2. Mr. ShieldsRegents Chemistry U05 L08

3. Avogadro’s Law Amedeo Avogadro (1776 – 1856) V/n = k (Constant T and P) -Related the volume of a gas to the number of molecules (moles) present of molecules (moles) present

4. Do you remember what Avogadro’s Principal stated? -Avogadro’s Principal (1811) - Equal volumes of gas contains EQUAL nos. of molecules nos. of molecules - 1 mole of any gas = Avogadro’s no. (N A ) 6.02 x 10 23 particles

5. Avogadro’s Principal Remember: The volume a gas occupies is independent of the Gas molecule itself. 1 mol of ANY gas = 22.4L This is called the Molar volume (V m ) And 1 molar volume = 6.02 x 10 23 particles So 0.25mol of CO 2 and 0.25mol He and 0.25mol of O 2 all Have exactly the same number of particles (1.5x10 23 ) and exactly the same Volume (5.6L)

6. Equal Volumes of gas at the same temperature and pressure contain an equal number of particles If we held P and T constant the only way to change V is to change n, the number of particles (moles) But increasing n should increase pressure, which we want to hold constant, since the frequency of collisions with the container wall increases. So how do we increase n but hold P constant?

7. If I want to keep the pressure constant while adding molecules then we need to increase the volume of the Container at the same proportional rate. This will… - reduce the # of collisions per unit area - reduce the # of collisions per unit time In mathematical terms Avogadro’s Hypothesis states V/n = k (Const. P, T) And Like Charles and Gay-Lussac’s Laws in which there Is a direct Relationship between variables, the relationship between V and n is also a direct relationship.

8. The format for Avogadro’s law that we will use to Solve problems is: V1/n1 = V2/n2

9. Let’s do a problem: 0.25mol of Hydrogen are added to 0.10mol of hydrogen To yield 0.35mol in a 15 ml container at 25 deg. C at a pressure of 1.5 atm. What’s the new volume of Hydrogen if the pressure and temperature do not change. Solution: n,V are variables;P and T are constant V 1 /n 1 = V 2 /n 2 15 ml / 0.1 mol = V 2 / 0.35 mol V 2 = 150 x.35 V 2 = 52.5 ml

11. Combined Gas Law We now have all the relationships we need to Explain gas behavior: PV = kP/T = kV/T = kV/n = k If we combine these terms we end up with what Is called the Combined Gas Law (i.e. CGL) PV/nT = k No matter how P, V, n, or T change, k is constant Therefore:P1V1/n1T1 = P2V2/n2T2 (must be used when more than 2 variables are changing)

12. Since P 1 V 1 /n 1 T 1 = P 2 V 2 /n 2 T 2 If T and n are constant (don’t change) then n 1 T 1 = n 2 T 2 And P 1 V 1 x n 2 T 2 = P 2 V 2 n 1 T 1 n 1 T 1 But n 2 T 2 = 1 n 1 T 1 n 1 T 1 So the CGL reduces to P 1 V 1 = P 2 V 2 (Boyles law!)

13. And… If P and n are constant the CGL reduces to V1/T1 = V2/T2(Charles law) If V and n are constant the CGL reduces to P1/T1 = P2/T2(Gay-lussac’s law) And if P and T are constant the CGL reduces to V1/n1 = V2/n2(Avogadro’s law)

14. Let’s try a CGL problem. 0.5 mol of Nitrogen gas in 1.5L has a temperature of 25 deg. C and a pressure of 1.2 atm. If the volume of the container is increased to 2.25L, the temperature increased to 75 deg. C and the amount of nitrogen is increased to 1.3 mol what is the new pressure? P1V1/n1T1 = P2V2/n2T2 1.2atm(1.5L)/0.5mol(298K) = P2(2.25L)/1.3mol(348K) (1.8/149)(452.4)/2.25 = P2 5.47/2.25 = P2 2.43 atm = P2 CGL Problem

15. Let’s try another one… Argon at a temperature -10 deg C is held in a 2.5L tank at Standard Pressure. It is later transferred to a 4.0 L tank And warmed to 85 deg C. What’s the new Pressure in atm? What’s constant in this problem? So, P1V1/n1T1 = P2V2/n2T2 1atm(2.5L)/263K = P2(4.0L)/358K (2.5x358)/(263x4.0) = P2 0.851 atm = P2 CGL Problem