Transcript Gases - Schoolwires

Gases
Chapter 13
Some basics
Gases have properties that are very
different from solids and liquids.
 They are very sensitive to changes in
pressure and temperature and have low
density.
 Gases do have mass( although small).
 Think of properties of gases when your
ears pop while driving to Tahoe , deep-sea
divers or your kid sister’s birthday
balloons.

Pressure and Temperature

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Pressure of gases is
usually denoted in
atm or mmHg.
1atm =760 mmHg
1atm =760 torr
Other units include
Pascal and psi.
Measured by a
barometer




Temperature of gases
is measured in Kelvin
Kelvin = °C +273
Other unit for
temperature is
Fahrenheit.
Measured by a
thermometer
Kinetic Molecular Theory

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
Volume of the molecules is negligible (zero)
compared to the volume of the container.
The molecules are colliding with the walls of the
container and this causes the pressure of the
gas.
The molecules do not attract or repel each
other.
The average kinetic energy of the molecules is
directly proportional to the Kelvin temperature.
Boyle’s Law


When pressure of a gas is increased the volume of a
gas decreases provided the temperature remains the
same.
PV = k (constant temperature)
Pressure and volume are inversely proportional

Or as P goes up, V must go down to keep the above
equation correct.

P1V1 =P2V2 is a relationship that helps us solve
problems when conditions are changed for a gas.
Sample problem

What is the new pressure if 0.500L of oxygen at
pressure 0.87atm is changed to 0.750L at constant
temperature?
Solution
 Since this is at constant temperature, Boyle’s law
applies,
P1V1 =P2V2
P1= 0.87atm, V1 = 0.500L, P2 = ?,V2 = 0.750L

P2 = P1V1 = 0.87atm X 0.500 L = 0.58atm
V2
0.750 L
Practice problems

The volume of a sample of hydrogen is 250.mL at 3.5
atm pressure. What will be the volume when the
pressure is reduced to 0.75atm, assuming that
temperature remains constant?
1200mL or 1.2 L

The pressure of a 2.34L sample of helium is 785 torr.
Calculate the pressure in atm if volume is decreased to
2.04L and temperature kept constant.
1.18 atm
Charles’s Law


According to Charles’s Law, the volume of a gas
increases when temperature increases, provided the
pressure is kept constant.
V = k or when temperature increases, volume must
T
increase as well
V1 = V2 all temperatures have to be in KELVIN scale!!!!
T1
T2
This law can now be used to solve problems
Sample problem
The volume of a gas at 25ºC is 234mL. What will its
volume be at 50ºC if pressure is kept constant?
 Solution
25ºC + 273 = 298K = T1 , V1 = 234mL
50ºC + 273 = 323K = T2 , V2 = ?

V1 = V2
T1 T2
, V2 = V1 X T2 = 234mL X 323K = 254mL
T1
298K
Practice problems
A gas occupies 670mL at 45ºC. At what
temperature (in ºC) will it occupy 750mL if
pressure is kept unchanged?
 83ºC

If methane gas occupies 58.0L at 17ºC,
what volume will it occupy at 27ºC if
pressure is left the same?
 60.L

Avogadro’s Law


When the number of moles of gas
increases, the volume increases too
(duh!!)
V = k (at constant temperature and
n
pressure)
V1= V2
n 1 n2
This above equation can be used to
solve problems.
Problems
If 6 moles of oxygen occupies 23L how
many L will 3.67moles occupy at constant
temperature and pressure?
 V1 = V2
n1
n2
V2 = V1 X n2 = 23L X 3.67moles = 14L
n1
6 moles

Ideal Gas
An ideal gas is one whose molecules are
not attracted to each other and the volume
occupied by each molecule is too small to
matter.
 Most gases behave ideally at high
temperatures and low pressures.
 Gases are not ideal when compressed or
cooled down.

Ideal Gas Law
If we combine all the three gas laws
(Boyle’s, Charles’s and Avogadro's) we get
the ideal gas law:
 PV =nRT where R =0.08206 L.atm
K.mol
If any three of the properties are known the
fourth can be calculated.

This equation has some limits, can be used only in
low pressures and high temperatures. If the
pressure goes up or temperature decreases,
corrections need to be applied.
Problems
What volume is occupied by 0.250mol of
CO2 at 25°C and 371torr?
 PV =nRT
P = 371/760 = 0.488atm
T = 298K R =0.08206 L.atm
K.mol
V= nRT = .250mol X 298K X .08206 L.atm
P
0.488 atm
K.mol
= 12.5L

Practice
A 1.5mol of radon gas has a volume of 21.0L
at 33°C. What is the pressure of the gas?
 1.8atm
 What is the mass of oxygen needed to fill a
tank of volume 22.7L at temperature 34°C
and pressure 1.5 atm?
 43.3g

Combined Gas Law
We can use the combined gas law if the
conditions of pressure, volume or
temperature for the same amount of gas
are altered.
 P1V1 = P2V2
T1
T2
Temperature must be in Kelvin !!!!

Sample problem
A sample of neon gas has a volume of 27.5mL
at 22.0°C and 740. torr pressure. What will its
volume be at temperature 15.0°C and pressure
755 torr?
 P1V1 = P2V2
T1
T2
 740.torr X 27.5mL = 755torr X V2
295K
288K
V2=26.3mL

Dalton’s Law of partial pressures
According to Dalton’s law the pressure of
a mixture of gases is related to the number
of moles of the gas in the mixture.
 Ptotal = p1 +p2 + p3 +…..
 ntotal = n1 + n2 + n3 + …..
 Ptotal V =ntotalRT

Problem

A 6.00L tank contains 32.0g of oxygen gas and
18.02g of water vapor. At 20°C what is the total
pressure of the tank?

32.0g = 1.00mol O2; 18.02g = 1.00mol H2O
ntotal = 1.00 + 1.00 = 2.00mol


Ptotal = ntotalRT = 2.00* 0.0821* 293K = 8.01atm
V
6.00L
Practice
If a 23.0L tank contains 3.00mol of H2,
2.00mol of He and 1.00mol of Ne gases,
calculate the total pressure of the tank at
40°C? Calculate the partial pressure of
each gas as well.
 Ptotal = 6.7atm
 pH2 = 3.35atm
 pHe = 2.23atm
 pNe = 1.12atm
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Gas Stoichiometry
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First write a balanced chemical equation.
Calculate moles of given
Calculate moles of asked
Calculate volume of gas produced
V=nRT
P
Mole ratio
MOLES OF GIVEN
Grams of Given
MOLES OF ASKED
n
Sample problem

Calculate the volume of NH3 produced when 43g
of N2 reacts with excess H2 at 29°C and 0.997atm
pressure.

N2 + 3H2
2NH3
43g X 1molN2 X 2mol NH3 = 3.07mol NH3
28.02g
1molN2
V = nRT = 3.07mol* 0.0821* 302K = 76. L
P
0.997atm

Practice problem
Calculate the volume of oxygen produced
from the decomposition of 6.00g of KClO3
to KCl and O2 at 30.0°C and 100.kPa
pressure.
 1.85L

Graham’s Law of Diffusion

According to Graham’s law, the rates of
diffusion of two gases is inversely proportional to
the square root of their molar masses. In other
words, lighter gases travel faster under similar
conditions of temperature.

RateA = √molar massB
RateB √ molar massA
Practice
Find the molar mass of a gas that diffuses
0.31 times as fast as oxygen gas.
 If oxygen travels at a rate of 1, then the
other gas travels at 0.31,
 Rate O = √molar mass gas
2
Rate gas √molar mass O2
1
= √molar mass gas
0.31
√32.0g/mol
molar mass gas = 330 g/mol

Suppose a gas diffuses 1.41 times faster
than sulfur dioxide. What is the molar
mass of this gas?
 Rategas =√64.0g/mol
RateSO2 √molar massgas

= √64.0g/mol
√ molar massgas
molar mass gas = 32.2g/mol
1.41