John Dalton, in his experiment and study of gases, observed that each gas in the mixture exerted its own pressure on the walls of their container as if no other gas was present in the container.

In 1801, his observation was put forward as Dalton’s law of Partial Pressure which states that:

In a mixture of gases that do not react chemically together, the total pressure exerted by the mixture is equal to the sum of the partial pressures of the individual gases present in the mixture.

This implies that if gases a, b, c ….n are in a vessel, the total pressure, P_{T}, exerted on the walls of the vessel can be determined as:

P_{T} = P_{A} + P_{B} + P_{C} = ………….Pn

Where P_{A}, P_{B}, P_{C} …………….Pn are the partial pressure of individual gases a,b,c, ……n respectively.

**Example**

1. 200cm^{3} of nitrogen gas at a pressure of 500mmHg and 100cm^{3} of carbon (iv) oxide at a pressure of 50mmHg were introduced into a 150cm^{3} vessel. What is the total pressure in the vessel?

**Solution**

N_{2} | C0_{2} |

Vol = 200cm^{3} | Vol = 100cm^{3} |

Pressure = 500mmHg | Pressure = 50mmHg |

Volume of the vessel = 150cm^{3}

\( \frac{Volume \; of \; the \; gas}{Volume \; of \; the \; vessel} \; \times \; \frac{Pressure \; of \; the \; gas}{1} \)

N_{2} | C0_{2} |

\( \frac{200}{150} \; \times \; \frac{500}{1} \) | \( \frac{100}{150} \; \times \; \frac{50}{1} \) |

666.67mmHg | 33.33mmHg |

According to Daltons law of Partial Pressure

P_{Total} = Pa + Pb + Pc ………..+ Pn

P_{Total} = PN_{2} + PCO_{2}

P_{Total} = 666.67mmHg + 33.33mmHg

P_{Total} = 700mmHg

2. 50cm^{3} of a gas x and 30cm^{3} of a gas y occupies a container with a capacity of 80cm^{3}. If the total pressure of the gas mixture is 760mmHg. Calculate the partial pressures of gases x and y respectively.

**Solution**

Volume of gas y = 50cm^{3}

Volume of gas x = 30cm^{3}

Total Volume of the container = 80cm^{3}

Total Pressure of the gas = 760mmHg

Partial Pressure of gas x = \( \frac{50}{80} = \frac{760}{1} \)

Partial Pressure of gas x = 475mmHg

Partial Pressure of gas y = \( \frac{30}{80} \; \times \; \frac{760}{1} \)

Partial Pressure of gas y = 285mmHg

P_{Total }= PX + PY

i.e 475 + 285 = 760mmHg.

3. Calculate the pressure of the dry gas collected over water at 6^{0}C and 765mmHg. Given that the vapour pressure of water at 6^{0}C is 7mmHg.

**Solution:**

At 6^{0}C, pressure of the dry gas

= Pressure of wet gas – vapour pressure of water

765 – 7 = 758mmHg

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