Grahams-Law-of-Diffusion-(Gas)final.pptx

Grahams-Law-of-Diffusion-final
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1. Graham’s Law of Diffusion Penny Roche

2. Introduction Have you ever sprayed an air freshener in a room? At first, you can only smell it directly where you sprayed, but eventually, the particles will spread across the room and can be smelt anywhere.

3. Graham’s Law of Diffusion, named after the Scottish Chemist THOMAS GRAHAM in 1848, known for his pioneering work in dialysis and the diffusion of gases. He is one of the founders of Colloidal Chemistry.

4. Graham’s Law of Diffusion is a fundamental principle in the field of inorganic chemistry that relates to the rates at which different gases diffuse or spread through a medium. It provides insight into the behavior o f gas particles and their movement. The concept of diffusion refers to the spontaneous movement of particle from an area of higher concentration to all area of lower concentration.

5. From the intro .

6. The purpose of diffusion is to create an equilibrium. Equilibrium is essentially a state of balance.

7. Some examples of diffusion that occurs in our daily life: The smell of perfume Opening the soda / cold drinks bottle and the CO2 diffuses in the air Dipping the tea bags in hot water will diffuse into water and change its color Small dust particles or smoke diffuse into the air and cause air pollution Sugar get dissolve evenly and sweetens the water w/o having stir it.

8. By definition Graham’s Law, states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass

9. Rate of Diffusion Equation:   Refers to the speed at which gas spread Mass of one mole of gas particles

10. EXAMPLES

11. Example #1 Two gases, Gas A and Gas B are released simultaneously from two separate container into a room. Gas A has a molar mass of 32g/ mol , Gas B has a molar mass of 64g/mol. If both gases are at the same temperature and pressure, Calculate the ratio of diffusion rates.

12. Given: Gas A = 32g/ mol Gas B= 64g/ mol Required: diffusion rates Solution:        

13. Example #2 Find the relative diffusion rate of water (molar mass=18.0152 g/ mol ) as compared to hard water (molar mass=20.0276 g/ mol ).

14. Given: The molar mass of water(M A )= 18.0152 g/ mol The molar mass of hard water(M B )= 20.0276 g/ mol Required : diffusion rates Solution:        

15. Example #3 Calculate the molar mass of a given gas whose diffusion rate is 2.92 times the diffusion rate of NH₃. 

16. Given: T he diffusion rate is 2.92 times of ammonia; hence we understand that the ratio of diffusion rates of the given gases should be 1/2.92 Required: Molar mass Solution:     ( squaring both sides ) ( squaring both sides )          

17. THANK YOU FOR LISTENING! HAVE A NICE DAY!