For example 1 above when we calculated for H2's Pressure, why did we use 300L as Volume? Isn't that the volume of "both" gases? Why didn't we use the volume that is due to H2 alone?

One of the assumptions of ideal gases is that they don't take up any space. In addition, (at equilibrium) all gases (real or ideal) are spread out and mixed together throughout the entire volume. Even in real gasses under normal conditions (anything similar to STP) most of the volume is empty space so this is a reasonable approximation.

Oxygen and helium are taken in equal weights in a vessel . The pressure exerted by helium in the mixture is

Since oxygen is diatomic, one molecule of oxygen would weigh 32 amu, or eight times the mass of an atom of helium. If you have equal amounts, by mass, of these two elements, then you would have eight times as many helium particles as oxygen particles. Therefore, the pressure exerted by the helium would be eight times that exerted by the oxygen.

In the first question, I tried solving for each of the gases' partial pressure using Boyle's law P_1*V_1 = P_2*V_2 For Nitrogen : P2 = P_N2 = P1*V1/V2 = 2*24/10 = 4.80atm For Oxygen: P2 = P_O2 = P1*V1/V2 = 2*12/10 = 2.40atm P_total = 4.80+2.40 = 7.20atm which is pretty close to the 7.19atm calculated here. Can anyone explain what is happening lol EDIT: Is it because the temperature is not constant but changes a bit with volume, thus causing the error in my calculation?

The minor difference is just a rounding error in the article (probably a result of the multiple steps used) - nothing to worry about. The temperature is constant at 273 K.

In question 2 why didn't the addition of helium gas not affect the partial pressure of radon? please explain further.

The pressures are independent of each other. In other words, if the pressure from radon is X then after adding helium the pressure from radon will still be X even though the _total_ pressure is now higher than X. The total pressure is the sum of the individual partial pressures.

In step 1, we were asked to find the total pressure. The pressure of nitrogen is 2atm and that of oxygen is 2atm. Why could we not say that total pressure is equal to pressure of the individual gases(nitrogen and oxygen). Ptotal= Pn + Po. Ptotal= 2+2 which is equivalent to 4atm. I understand the fact that we`ve the pressure of nitrogen as P1, P1 is equivalent to n1RT/V and P total is equivalent to P1 + P2. P2= n2RT/V. So we find P1 and P2 and add them together using n1RT/V + n2RT/V will give the total pressure.

Using Dalton’s law as you’re suggesting only works if all the gases are already mixed together in the same container where the volume and temperature are the same. For the nitrogen and oxygen gas example, they begin in two separate containers with different volumes. And finally we mix them together in a third container with a different volume from the previous two containers. The 2atm pressures represent the pressures of the gases in their original containers before mixing. Their sum, 4atm, isn’t automatically the pressure in the third container when they are mixed because of the different volumes. Using the idea gas to create new equations to substitute for the partial pressures in Dalton’s law like you’ve done is how they solved the problem in method 1.

idk if this is a partial pressure question but a sample of oxygen of mass 30.0 g is confined in a vessel at 8°C and 3000. torr. Then 2.00 g of hydrogen is pumped into the vessel at constant temperature. What will be the final pressure in the vessel? (no reaction just mixing) how would you approach this question?

I initially solved the problem this way: You know the final total pressure is going to be the partial pressure from the O2 plus the partial pressure from the H2. And you know the partial pressure oxygen will still be 3000 torr when you pump in the hydrogen, but you still need to find the partial pressure of the H2. You can find the volume of the container using PV=nRT, just use the numbers for oxygen gas alone (convert 30.0g to moles of O2 first). Once you know the volume, you can solve to find the pressure that hydrogen gas would have in the container (again, finding n by converting from 2g to moles of H2 using the molar mass). Then the total pressure is just the sum of the two partial pressures. But then I realized a quicker solution-you actually don't need to use partial pressure at all First, calculate the number of moles you have of each gas, and then add them to find the total number of particles in moles. Then, since volume and temperature are constant, just use the fact that number of moles is proportional to pressure. Set up a proportion with (original pressure)/(original moles of O2) = (final pressure) / (total number of moles)

Is there a way to calculate the partial pressures of different reactants and products in a reaction when you only have the total pressure of the all gases and the number of moles of each gas but no volume?

Yes. As has been mentioned in the lesson, partial pressure can be calculated as follows: P(gas 1) = x(gas 1) * P(Total); where x(gas 1) = no of moles(gas 1)/ no of moles(total). As you can see the above formulae does not require the individual volumes of the gases or the total volume. Of course, such calculations can be done for ideal gases only.

Main content

Dalton's law of partial pressure

Google Classroom

Definition of partial pressure and using Dalton's law of partial pressures

Key points

The pressure exerted by an individual gas in a mixture is known as its partial pressure.
Assuming we have a mixture of ideal gases, we can use the ideal gas law to solve problems involving gases in a mixture.
Dalton's law of partial pressures states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases:

P_{Total} = P_{gas 1} + P_{gas 2} + P_{gas 3} …

Dalton's law can also be expressed using the mole fraction of a gas, $x$ ‍:

P_{gas 1} = x_{1} P_{Total}

Introduction

In day-to-day life, we measure gas pressure when we use a barometer to check the atmospheric pressure outside or a tire gauge to measure the pressure in a bike tube. When we do this, we are measuring a macroscopic physical property of a large number of gas molecules that are invisible to the naked eye. On the molecular level, the pressure we are measuring comes from the force of individual gas molecules colliding with other objects, such as the walls of their container.

Let's take a closer look at pressure from a molecular perspective and learn how Dalton's Law helps us calculate total and partial pressures for mixtures of gases.

Ideal gases and partial pressure

In this article, we will be assuming the gases in our mixtures can be approximated as ideal gases. This assumption is generally reasonable as long as the temperature of the gas is not super low (close to

0 K

), and the pressure is around

1 atm

This means we are making some assumptions about our gas molecules:

We assume that the gas molecules take up no volume.
We assume that the molecules have no intermolecular attractions, which means they act independently of other gas molecules.

Based on these assumptions, we can calculate the contribution of different gases in a mixture to the total pressure. We refer to the pressure exerted by a specific gas in a mixture as its partial pressure. The partial pressure of a gas can be calculated using the ideal gas law, which we will cover in the next section, as well as using Dalton's law of partial pressures.

Example 1: Calculating the partial pressure of a gas

Let's say we have a mixture of hydrogen gas,

H_{2} (g)

, and oxygen gas,

O_{2} (g)

. The mixture contains

6.7 mol

hydrogen gas and

3.3 mol

oxygen gas. The mixture is in a

300 L

container at

273 K

, and the total pressure of the gas mixture is

0.75 atm

The contribution of hydrogen gas to the total pressure is its partial pressure. Since the gas molecules in an ideal gas behave independently of other gases in the mixture, the partial pressure of hydrogen is the same pressure as if there were no other gases in the container. Therefore, if we want to know the partial pressure of hydrogen gas in the mixture,

P_{H_{2}}

, we can completely ignore the oxygen gas and use the ideal gas law:

P_{H_{2}} V = n_{H_{2}} RT

Rearranging the ideal gas equation to solve for

P_{H_{2}}

, we get:

\begin{aligned} P_{H_{2}} & = \frac{n_{H_{2}} RT}{V} \\ = \frac{(6.7 mol) (0.08206 \frac{atm \cdot L}{mol \cdot K}) (273 K)}{300 L} = 0.50 atm \end{aligned}

Thus, the ideal gas law tells us that the partial pressure of hydrogen in the mixture is

0.50 atm

. We can also calculate the partial pressure of hydrogen in this problem using Dalton's law of partial pressures, which will be discussed in the next section.

Dalton's law of partial pressures

Dalton's law of partial pressures states that the total pressure of a mixture of gases is the sum of the partial pressures of its components:

P_{Total} = P_{gas 1} + P_{gas 2} + P_{gas 3} …

where the partial pressure of each gas is the pressure that the gas would exert if it was the only gas in the container. That is because we assume there are no attractive forces between the gases.

Dalton's law of partial pressure can also be expressed in terms of the mole fraction of a gas in the mixture. The mole fraction of a gas is the number of moles of that gas divided by the total moles of gas in the mixture, and it is often abbreviated as

x

x_{1} = mole fraction of gas 1 = \frac{moles of gas 1}{total moles of gas}

Dalton's law can be rearranged to give the partial pressure of gas 1 in a mixture in terms of the mole fraction of gas 1:

P_{gas 1} = x_{1} P_{Total}

Both forms of Dalton's law are extremely useful in solving different kinds of problems including:

Calculating the partial pressure of a gas when you know the mole ratio and total pressure
Calculating moles of an individual gas if you know the partial pressure and total pressure
Calculating the total pressure if you know the partial pressures of the components

Example 2: Calculating partial pressures and total pressure

Let's say that we have one container with

24.0 L

of nitrogen gas at

2.00 atm

, and another container with

12.0 L

of oxygen gas at

2.00 atm

. The temperature of both gases is

273 K

If both gases are mixed in a $10.0 L$ ‍ container, what are the partial pressures of nitrogen and oxygen in the resulting mixture? What is the total pressure?

Step 1: Calculate moles of oxygen and nitrogen gas

Since we know

P

V

,and

T

for each of the gases before they're combined, we can find the number of moles of nitrogen gas and oxygen gas using the ideal gas law:

n = \frac{PV}{RT}

Solving for nitrogen and oxygen, we get:

n_{N_{2}} = \frac{(2 atm) (24.0 L)}{(0.08206 \frac{atm \cdot L}{mol \cdot K}) (273 K)} = 2.14 mol nitrogen

n_{O_{2}} = \frac{(2 atm) (12.0 L)}{(0.08206 \frac{atm \cdot L}{mol \cdot K}) (273 K)} = 1.07 mol oxygen

Step 2 (method 1): Calculate partial pressures and use Dalton's law to get $P_{Total}$ ‍

Once we know the number of moles for each gas in our mixture, we can now use the ideal gas law to find the partial pressure of each component in the

10.0 L

container:

P = \frac{nRT}{V}

P_{N_{2}} = \frac{(2.14 mol) (0.08206 \frac{atm \cdot L}{mol \cdot K}) (273 K)}{10 L} = 4.79 atm

P_{O_{2}} = \frac{(1.07 mol) (0.08206 \frac{atm \cdot L}{mol \cdot K}) (273 K)}{10 L} = 2.40 atm

Notice that the partial pressure for each of the gases increased compared to the pressure of the gas in the original container. This makes sense since the volume of both gases decreased, and pressure is inversely proportional to volume.

We can now get the total pressure of the mixture by adding the partial pressures together using Dalton's Law:

\begin{aligned} P_{Total} & = P_{N_{2}} + P_{O_{2}} \\ = 4.79 atm + 2.40 atm = 7.19 atm \end{aligned}

Step 2 (method 2): Use ideal gas law to calculate $P_{Total}$ ‍ without partial pressures

Since the pressure of an ideal gas mixture only depends on the number of gas molecules in the container (and not the identity of the gas molecules), we can use the total moles of gas to calculate the total pressure using the ideal gas law:

\begin{aligned} P_{Total} & = \frac{(n_{N_{2}} + n_{O_{2}}) RT}{V} \\ = \frac{(2.14 mol + 1.07 mol) (0.08206 \frac{atm \cdot L}{mol \cdot K}) (273 K)}{10 L} \\ = \frac{(3.21 mol) (0.08206 \frac{atm \cdot L}{mol \cdot K}) (273 K)}{10 L} \\ = 7.19 atm \end{aligned}

Once we know the total pressure, we can use the mole fraction version of Dalton's law to calculate the partial pressures:

P_{N_{2}} = x_{N_{2}} P_{Total} = (\frac{2.14 mol}{3.21 mol}) (7.19 atm) = 4.79 atm

P_{O_{2}} = x_{O_{2}} P_{Total} = (\frac{1.07 mol}{3.21 mol}) (7.19 atm) = 2.40 atm

Luckily, both methods give the same answers!

You might be wondering when you might want to use each method. It mostly depends on which one you prefer, and partly on what you are solving for. For instance, if all you need to know is the total pressure, it might be better to use the second method to save a couple calculation steps.

Summary

The pressure exerted by an individual gas in a mixture is known as its partial pressure.
Assuming we have a mixture of ideal gases, we can use the ideal gas law to solve problems involving gases in a mixture.
Dalton's law of partial pressures states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases:

P_{Total} = P_{gas 1} + P_{gas 2} + P_{gas 3} …

Dalton's law can also be expressed using the mole fraction of a gas, $x$ ‍:

P_{gas 1} = x_{1} P_{Total}

Try it: Evaporation in a closed system

Part 1

A closed system with a volume of

2.0 L

contains radon gas and liquid water, and the container is allowed to equilibrate at

27^{\circ} C

until the total pressure is constant.

What is the partial pressure of radon if the total pressure is $780 torr$ ‍ and the water vapor partial pressure is $1.0 atm$ ‍?

atm

Part 2

Some helium gas is added to the system, and the total pressure increases to

1.20 atm

What is the new partial pressure of radon?

atm

The partial pressure of radon is independent of other gases in the container, so adding another gas to the mixture doesn't change its partial pressure (although it does increase the total pressure). The partial pressure of radon depends only on the number of moles of radon, the volume of the container, and the temperature, none of which are affected by adding the helium gas.

Therefore, the partial pressure of the radon gas is still

0.03 atm

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Isabel Zimmerman
Posted 8 years ago. Direct link to Isabel Zimmerman's post “In the very first example...”
In the very first example, where they are solving for the pressure of H2, why does the equation say 273L, not 273K?
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(20 votes)
Answer
AJ
Posted 7 years ago. Direct link to AJ's post “For example 1 above when ...”
For example 1 above when we calculated for H2's Pressure, why did we use 300L as Volume? Isn't that the volume of "both" gases? Why didn't we use the volume that is due to H2 alone?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- tyersome
  Posted 7 years ago. Direct link to tyersome's post “One of the assumptions of...”
  One of the assumptions of ideal gases is that they don't take up any space. In addition, (at equilibrium) all gases (real or ideal) are spread out and mixed together throughout the entire volume.
  
  Even in real gasses under normal conditions (anything similar to STP) most of the volume is empty space so this is a reasonable approximation.
  Button navigates to signup page
  (8 votes)
Rebecca C
Posted 7 years ago. Direct link to Rebecca C's post “Under the heading "Ideal ...”
Under the heading "Ideal gases and partial pressure," it says the temperature should be close to 0 K at STP. Shouldn't it really be 273 K?
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(3 votes)
Answer
- Deb Argha Saha
  Posted 6 years ago. Direct link to Deb Argha Saha's post “"This assumption is gener...”
  "This assumption is generally reasonable as long as the temperature of the gas is not super low (close to 0 K), and the pressure is around 1 atm."
  The sentence means not super low that is not close to 0 K.
  Button navigates to signup page
  (4 votes)
kaps1612003
Posted 8 years ago. Direct link to kaps1612003's post “Oxygen and helium are tak...”
Oxygen and helium are taken in equal weights in a vessel . The pressure exerted by helium in the mixture is
Button navigates to signup pageComment on kaps1612003's post “Oxygen and helium are tak...”
(3 votes)
Answer
- Esther Dickey
  Posted 8 years ago. Direct link to Esther Dickey's post “Since oxygen is diatomic,...”
  Since oxygen is diatomic, one molecule of oxygen would weigh 32 amu, or eight times the mass of an atom of helium. If you have equal amounts, by mass, of these two elements, then you would have eight times as many helium particles as oxygen particles. Therefore, the pressure exerted by the helium would be eight times that exerted by the oxygen.
  Comment on Esther Dickey's post “Since oxygen is diatomic,...”
  (3 votes)
😊
Posted 7 years ago. Direct link to 😊's post “In question 2 why didn't ...”
In question 2 why didn't the addition of helium gas not affect the partial pressure of radon? please explain further.
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(2 votes)
Answer
- RogerP
  Posted 7 years ago. Direct link to RogerP's post “The pressures are indepen...”
  The pressures are independent of each other. In other words, if the pressure from radon is X then after adding helium the pressure from radon will still be X even though the total pressure is now higher than X. The total pressure is the sum of the individual partial pressures.
  Button navigates to signup page
  (5 votes)
Abhay Bhardwaj
Posted 6 years ago. Direct link to Abhay Bhardwaj's post “In the first question, I ...”
In the first question, I tried solving for each of the gases' partial pressure using Boyle's law
P_1*V_1 = P_2*V_2

For Nitrogen :
P2 = P_N2 = P1*V1/V2 = 2*24/10 = 4.80atm
For Oxygen:
P2 = P_O2 = P1*V1/V2 = 2*12/10 = 2.40atm

P_total = 4.80+2.40 = 7.20atm which is pretty close to the 7.19atm calculated here.

Can anyone explain what is happening lol
EDIT: Is it because the temperature is not constant but changes a bit with volume, thus causing the error in my calculation?
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(3 votes)
Answer
- RogerP
  Posted 6 years ago. Direct link to RogerP's post “The minor difference is j...”
  The minor difference is just a rounding error in the article (probably a result of the multiple steps used) - nothing to worry about. The temperature is constant at 273 K.
  Button navigates to signup page
  (3 votes)
Tessie
Posted a year ago. Direct link to Tessie's post “In step 1, we were asked ...”
In step 1, we were asked to find the total pressure. The pressure of nitrogen is 2atm and that of oxygen is 2atm. Why could we not say that total pressure is equal to pressure of the individual gases(nitrogen and oxygen). Ptotal= Pn + Po. Ptotal= 2+2 which is equivalent to 4atm. I understand the fact that we`ve the pressure of nitrogen as P1, P1 is equivalent to n1RT/V and P total is equivalent to P1 + P2. P2= n2RT/V. So we find P1 and P2 and add them together using n1RT/V + n2RT/V will give the total pressure.
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Richard
  Posted a year ago. Direct link to Richard's post “Using Dalton’s law as you...”
  Using Dalton’s law as you’re suggesting only works if all the gases are already mixed together in the same container where the volume and temperature are the same. For the nitrogen and oxygen gas example, they begin in two separate containers with different volumes. And finally we mix them together in a third container with a different volume from the previous two containers.
  
  The 2atm pressures represent the pressures of the gases in their original containers before mixing. Their sum, 4atm, isn’t automatically the pressure in the third container when they are mixed because of the different volumes.
  
  Using the idea gas to create new equations to substitute for the partial pressures in Dalton’s law like you’ve done is how they solved the problem in method 1.
  Button navigates to signup page
  (3 votes)
konekocommander
Posted 7 years ago. Direct link to konekocommander's post “idk if this is a partial ...”
idk if this is a partial pressure question but a sample of oxygen of mass 30.0 g is confined in a vessel at 8°C and 3000. torr. Then 2.00 g of hydrogen is pumped into the vessel at constant temperature. What will be the final pressure in the vessel? (no reaction just mixing) how would you approach this question?
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(1 vote)
Answer
- oliver krefta
  Posted 7 years ago. Direct link to oliver krefta's post “I initially solved the pr...”
  I initially solved the problem this way:
  You know the final total pressure is going to be the partial pressure from the O2 plus the partial pressure from the H2. And you know the partial pressure oxygen will still be 3000 torr when you pump in the hydrogen, but you still need to find the partial pressure of the H2.
  You can find the volume of the container using PV=nRT, just use the numbers for oxygen gas alone (convert 30.0g to moles of O2 first). Once you know the volume, you can solve to find the pressure that hydrogen gas would have in the container (again, finding n by converting from 2g to moles of H2 using the molar mass). Then the total pressure is just the sum of the two partial pressures.
  
  But then I realized a quicker solution-you actually don't need to use partial pressure at all
  First, calculate the number of moles you have of each gas, and then add them to find the total number of particles in moles. Then, since volume and temperature are constant, just use the fact that number of moles is proportional to pressure. Set up a proportion with (original pressure)/(original moles of O2) = (final pressure) / (total number of moles)
  Comment on oliver krefta's post “I initially solved the pr...”
  (3 votes)
BlackBird
Posted 8 years ago. Direct link to BlackBird's post “How did he get 3.21 moles...”
How did he get 3.21 moles in the mole fraction section.
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(1 vote)
Answer
ethanthompsonroland
Posted 7 years ago. Direct link to ethanthompsonroland's post “Is there a way to calcula...”
Is there a way to calculate the partial pressures of different reactants and products in a reaction when you only have the total pressure of the all gases and the number of moles of each gas but no volume?
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(1 vote)
Answer
- Deb Argha Saha
  Posted 6 years ago. Direct link to Deb Argha Saha's post “Yes. As has been mentione...”
  Yes. As has been mentioned in the lesson, partial pressure can be calculated as follows:
  P(gas 1) = x(gas 1) * P(Total); where x(gas 1) = no of moles(gas 1)/ no of moles(total).
  As you can see the above formulae does not require the individual volumes of the gases or the total volume. Of course, such calculations can be done for ideal gases only.
  Button navigates to signup page
  (3 votes)

Course: Class 11 Physics (India) > Unit 17

Dalton's law of partial pressure

Key points

Introduction

Ideal gases and partial pressure

Example 1: Calculating the partial pressure of a gas

Dalton's law of partial pressures

Example 2: Calculating partial pressures and total pressure

Step 1: Calculate moles of oxygen and nitrogen gas

Step 2 (method 1): Calculate partial pressures and use Dalton's law to get $P_{Total}$ ‍

Step 2 (method 2): Use ideal gas law to calculate $P_{Total}$ ‍ without partial pressures

Summary

Attributions

Try it: Evaporation in a closed system

Part 1

Part 2

Want to join the conversation?

Class 11 Physics (India)

Course: Class 11 Physics (India) > Unit 17

Key points

Introduction

Ideal gases and partial pressure

Example 1: Calculating the partial pressure of a gas

Dalton's law of partial pressures

Example 2: Calculating partial pressures and total pressure

Step 1: Calculate moles of oxygen and nitrogen gas

Step 2 (method 1): Calculate partial pressures and use Dalton's law to get PTotal‍

Step 2 (method 2): Use ideal gas law to calculate PTotal‍ without partial pressures

Summary

Try it: Evaporation in a closed system

Part 1

Part 2

Want to join the conversation?

Step 2 (method 1): Calculate partial pressures and use Dalton's law to get $P_{Total}$ ‍

Step 2 (method 2): Use ideal gas law to calculate $P_{Total}$ ‍ without partial pressures