Question

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?

Answer 1

`"6 atm"`.

Explanation:

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

• The ideal gas law, or `pV = nRT`.
• An expression of Dalton's law of partial pressures, or
  `P_a = x_a xx P_"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 `"300 K"`, `"2.3 atm"`, and stored in a `"1.00 L"` container.
So, we'll just plug these values into the equation, using `"0.08206 L atm/K mol"` as our gas constant because it fits our units the best:

`pV = nRT`

`n = (pV)/(RT) = ("2.3 atm" xx "1.00 L")/("0.08206 L atm/K mol" xx "300 K")`

`n = "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 `"0.30 mol"` of `O_2` was added into the container and then `"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 `"0.040 mol"`, `"0.020 mol"` of `N_2` left and `"0.020 mol"` of `O_2` left.

This would leave us with `"0.09 mol" - "0.020 mol" = "0.07 mol"` of `N_2` gas and `"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) = (O_2)/(O_2 + N_2) = ("0.26 mol")/("0.26 mol" + "0.07 mol") = 0.8`

Then, we need to find the total pressure of both `N_2` and `O_2`.
Since the container is still `"1.00 L"`, (assuming it's) still at `"300 K"`, and now has `"0.26 mol" + "0.07 mol" = "0.3 mol"` of gas in it, we can solve for pressure using the ideal gas law:

`pV = nRT`

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

`p = "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 xx P_"total"`

`P_(O_2) = x_(O_2) xx P_"total"`

`P_a = 0.8 xx "7 atm" = "6 atm"`

Answer 2

`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.

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

We can rearrange this equation to get

`n = (pV)/(RT)`

In this problem,

`pcolor(white)(l) = "2.3 atm"`
`V = "1.00 L"`
`R = "0.082 06 L·atm·K"^"-1""mol"^"-1"`
`T = "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

After adding oxygen, we have

`"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

`"(0.3934 - 0.040) mol gas = 0.3534 mol gas"`

`" 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 = (nRT)/V`

This time,

`n = "0.270 mol"`
`R = "0.082 06 L·atm·K"^"-1""mol"^"-1"`
`T = "300 K"`
`V = "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"`