Avogadro's Law

1. Avogadro's Law And the combined Gas Law Mr. Shields Regents Chemistry U05 L08

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

3. Do you remember what Avogadro’s Principal stated? • Avogadro’s Principal (1811) • - Equal volumes of gas contains EQUAL • nos. of molecules • - 1 mole of any gas = Avogadro’s no. (NA) 6.02 x 1023 particles

4. 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 (Vm) And 1 molar volume = 6.02 x 1023 particles So 0.25mol of CO2 and 0.25mol He and 0.25mol of O2 all Have exactly the same number of particles (1.5x1023) and exactly the same Volume (5.6L)

5. 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?

6. 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.

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

8. 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 V1/n1 = V2/n2 15 ml / 0.1 mol = V2 / 0.35 mol V2 = 150 x .35 V2 = 52.5 ml

9. Combined Gas Law

10. Combined Gas Law We now have all the relationships we need to Explain gas behavior: PV = k P/T = k V/T = k V/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)

11. Since P1V1/n1T1 = P2V2/n2T2 If T and n are constant (don’t change) then n1T1= n2T2 And P1V1 x n2T2 = P2V2 n1T1 But n2T2 = 1 n1T1 So the CGL reduces to P1V1 = P2V2 (Boyles law!)

12. 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)

13. CGL Problem 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

14. CGL Problem 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