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

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slide1

Avogadro\'s Law

And the combined Gas Law

Mr. Shields Regents Chemistry U05 L08

avogadro s law
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
slide3

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

avogadro s principal
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)

slide5

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?

slide6

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.
slide8

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

combined gas law
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)

slide11

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!)

slide12

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)

cgl problem
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

cgl problem1
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

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