Avogadro’s Law.pptx
- 3. Suppose the amount of a gas is increased, we can say that there are more gas molecules. An increase in the number of molecules will cause an increase in the number of impacts that these molecules will make on the walls of the container. If this happens, the pressure of the gas inside the container will increase. The pressure inside the container becomes greater than the pressure outside the walls. This causes the walls to move outward. Since there is more wall space, the impact will lessen and the pressure will return to its original value. Through this simple demonstration, Amedeo Avogadro was able to establish the relationship between the number of moles and the volume given that the pressure and temperature are held constant. Avogadro’s law states that the number of moles is directly proportional to the volume of gas at a constant temperature and pressure.
- 4. Mathematically, we can express this relationship as: V ∝ n V = k4n Where k4 is the constant proportionality. By simultaneously finding for the formula, k4 = 𝑉 𝑛 For a given sample of gas under two different sets of condition, at a constant temperature and pressure, 𝑉 𝑛1 = k4 = 𝑉2 𝑛2
- 5. Finally, Avogadro’s law is expressed as: 𝑽𝟏 𝒏𝟏 = 𝑽𝟐 𝒏𝟐 The graph of Avogadro’s law is the same as the graph of Charles’ and Gay- Lussac’s Law, since the volume and the number of moles are directly proportional to each other.
- 6. Example 1. If 0.30 mol of Argon gas occupies a volume of 75 mL at a particular temperature and pressure, what volume would 0.50 mol ofArgon have under the same condition? Given: V1 = 75 mL V2 = ? n1 = 0.30 mol n2 = o.50 mol
- 7. Solution: 𝑽𝟏 𝒏𝟏 = 𝑽𝟐 𝒏𝟐 V2 = 𝑉1 𝑛2 𝑛1 V2 = (75𝑚𝐿)(0.50 𝑚𝑜𝑙) (0.30 𝑚𝑜𝑙) = 125 mL Interpretation: As stated, the number of volume is directly proportional to the moles.Thus, the number of mole increases from 0.30 mol to 0.50 mol, the volume also increases from 75 mL to 125 mL.
- 8. Example 2. A 2. 0 moles of gas occupies an adjustable container with a volume of 50 L at a particular temperature and pressure. If 0. 8 mole of gas leaks out, what volume occupy if the temperature and pressure remain constant? Given: V1 = 50 L V2 = ? n1 = 2.0 mol n2 = 2.0 mol – 0.8 mol (gas leaks) = 1.2 mol
- 9. Solution: V2 = 𝑉1 𝑛2 𝑛1 V2 = (5𝑜 𝐿)(1.2 𝑚𝑜𝑙) (2.0 𝑚𝑜𝑙) = 60 𝐿.𝑚𝑜𝑙 2.0 𝑚𝑜𝑙 = 30 L