Dalton’s Law of Partial Pressure

Dalton’s Law Core Concepts

In 1801, English chemist John Dalton made observations about steam and air, that is published in 1802 and eventually because Dalton’s law of partial pressure. In this tutorial, you will learn what partial pressure is, how to find the partial pressure of a gas, and how Dalton’s Law relates it to mole fraction.

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Multiple Gases

Consider a tank containing 1 mole of neon gas. Let the pressure be 1-atmosphere. Now, consider what would happen if you added another mole of neon gas.

From our article on pressure, What is Pressure, we know that pressure is caused by the particles of gas colliding against the container walls. Because doubling the amount of moles of gas would double the amount of gas particles, we know that adding this second mole of gas would double the pressure.

Now, consider what would happen if instead of adding another mole of neon gas, we added a mole of helium. The amount of particles in a mole is the same no matter the substance, so we know that the amount of particles we’re adding is the same as before.

Thus, we can conclude that by adding more of a different gas, the pressure of the tank will increase.

Partial Pressure

In the tank from before containing neon and helium, the pressure would be 2-atmospheres, because the amount total amount of moles is 2 moles, and each mole exerts one atmosphere.

From this conclusion, we can see that the amount of pressure exerted by each species of gas is not the full 2 atmospheres, but is instead a portion of the entire pressure.

Dalton’s Law of Partial Pressure

Dalton’s Law of Partial Pressure states that the sum of these portions add up to the entire pressure of the container — i.e., the sum of the pressures of the neon and helium from the tank before add up to the total pressure of the tank.

From this, we can see the equation arise: P_{tot}=P_1+P_2+P_3+\hdots+P_n. This equation is Dalton’s Law, and it can be used to determine the total pressure of a system.

Mole Fractions and Partial Pressure

Before, we were adding in equal portions of gas. What would happen if we added unequal portions instead?

Consider the same tank from before, except instead of 1 mole of neon gas, we add 3 moles originally. From Avogadro’s Law, we know that the pressure will be proportional to the amount of moles, so in this case the pressure of the neon gas will be 3 moles.

Next, we add 1 mole of helium gas. Now, the total amount of moles of gas isn’t doubled, but is instead increased by 33%. Because of this, there are only 33% more collisions with the container walls, and thus the pressure only increases by 1 atm.

From this, we can see that the overall pressure becomes 4 atm, and the total moles becomes 4 moles. We can take the mole fraction of the helium gas, \frac{N_{\text{He}}}{N_{tot}}, and see that it is equal to 0.25. This also is the fraction of pressure that is exerted by the helium gas.

Because of this, the partial pressure of a gas can be determined by its mole fraction through the equation P_1=X_1P_{tot}, where x_1 is the mole fraction of gas 1.

Dalton’s Law Example Problem

How to find the Partial Pressure

Here’s an example of how you can solve a problem about Dalton’s Law of Partial Pressure, and how to find the partial pressure of a gas.

A tank is filled with 32 g of oxygen gas, and 12 g of helium gas. The total pressure in the tank is 4 atm. What are the partial pressures of the oxygen and helium gas?

First, determine the amount of moles of both the oxygen and helium gas.

     \begin{align*} \text{O$_2$: }\frac{32\text{ g}}{32.00\text{ g/mol}}=1\text{ mol}\\ \text{He: }\frac{12\text{ g}}{\text{ g/mol}}=3\text{ mol} \end{align*}

Then, determine the mole fractions of both the oxygen and helium gas.

     \begin{align*} \text{O$_2$: }\frac{N_{\text{O$_2$}}}{N_{tot}}=\frac{1\text{ mol}}{4\text{ mol}}=0.25\\ \text{He: }\frac{N_{He}}{N_{tot}}=\frac{3\text{ mol}}{4\text{ mol}}=0.75 \end{align*}

Finally, use the mole fraction to determine the partial pressure.

    \begin{align*}\text{O$_2$: }x_{\text{O$_2$}}P_{tot}=0.25\cdot4\text{ atm}=\boxed{1\text{ atm}}\\\text{He: }x_{\text{He}}P_{tot}=0.75\cdot4\text{ atm}=\boxed{3\text{ atm}}\end{align*}

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