Dalton’s law of partial pressure is by far one of the most important laws in the field of thermodynamics. This empirical law was proposed by John Dalton in 1801 and published in 1802.

In this exclusive article, I am going to explain what is dalton’s law of partial pressure? What is the essence of this law in the field of thermodynamics!!!

Most importantly, I will be solving some of the numerical problems related to this partial pressure law formula. Therefore, I would suggest you stick with me till the end. Here we go!!!

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## What is Dalton’s Law of Partial Pressure?

Dalton’s law of partial pressure states that at constant volume and temperature, the total pressure of a mixture of gases is equal to the sum of the partial pressures of each individual gas in the mixture. A typical example to observe this partial pressure law would be the air that is nothing but a mixture of several different gases. See the above image for proper understanding.

Not to mention, like all the other Ideal Gas Laws, this law describes the behavior of an Ideal Gas. However, we can also apply this partial pressure of a gas law to all the real gases, of course, at normal temperatures and low pressure. In fact, with some mathematical tricks, we can easily derive this law using the ideal gas law.

## Dalton’s Law Formula

According to Dalton’s law definition, the ideal gas law formula can be represented in two different forms. Let us take a look at both equations one by one:

**In Terms of Pressure**

In terms of pressure, Dalton’s law is mathematically expressed as:

or,

where,

n = number of participating gases.

P_{total} = is the total pressure exerted by the mixture of gases.

where p_{1}, p_{2}, …, p_{n} represent the partial pressures of each component of the participating gas.

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**In Terms of Mole Fraction**

where,

X_{i} = mole fraction of participating gases.

n = number of moles.

P = Pressure of participating gases.

V = Volume of participating gases.

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## Dalton’s Law of Partial Pressure Example Problems

**Example Problem 1**

**A mixture of nitrogen and sodium gas exerts a total pressure of 3.0 atm on the walls of its container. If the partial pressure of nitrogen gas is 1 atm, what is the mole fraction of sodium gas in the mixture?**

**Ans:** As you can see, in the above question, we have to find the mole fraction of sodium gas in the mixture. Therefore, for this type of numerical, we should be using both of Dalton’s law formulas. Therefore,

Given data:

P_{total} = 3 atm, P_{nitrogen} = 1 atm, P_{sodium} = ?

Applying Dalton’s law formula, and putting the above values in the formula given below, we get,

P_{total} = P_{nitrogen} + P_{sodium}

On solving, P_{sodium} = 2 atm.

Now, in order to find the molar fraction of sodium gas, we have to use Dalton’s law formula in terms of mole fraction.

Therefore, by rewriting the formula in terms of sodium, we get,

X_{sodium} = (P_{sodium}/P_{total})

that gives,

X_{sodium} = 2/3

On solving, the mole fraction of sodium is 0.33

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**Example Problem 2**

**At a temperature of 272 k, 30 L of gas A is kept in one container at 2 atm, and 40 L of gas B is kept in another container at 3 atm. Calculate the total pressure inside the container and the partial pressures of gas A and gas B if both gases are mixed into an empty 20 L container.**

**Ans:** As you can see that in the above question, we have to calculate the total pressure as well as the partial pressure of gas A and B after mixing them into an empty container. Therefore, in order to calculate this, let us divide the question into two sections:

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**Before Mixing:**

Using Ideal Gas Equation, we can find the number of moles of gas A and gas B. And, later, we will use it to find the partial pressure of gas A and Gas B after mixing.

Given that,

Pressure of gas A (P) = 2 atm, Volume of gas A (V) = 30 L

Pressure of gas B (P) = 3atm, Volume of gas B (V) = 40 L

Universal gas constant = 0.08206 atm.L.mol^{-1}.K^{-1}

Temperature = 272 k

To find:

(n_{A}) = ?

(n_{B}) = ?

As we know PV = nRT, after rearranging:

n = PV/RT

To calculate the number of moles of gas A and gas B, putting the above values side by side, we get:

Number of moles of gas A (n_{A}) = 2atm*30L/0.08206 atm.L.mol^{-1}.K^{-1}*272k = **2.67 mol**.

Number of moles of gas B (n_{B}) = 3atm*40L/0.08206 atm.L.mol^{-1}.K^{-1}*272k = **5.35 mol**.

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**After Mixing:**

As of now, we know the number of moles of gas A and gas B. Now we will use these two values to calculate the partial pressure of gas A and gas B after mixing by using the same ideal gas law formula.

Given that,

Volume of Container = 20 L

Number of moles of gas A (n_{A}) = 2.67 mol

Number of moles of gas B (n_{B}) = 5.35 mol

Universal gas constant = 0.08206 atm.L.mol^{-1}.K^{-1}

Temperature = 272 k

To find:

Partial pressure of gas A (P_{A}) = ?

Partial pressure of gas B (P_{B}) = ?

As we know PV = nRT, after rearranging:

P = nRT/V

P_{A} = 2.67mol*0.08206atm.L.mol^{-1}.K^{-1}*272k/20L = **2.99 atm**.

P_{B} = 5.35mol*0.08206atm.L.mol^{-1}.K^{-1}*272k/20L = **5.99 atm**.

At last, we will use Dalton’s law of partial pressure formula to calculate the total pressure inside the container after the mixing of gas.

P_{total} = P_{A} + P_{B} = 2.99atm +5.99atm = **8.98atm**.

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