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

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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?

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Answer 1 · 2018-06-16T15:03:07

$6 atm$ $"6 atm"$ .

Explanation:

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

The ideal gas law, or $p V = n R T$ $pV = nRT$ .
An expression of Dalton's law of partial pressures, or
$P_{a} = x_{a} \times P_{total}$ $P_a = x_a xx P_"total"$ ,
where $a$ $a$ is a gas and $x_{a}$ $x_a$ 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 $N_{2}$ $N_2$ gas is at $300 K$ $"300 K"$ , $2.3 atm$ $"2.3 atm"$ , and stored in a $1.00 L$ $"1.00 L"$ container.
So, we'll just plug these values into the equation, using $0.08206 L atm/K mol$ $"0.08206 L atm/K mol"$ as our gas constant because it fits our units the best:

$p V = n R T$ $pV = nRT$

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

$n = 0.09 mol$ $n = "0.09 mol"$

Now, we'll start gathering the values needed for the second equation to solve for the partial pressure of $O_{2}$ $O_2$ .
We need to find the mole fraction of $O_{2}$ $O_2$ first—we know that $0.30 mol$ $"0.30 mol"$ of $O_{2}$ $O_2$ was added into the container and then $0.040 mol$ $"0.040 mol"$ of gas molecules leave after that.

Let's assume that an equal amount of $N_{2}$ $N_2$ and $O_{2}$ $O_2$ left; that, out of the $0.040 mol$ $"0.040 mol"$ , $0.020 mol$ $"0.020 mol"$ of $N_{2}$ $N_2$ left and $0.020 mol$ $"0.020 mol"$ of $O_{2}$ $O_2$ left.

This would leave us with $0.09 mol - 0.020 mol = 0.07 mol$ $"0.09 mol" - "0.020 mol" = "0.07 mol"$ of $N_{2}$ $N_2$ gas and $0.30 mol - 0.040 mol = 0.26 mol$ $"0.30 mol" - "0.040 mol" = "0.26 mol"$ of $O_{2}$ $O_2$ gas.

So, the mole fraction of $O_{2}$ $O_2$ gas would be:

$x_{O_{2}} = \frac{O_{2}}{O_{2} + N_{2}} = \frac{0.26 mol}{0.26 mol + 0.07 mol} = 0.8$ $x_(O_2) = (O_2)/(O_2 + N_2) = ("0.26 mol")/("0.26 mol" + "0.07 mol") = 0.8$

Then, we need to find the total pressure of both $N_{2}$ $N_2$ and $O_{2}$ $O_2$ .
Since the container is still $1.00 L$ $"1.00 L"$ , (assuming it's) still at $300 K$ $"300 K"$ , and now has $0.26 mol + 0.07 mol = 0.3 mol$ $"0.26 mol" + "0.07 mol" = "0.3 mol"$ of gas in it, we can solve for pressure using the ideal gas law:

$p V = n R T$ $pV = nRT$

$p = \frac{n R T}{v} = \frac{0.3 mol \times 0.08206 L atm/K mol \times 300 K}{1.00 L}$ $p = (nRT)/v = ("0.3 mol" xx "0.08206 L atm/K mol" xx "300 K")/"1.00 L"$

$p = 7 atm$ $p = "7 atm"$

Now, we can use an expression of Dalton's law of partial pressures to solve for the partial pressure of $O_{2}$ $O_2$ :

$P_{a} = x_{a} \times P_{total}$ $P_a = x_a xx P_"total"$

$P_{O_{2}} = x_{O_{2}} \times P_{total}$ $P_(O_2) = x_(O_2) xx P_"total"$

$P_{a} = 0.8 \times 7 atm = 6 atm$ $P_a = 0.8 xx "7 atm" = "6 atm"$

Answer 2 · 2018-06-16T23:17:34

$p_text(O₂) = "6.6 atm"$

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,

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

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

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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?

2 Answers

Explanation:

Explanation:

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Impact of this question

Science

Math

Humanities

... and beyond

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?

2 Answers

Explanation:

Explanation:

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Impact of this question

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?