Mr. ShieldsMr. Shields Regents Chemistry Regents Chemistry U05 L08 U05 L08

Avogadro’s LawAvogadro’s Law

Amedeo Avogadro (1776 – 1856)Amedeo Avogadro (1776 – 1856)

V/n = kV/n = k(Constant T and P)(Constant T and P)

-Related the Related the volumevolume of a gas to the number of a gas to the number of molecules (of molecules (molesmoles) present) present

Do you remember what Avogadro’s Principal stated?Do you remember what Avogadro’s Principal stated?

-Avogadro’s Principal (1811)Avogadro’s Principal (1811)

- Equal volumes of gas contains - Equal volumes of gas contains EQUALEQUAL nos. of moleculesnos. of molecules

- 1 mole of any gas = Avogadro’s no. (N- 1 mole of any gas = Avogadro’s no. (NAA))

6.02 x 106.02 x 102323 particles particles

Avogadro’s PrincipalAvogadro’s Principal

Remember: The volume a gas occupies is Remember: The volume a gas occupies is independentindependent of the Gas molecule itself. of the Gas molecule itself.

1 mol of 1 mol of ANYANY gas = 22.4L gas = 22.4L

This is called the MolarThis is called the Molarvolume (volume (VVmm))

And 1 molar volume =And 1 molar volume =6.02 x 106.02 x 102323 particles particles

So 0.25mol of COSo 0.25mol of CO22 and 0.25mol He and 0.25mol of O and 0.25mol He and 0.25mol of O2 2 allallHave exactly the same number of particles (Have exactly the same number of particles (1.5x101.5x102323))and exactly the same Volume (and exactly the same Volume (5.6L5.6L))

Equal Volumes of gas at the same temperature andEqual Volumes of gas at the same temperature andpressure contain an equal number of particlespressure contain an equal number of particles

If we held P and T constant the only way to change VIf we held P and T constant the only way to change Vis to change n, the number of particles (moles)is to change n, the number of particles (moles)

But increasing n should increase pressure,But increasing n should increase pressure,which we want to hold constant, since thewhich we want to hold constant, since thefrequency of collisions with the container wall increases.frequency of collisions with the container wall increases.

So how do we increase n but hold So how do we increase n but hold PP constant? constant?

If I want to keep the pressure constant while adding If I want to keep the pressure constant while adding molecules then we need to increase the volume of themolecules then we need to increase the volume of theContainer at the same proportional rate. This will…Container at the same proportional rate. This will…

- reduce the # of collisions per unit area- reduce the # of collisions per unit area- reduce the # of collisions per unit time- reduce the # of collisions per unit time

In mathematical terms Avogadro’s Hypothesis statesIn mathematical terms Avogadro’s Hypothesis states

V/n = k (Const. P, T)V/n = k (Const. P, T)

And Like Charles and Gay-Lussac’s Laws in which thereAnd Like Charles and Gay-Lussac’s Laws in which thereIs a Is a direct Relationshipdirect Relationship between variables, the between variables, therelationship between relationship between VV and and nn is also a is also a directdirectrelationship.relationship.

The format for Avogadro’s law that we will use to The format for Avogadro’s law that we will use to Solve problems is:Solve problems is:

VV11/n/n11 = V = V22/n/n22

Let’s do a problem:Let’s do a problem:

0.25mol of Hydrogen are added to 0.10mol of hydrogen0.25mol of Hydrogen are added to 0.10mol of hydrogenTo yield 0.35mol in a 15 ml container at 25 deg. C 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 ofat a pressure of 1.5 atm. What’s the new volume ofHydrogen if the pressure and temperature do not change.Hydrogen if the pressure and temperature do not change.

Solution:Solution:n,V are variables;n,V are variables; P and T are constantP and T are constant

VV11/n/n11 = V = V22/n/n22

15 ml / 0.1 mol = V15 ml / 0.1 mol = V22 / 0.35 mol / 0.35 molVV22 = 150 x .35 = 150 x .35

VV22 = 52.5 ml = 52.5 ml

Combined Gas LawCombined Gas Law

We now have all the relationships we need toWe now have all the relationships we need toExplain gas behavior:Explain gas behavior:

PV = kPV = k P/T = kP/T = k V/T = k V/T = k V/n = kV/n = k

If we combine these terms we end up with whatIf we combine these terms we end up with whatIs called the Is called the Combined Gas Law (i.e. CGL)Combined Gas Law (i.e. CGL)

PV/nT = kPV/nT = k

No matter how P, V, n, or T change, k is constantNo matter how P, V, n, or T change, k is constant

Therefore:Therefore: PP11VV11/n/n11TT1 1 = P= P22VV22/n/n22TT22 (must be used when (must be used when more than 2 variables are changing)more than 2 variables are changing)

Since PSince P11VV11/n/n11TT11 = P = P22VV22/n/n22TT22

If T and n are constant (don’t change) then If T and n are constant (don’t change) then

nn11TT11= n= n22TT22

And And PP11VV11 x x nn22TT22 = P = P22VV22

nn11TT11

But But nn22TT22 = 1 = 1 nn11TT11

So the CGL reduces toSo the CGL reduces to

PP11VV11 = P = P22VV22 (Boyles law!)(Boyles law!)

And…And…

If P and n are constant the CGL reduces toIf P and n are constant the CGL reduces to

V1/T1 = V2/T2V1/T1 = V2/T2 (Charles law)(Charles law)

If V and n are constant the CGL reduces toIf V and n are constant the CGL reduces to

P1/T1 = P2/T2P1/T1 = P2/T2 (Gay-lussac’s law)(Gay-lussac’s law)

And if P and T are constant the CGL reduces toAnd if P and T are constant the CGL reduces to

V1/n1 = V2/n2V1/n1 = V2/n2 (Avogadro’s law)(Avogadro’s law)

Let’s try a CGL problem.Let’s try a CGL problem.

0.5 mol of Nitrogen gas in 1.5L has a temperature of0.5 mol of Nitrogen gas in 1.5L has a temperature of25 deg. C and a pressure of 1.2 atm. If the volume of the25 deg. C and a pressure of 1.2 atm. If the volume of thecontainer is increased to 2.25L, the temperature increasedcontainer is increased to 2.25L, the temperature increasedto 75 deg. C and the amount of nitrogen is increased toto 75 deg. C and the amount of nitrogen is increased to1.3 mol 1.3 mol what is the new pressure?what is the new pressure?

P1V1/n1T1 = P2V2/n2T2P1V1/n1T1 = P2V2/n2T2

1.2atm(1.5L)/0.5mol(298K) = P2(2.25L)/1.3mol(348K)1.2atm(1.5L)/0.5mol(298K) = P2(2.25L)/1.3mol(348K)

(1.8/149)(452.4)/2.25 = P2(1.8/149)(452.4)/2.25 = P25.47/2.25 = P25.47/2.25 = P22.43 atm = P22.43 atm = P2

CGL ProblemCGL Problem

Let’s try another one…Let’s try another one…

Argon at a temperature -10 deg C is held in a 2.5L tank atArgon at a temperature -10 deg C is held in a 2.5L tank atStandard Pressure. It is later transferred to a 4.0 L tank 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?And warmed to 85 deg C. What’s the new Pressure in atm?

What’s constant in this problem? What’s constant in this problem? So,So,P1V1/n1T1 = P2V2/n2T2P1V1/n1T1 = P2V2/n2T2

1atm(2.5L)/263K = P2(4.0L)/358K1atm(2.5L)/263K = P2(4.0L)/358K(2.5x358)/(263x4.0) = P2(2.5x358)/(263x4.0) = P2 0.851 atm = P20.851 atm = P2

CGL ProblemCGL Problem