To understand the partial pressures, let’s suppose a certain amount of **gas “A”** is added to the tank and extorts **5 atm** pressure, and in another tank with **gas “B”** the pressure is **3 atm**:

Now, it turns out that if we mix these amounts of gas A and B together, the total pressure is 8 atm. And this intuitive observation was first formulated by John Dalton and is known as the **Dalton’s law of partial pressures**.

It states that the total pressure of a gas mixture is equal to the sum of the pressures that each gas would exert if it *were present alone:*

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* *The individual pressure of each gas component is called a **partial pressure**, so the total pressure is the sum of the partial pressures:

**P _{Total} = P_{A} + P_{B}**

This formula is true for any number of gas types in the, and in a more general way, we can write:

**P _{Total} = P_{1} + P_{2 }+ … + Pn**

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**For example**,

What is the partial pressure of nitrogen in a mixture of N_{2}, SO_{2} and CO_{2} that has a total pressure of 6.84 atm. The partial pressure of SO_{2} is 2.10 atm, and the partial pressure of CO_{2 }is 1.74 atm.

The total pressure is the sum of the partial pressures of all the gasses in the mixture:

P_{T} = P(N_{2}) + P(SO_{2}) + P(CO_{2}) = 6.84 atm

P(N_{2}) = P_{T }– [P(SO_{2}) + P(CO_{2})]

P(N_{2}) = 6.84 atm_{ }– [2.10 atm + 1.74 atm]

**P(N _{2}) = 3.00 atm**

**Another example**,

13.2 grams of CO_{2 }and 6.00 grams of He are mixed in a 4.00 L container at 300. K. Calculate the partial pressure of both gases and the total pressure of the mixture.

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First, we need to determine the moles of the gases using the ideal gas law equation:

\[{\rm{P}}\;{\rm{ = }}\;\frac{{{\rm{nRT}}}}{{\rm{V}}}\]

\[{\rm{P}}\;{\rm{(C}}{{\rm{O}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;\frac{{{\rm{nRT}}}}{{\rm{V}}}\]

\[{\rm{P}}\;{\rm{(C}}{{\rm{O}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.300}}\;\cancel{{{\rm{mol}}}}{\rm{ \times }}\;{\rm{0}}{\rm{.08206}}\;\cancel{{\rm{L}}}\;{\rm{atm}}\;\cancel{{{{\rm{K}}^{{\rm{ – 1}}}}}}\;{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}\;{\rm{300}}{\rm{.}}\;\cancel{{\rm{K}}}}}{{{\rm{4}}{\rm{.00}}\;\cancel{{\rm{L}}}}}\; = \;{\rm{1}}{\rm{.85}}\;{\rm{atm}}\]

\[{\rm{P}}\;{\rm{(He)}}\;{\rm{ = }}\;\frac{{{\rm{1}}{\rm{.50}}\;\cancel{{{\rm{mol}}}}{\rm{ \times }}\;{\rm{0}}{\rm{.08206}}\;\cancel{{\rm{L}}}\;{\rm{atm}}\;\cancel{{{{\rm{K}}^{{\rm{ – 1}}}}}}\;{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}\;{\rm{300}}{\rm{.}}\;\cancel{{\rm{K}}}}}{{{\rm{4}}{\rm{.00}}\;\cancel{{\rm{L}}}}}\; = \;{\rm{9}}{\rm{.23}}\;{\rm{atm}}\]

These are the partial pressure of CO_{2} and He, so to find the total pressure, we add them up:

**P _{total} = 1.85 + 9.23 = 11.1 atm**

**Check Also**

- Ideal-Gas Laws
- Combined Gas Law Equation
- How to Know Which Gas Law Equation to Use
- Molar Mass and Density of Gases
- Graham’s Law of Effusion and Diffusion
- Mole Fraction and Partial Pressure of the Gas
- Gases in Chemical Reactions
- Gases – Practice Problems

#### Practice

A 3.00-L bulb containing N_{2} at 1.80 atm pressure is connected to a 2.00-L bulb filled with H_{2} at 3.50 atm pressure. What is the final pressure of the system when the valve is opened? *P*N_{2} = 1.80 atm * P*H_{2} = 3.50 atm *P*total = ? atm

Thank so kindly for sharing such great knowledge in chemistry. Kudos

Glad it was helpful. Thanks.