- By
**ceana** - Follow User

- 489 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about ' Avogadro's Law' - ceana

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

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

- 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

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)

- 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?

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.

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

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

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)

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!)

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

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

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

Download Presentation

Connecting to Server..