Avogadro’s Law
What is Avogadro’s Law
• Avogadro’s Principle – equal volumes of gases at the
same temperature and pressure contain equal numbers
of particles
Avogadro’s Formula
Formula:
V1 = V2
n1 n2
* n represents the amount of gas
• This is a direct relationship!
• So if the amount of gas increases, then the volume will
___________.
If the amount of gas decreases, then the
increase
volume will __________.
decrease
Graph for Avogadro’s Law
What Laws have we learned were also direct relationships, in which their
graphs were similar to Avogadro’s?
Avogadro’s Law
Avogadro’s Law
• Molar Volume – for a gas, this is the volume that one mole of
that gas occupies at STP
(STP is standard temperature and pressure which is 0oC and 1 atm)
• Avogadro showed experimentally that 1 mole of any gas will occupy
a volume of 22.4L at STP
**Conversion Factor: 1 mol (any gas) = 22.4 L at STP **
(This is on your formula chart under constants and conversions!)
Avogadro’s Law: Example 1
Calculate the volume that 0.881 moles of oxygen
gas at STP will occupy.
(This can be solved using the Avogadro’s law formula or using
the dimensional analysis method)
Avogadro’s Law:
Example 1 Answer
• Formula:
V1 = V2
n1 n 2
• What do we know?
V1 = ? L
Remember, at STP:
n1 = 0.881 mol O2
22.4 L/mol
V2 = 22.4 L
n2 = 1 mol O2
Second way to solve, using the
First way to solve, using the formula:
dimensional analysis:
V1
= 22.4 L
.881 mol O2
1 mol O2
--------.881 mol
O2 | 22.4 L = 19.73 L
| 1 --------mol O2
After cross multiplying you end up with
V1(1---------mol O2) = (22.4 L)(.881 ---------mol O2)
1----------mol O2
1 ---------mol O2
19.73 L
V1 = ___________
Avogadro’s Law: Example 2
How many grams of N2 will be contained in a 2.0 L
flask at STP?
Avogadro’s Law:
Example 2 Answer
• Formula:
V1 = V2
n1 n2
• What do we know?
V1 = 2.0 L
n1 = ? g N 2
V2 = 22.4 L
n2 = 1 mol N2
Remember, at STP:
22.4 L/mol
First , solve for the number of moles of N2:
2 L N2 | 1 mol N2
| 22.4 L N2
= .089 mol N2
Then, use dimensional analysis to convert from
moles of N2 to grams of N2:
.089 mol N2 | 28.014 g N2
| 1 mol N2
= 2.50 g N2
Ideal Gas Law
What is the Ideal Gas Law?
It is a good approximation to the behavior of many
gases under many conditions
Ideal Gas Law Formula
Ideal Gas Law Formula
• Formula: PV = nRT
(called Piv-Nert formula)
• R is called the Ideal Gas Constant
•
•
•
•
R is dependent on the units of the variables for P, V, and T
Temperature is always in Kelvin
Volume is always in Liters
Pressure is either in atm, mmHg, or kPa. Because of the different
pressure units, there are 3 possibilities of the ideal gas constant (refer to
the EOC Chart under constants and conversions)
• Example: R = 0.0821
(Liters)(atm)
(moles)(Kelvin)
*We would use this value for R if the given pressure’s units are in atm*
Ideal Gas Law Simulation
http://phet.colorado.edu/en/simulation/gas-properties
Ideal Gas Law Example 1
(using moles)
If the pressure by a gas at 30oC in a volume of
.05 L is 3.52 atm, how many moles of the gas is
present?
**To know which R value to use, look at what units pressure is in**
Ideal Gas Law Example 1 Ans
(finding moles)
• Formula:
PV = nRT
• What do we know?
P = 3.52 atm
What R value will we use?
(Hint: Look at the units for pressure)
V = .05 L
n = ? moles
R = 0.0821 (L*atm)/(mol*K)
T = 30oC + 273 = 303K
Now lets plug in the information and solve:
P * V = n* R * T
-------- ------(3.52)(.05) = n(.0821)(303)
-------- ----(.0821)(303) (.0821)(303)
.0071 mol
n = ______________
Ideal Gas Law Example 2
(finding grams)
Avogadro’s Law allows us to write a gas law that is valid not
only for any P, V, or T but also for any mass of any gas!
Example:
Calculate the grams of N2 gas present in a 0.600 L sample kept
at 1.00 atm and a temperature of 22.0oC.
Ideal Gas Law Example 2 Answer
(using moles)
• Formula:
PV = nRT
• What do we know?
P = 1.00 atm
What R value will we use?
(Hint: Look at the units for pressure)
V = 0.600 L
n = ? g N2
Now lets plug in the information and solve:
R = 0.0821 (L*atm)/(mol*K)
P * V = n* R * T
o
T = 22.0 C + 273 = 295 K
(1.00)(.600) = n(.0821)(295)
------- ------(.0821)(295) (.0821)(295)
-------- ----n = ______________
.025 mol N2
But, they want the answer in grams, so we need to do
dimensional analysis using molar mass:
--------.694 g N2
.025 mol
N2 | 28.014 g N2 = __________
| 1 ---------mol N2
Ideal vs. Real Gas
Ideal vs. Real Gas
• An ideal gas obeys all the assumptions of the kinetic molecular
theory. (Atoms or molecules are non-interacting particles,
etc.)
• NO gas is IDEAL (ideal gas doesn’t exist). They all take up
space and interact with other molecules (attraction, repultion)
but most gases will behave like ideal gases at the right
conditions of temp. and pressure.
• Real gases do NOT behave like ideal gases at extremely high
pressures and extremely low temperatures.
Ideal vs. Real Gas
Explanation
• Ideal gases do not have molecular volume and show no
attraction between molecules at any distance
• Real gas molecules have volume and show attraction at short
distances. High P will bring the molecules very close together.
This causes more collisions and also allows the weak attractive
forces to come into play. With low temperatures (close it gas’s
liquefication point), the molecules do not have enough energy
to continue on their path to avoid the attraction.
Dalton’s Law of
Partial Pressure
Dalton’s Law Analogy
In other words, Dalton’s Law of Partial Pressure would look like this:
Partial pressure 1
+
Partial presure 2
+ Partial pressure 3 + Pratial pressure 4 = ?
Dalton’s Law Formula
Formula:
Pt = P1 + P2 + P3 + ...
Pt = the total pressure
P1 = Partial Pressure of gas ‘1”,
P2 = Partial Pressure of gas “2,”
etc.
**ALL Units, MUST be the same for each pressure!!!**
What is Dalton’s Law?
• Dalton’s Law of Partial Pressures – the total pressure exerted
on a container by several different gases is equal to the sum of
the pressures exerted on the container by each gas
• (Partial pressure of a gas in a mixture of gases is the pressure which that
gas would exert if it were the only gas present in the container)
• Dalton’s Law of Partial Pressure assumes each gas in the
mixture is behaving like an ideal gas
Dalton’s Law Example
Demo Video’s
Simulation (or demo - or show both):
Collecting a Gas Over Water
http://www.kentchemistry.com/links/GasLaws/dalton.htm
Demo - Gas collection over water
(or use Butane gas from a cigarette lighter)
http://www.youtube.com/watch?v=UQRCUxrLU1c&edufilter=
QMQXe2SzEucYVMXVK3i8VQ
A more exciting Demo  – Oxygen Collection Over water
http://www.youtube.com/watch?v=2Tj3Ir0brco&edufilter=Q
MQXe2SzEucYVMXVK3i8VQ
Gas Stoichiometry
Volume Ratios
All the gas laws you learned so far can be applied to calculate
the stoichiometry of reactions in which gases are reactants or
products.
For Example
2 C4H10(g) + 13 O2(g)  8 CO2(g) + 10 H2O(g)
• We already know that the coefficients in a balanced chemical
equation represent the # of moles of the reactants and
products in the reaction:
___2___moles of C4H10 reacts with __13__moles of O2 to produce
___8___ moles of CO2 and __10__moles of H2O
Remember
Avogadro’s principle states that equal volumes of gases at the
same temperature and pressure contain equal numbers (or
moles) of particles.
**1 mole of any gas occupies a volume of 22.4L at STP**
So, the coefficients in a balanced equation also show the
relationships among the volumes of any gaseous reactants or
products.
Ex: ___2___ L of C4H10 react with ___13___L of O2 to produce
____8____L of CO2 and ____10____L of H2O
Calculations Involving
Only Volume
(the molar ratios are also volume ratios for gases)
What volume of oxygen gas is needed to complete the
combustion of 4.0L of propane gas (C3H8) at STP?
1
5
3
4 2O(g)
_____C
3H8(g) + _____O
2(g)  _____CO
2(g) + _____H
__1___C3H8(g) + __5___O2(g)  __3___CO2(g) + __4___H2O(g)
4L
?L
Start with what you are given, then use Dimensional Analysis to get the answer:
4 --------L C3H8 | 5 L O2 = 20 L O2
| 1 L--------C3H8
You Try!
What volume of oxygen is needed to react with solid sulfur to
form 3.5 L SO2? Assume constant pressure and temperature.
___O2 + ___S  ___SO2
Calculations Involving
Mass and Volume
(use the PT when converting grams to moles)
If 5.0 L of nitrogen gas reacts completely with hydrogen gas at
a temperature of 298K and a pressure of 3.0 atm, how many
grams of ammonia (NH3) are produced?
1
3
2
_____N
2(g) + _____H2(g)  _____NH3(g)
You Try!
Ammonia nitrate is a common ingredient in chemical
fertilizers. Use the reaction shown to calculate the mass of solid
ammonium nitrate that must be used to obtain 0.100 L of
dinitrogen oxide gas at STP.
___NH4NO3(s)  ___N2O(g) + ___H2O(g)