gas and ${x}_{a}$ is the mole fraction of that gas.

First, let's use the ideal gas law to find the number of moles of nitrogen in the container! :)

The question tells us that the ${N}_{2}$ gas is at $\text{300 K}$, $\text{2.3 atm}$, and stored in a $\text{1.00 L}$ container.
So, we'll just plug these values into the equation, using $\text{0.08206 L atm/K mol}$ as our gas constant because it fits our units the best:

$p V = n R T$

$n = \frac{p V}{R T} = \left(\text{2.3 atm" xx "1.00 L")/("0.08206 L atm/K mol" xx "300 K}\right)$

$n = \text{0.09 mol}$

Now, we'll start gathering the values needed for the second equation to solve for the partial pressure of ${O}_{2}$.
We need to find the mole fraction of ${O}_{2}$ first—we know that $\text{0.30 mol}$ of ${O}_{2}$ was added into the container and then $\text{0.040 mol}$ of gas molecules leave after that.

Let's assume that an equal amount of ${N}_{2}$ and ${O}_{2}$ left; that, out of the $\text{0.040 mol}$, $\text{0.020 mol}$ of ${N}_{2}$ left and $\text{0.020 mol}$ of ${O}_{2}$ left.

This would leave us with $\text{0.09 mol" - "0.020 mol" = "0.07 mol}$ of ${N}_{2}$ gas and $\text{0.30 mol" - "0.040 mol" = "0.26 mol}$ of ${O}_{2}$ gas.

So, the mole fraction of ${O}_{2}$ gas would be:

${x}_{{O}_{2}} = \frac{{O}_{2}}{{O}_{2} + {N}_{2}} = \left(\text{0.26 mol")/("0.26 mol" + "0.07 mol}\right) = 0.8$

Then, we need to find the total pressure of both ${N}_{2}$ and ${O}_{2}$.
Since the container is still $\text{1.00 L}$, (assuming it's) still at $\text{300 K}$, and now has $\text{0.26 mol" + "0.07 mol" = "0.3 mol}$ of gas in it, we can solve for pressure using the ideal gas law:

$p V = n R T$

p = (nRT)/v = ("0.3 mol" xx "0.08206 L atm/K mol" xx "300 K")/"1.00 L"

$p = \text{7 atm}$

Now, we can use an expression of Dalton's law of partial pressures to solve for the partial pressure of ${O}_{2}$:

${P}_{a} = {x}_{a} \times {P}_{\text{total}}$

${P}_{{O}_{2}} = {x}_{{O}_{2}} \times {P}_{\text{total}}$

${P}_{a} = 0.8 \times \text{7 atm" = "6 atm}$

Jun 16, 2018

p_text(O₂) = "6.6 atm"

#### Explanation:

Step 1. Calculate the moles of nitrogen

To answer this question, we'll need to use the Ideal Gas Law.

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

We can rearrange this equation to get

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

In this problem,

$p \textcolor{w h i t e}{l} = \text{2.3 atm}$
$V = \text{1.00 L}$
$R = \text{0.082 06 L·atm·K"^"-1""mol"^"-1}$
$T = \text{300 K}$

n = (2.3 color(red)(cancel(color(black)("atm"))) × 1.00 color(red)(cancel(color(black)("L"))))/("0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 300 color(red)(cancel(color(black)("K")))) = "0.0934 mol"

Step 2. Calculate the final moles of oxygen in the container

$\text{0.0934 mol N"_2 + "0.30 mol O"_2 = "0.3934 mol gas}$

After allowing 0.040 mol of gas to leave, we have

$\text{(0.3934 - 0.040) mol gas = 0.3534 mol gas}$

${\text{ Moles of O"_2 = 0.3534 color(red)(cancel(color(black)("mol gas"))) × "0.30 mol O"_2/(0.3934 color(red)(cancel(color(black)("mol gas")))) = "0.270 mol O}}_{2}$

Step 3. Calculate the partial pressure of the oxygen

We can again use the Ideal Gas Law.

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

This time,

$n = \text{0.270 mol}$
$R = \text{0.082 06 L·atm·K"^"-1""mol"^"-1}$
$T = \text{300 K}$
$V = \text{1.00 L}$

 p = (0.270 color(red)(cancel(color(black)("mol"))) × "0.082 06" color(red)(cancel(color(black)("L")))·"atm"·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 300 color(red)(cancel(color(black)("K"))))/(1.00 color(red)(cancel(color(black)("L")))) = "6.6 atm"