Name: Graham's Diffusion Law
Uploaded: 2019-07-19
Duration: 9 min 43 s
Description: Gases diffuse at a different rate. Watch this video to understand Graham's diffusion law, which deals with the diffusion of gases. Have fun learning!

Question

State and explain Dalton's law of partial pressure.

500 mL

of oxygen at a pressure of

700 mm

,

300 mL

of hydrogen at a pressure of

750 mm

and

700 mL

of Nitrogen at a pressure of

600 mm

are enclosed in a vessel of

1

litre capacity at

25 ° C

. Calculate the total pressure of the gas mixture.

Accepted Answer

The behaviour of a mixture of gases that do not react chemically with each other is frequently of interest. A law dealing with such behaviour was stated by John Dalton in $1801$ .

Statement:

The total pressure of the mixture of gases that do not react chemically with each other is equal to the sum of partial pressures of all the gases present If the total pressure be $P_{total}$ and partial pressures of gases present be $P_{A}, P_{B},$ etc.

$600 mm$

Partial Pressure:

It is defined as the pressure which each gas would exert if it is separately confined in the whole volume occupied by the mixture of gases.

$Partial Pressure = Mole fraction \times Total Pressure$ .

Mathematical deduction:

Let there be two chemically non-reacting gases $A$ and $B$ . Let us suppose that $n_{A}$ moles of gas $A$ are mixed with $n_{B}$ moles of gas $B$ .

Total no. of moles in the reaction mixture $= (n_{A} + n_{B})$ .

The ratio of no. of males of $A$ to the total no. of moles present is known as the mole fraction of ' $A$ '. It is represented by $X_{A}$ .

Thus, $X_{A} = \frac{n_{A}}{n_{A} + n_{B}}$ .

Similarly, mole fraction of ' $B$ ', $X_{B} = \frac{n_{B}}{n_{A} + n_{B}}$ .

Let $P_{Total}$ be the total pressure exerted by the mixture and $P_{A}, P_{B}$ be partial pressures of $A$ and $B$ respectively.

Partial pressure of gas ' $A$ ' will be its mole fraction times the total pressure.

i.e., $P_{A} = X_{A} \times P_{Total} = \frac{n_{A}}{n_{A} + n_{B}} \times P_{Total}$ .

Similarly, Partial pressure of gas ' $B$ ' will be its mole fraction times the total pressure.

i.e., $P_{B} = X_{B} \times P_{Total} = \frac{n_{B}}{n_{A} + n_{B}} \times P_{Total}$ .

Thus,

$P_{A} + P_{B} = (\frac{n_{A}}{n_{A} + n_{B}} \times P_{Total}) + (\frac{n_{B}}{n_{A} + n_{B}} \times P_{Total}) . P_{A} + P_{B} = (\frac{n_{A}}{n_{A} + n_{B}} + \frac{n_{B}}{n_{A} + n_{B}}) P_{Total} . P_{A} + P_{B} = P_{Total} .$

Given,

Pressure of oxygen, $P_{1} = 700 mm$ .

Pressure of hydrogen, $P_{2} = 750 mm$ .

Pressure of nitrogen, $P_{3} = 600 mm$ .

Volume of the vessel, $V = 1 L$ .

Volume of oxygen, $V_{1} = 500 mL$ .

Volume of hydrogen, $V_{2} = 300 mL$ .

Volume of nitrogen, $V_{3} = 700 mL$ .

The partial pressure of oxygen, $P_{O_{2}} = ?$

The partial pressure of hydrogen, $P_{H_{2}} = ?$

The partial pressure of nitrogen, $P_{N_{2}} = ?$

$P_{1} V_{1} = P_{O_{2}} V \Rightarrow P_{O_{2}} = \frac{P_{1} V_{1}}{V} = \frac{700 \times 500}{1000} = 350 mm . P_{2} V_{2} = P_{H_{2}} V \Rightarrow P_{H_{2}} = \frac{P_{2} V_{2}}{V} = \frac{750 \times 300}{1000} = 225 mm . P_{3} V_{3} = P_{N_{2}} V \Rightarrow P_{N_{2}} = \frac{P_{3} V_{3}}{V} = \frac{600 \times 700}{1000} = 420 mm . P_{Total} = P_{O_{2}} + P_{H_{2}} + P_{N_{2}} = (350 + 225 + 420) mm = 995 mm .$

Odisha Board Solutions for Chapter: States of Matter : Gases and Liquids, Exercise 1: QUESTIONS

Odisha Board Chemistry Solutions for Exercise - Odisha Board Solutions for Chapter: States of Matter : Gases and Liquids, Exercise 1: QUESTIONS

Questions from Odisha Board Solutions for Chapter: States of Matter : Gases and Liquids, Exercise 1: QUESTIONS with Hints & Solutions

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