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

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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?

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

#### Answer:

#### 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

So, we'll just plug these values into the equation, using

Now, we'll start gathering the values needed for the second equation to solve for the partial pressure of

We need to find the mole fraction of

Let's assume that an equal amount of

This would leave us with

So, the mole fraction of

Then, we need to find the total pressure of both

Since the container is still

Now, we can use an expression of Dalton's law of partial pressures to solve for the partial pressure of

#### Answer:

#### 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,

∴

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

After adding oxygen, we have

After allowing 0.040 mol of gas to leave, we have

**Step 3. Calculate the partial pressure of the oxygen**

We can again use the Ideal Gas Law.

#p = (nRT)/V#

This time,