Introduction to GasesCh 13HW: 1, 31, 39, 41Suggested: 42-46 10/21

About Gases

• Gases are the most understood form of matter

• Even though different gases have different chemical properties, they tend to exhibit similar physical properties

• This situation arises because gas molecules expand to fill a given space, and are relatively far apart from one another

• Thus, each molecule behaves as if the others are not there

The Kinetic Theory of Gases

• Kinetic Theory tells us the following:1. Gas molecules are ALWAYS in motion. They collide randomly

with each other and with the walls of the container

2. All collisions involving gas particles are ELASTIC. This means they simply “bounce off” of one another. No energy is lost, and the speed of the particles does not change.

3. The distance between gas molecules is large. So, gas mainly consists of empty space.

4. Interactions between gas molecules are negligible (no attraction or repulsion).

5. The kinetic energy (speed) of gas molecules is proportional to temperature.

Pressure

• The most readily measured properties of a gas are its temperature, volume, and pressure

• Pressure describes the force that a gas exerts on an area, A. P = F/A

• The image below shows gas molecules inside of a cubic container. The gas molecules strike against the walls of the container. These collisions are the source of the pressure.

Atmospheric Pressure

• You and I are currently experiencing an attractive force that pulls us toward the center of the earth (gravity).

• Gas molecules in the atmosphere also experience gravity.

• Because of their small masses and thermal energies, gas molecules can somewhat counteract gravity, which is why gases don’t just sit on the surface

• Nonetheless, gravity causes the gases in the atmosphere to “press down” on the surface. This is atmospheric pressure.

Atmospheric Pressure

• To calculate atmospheric pressure, we calculate the mass of a 1m2 column of air extending through the entire atmosphere.

• That column would have an approximate mass of 104 kg.

• The force exerted on that 1m2 area would be:

F= ma = (104kg)(9.8 ms-2) = 105 N

• Then, P = 105N/1m2 = 105 N/m2 = 105 Pa

• SI unit of pressure is the Pascal (Pa). Related units are bar, mmHg, and atmospheres.

Units of Pressure

1 atm = 760 mm Hg

Atmospheric Pressure

• The image to the bottom left shows a barometer.

• The tube is inverted into a dish that contains mercury. Some mercury flows into the because the atmospheric pressure forces it upward (pressure in tube is lower than atmosphere).

• The flow stops when the atmospheric pressure is equal to the pressure inside the tube (weight of Hg/area).

• The height h is proportional to the atmospheric pressure.

h

Closed, undervacuum

Open-end, exposed to atmosphere

Gas Laws: Boyle’s Law

• Robert Boyle was able to show that, at constant temperature, volume and pressure are inversely proportional.

V α

• In other words, if we decrease the volume of a gas, we increase its pressure. If we consider pressure to be caused by molecular collisions with the container walls, the figure below clearly indicates why this relationship exists.

Large volume

Small volume

Charles’s Law: A Demonstration

a. Liquid nitrogen (-196oC) poured on a balloonb. As air inside balloon is cooled, balloon compressesc. Balloon is allowed to warm back to room temperatured. Balloon re-inflates

Relationship between Volume and Temperature: Charles’s Law

• The average kinetic energy of a gas is given by:

Where R is the gas constant, 8.314 J mol-1 K-1 (Since kinetic energy is equal to , we can determine the average velocity of the molecules. This will be discussed later.)

• If we increase the temperature, the result will be molecules that are moving faster and causing higher energy collisions. This causes the gas to expand, increasing the volume.

kinetic theory

Charles’s Law

• So how can we predict the change in volume with temperature?

• Charles’s Law tells us that volume is directly proportional to temperature at constant pressure.

• If V = mT (m is just a constant)• Then Vi/Ti = m, where i stands for ‘initial’. And…

• Vf/Tf = m, where f stands for ‘final’. Thus…

Charles’s Law𝑽 𝒊

𝑻 𝒊=𝑽 𝒇

𝑻 𝒇

Example

• At 0 oC, the volume of the air trapped in a glass tube by a metal plug is 0.180 mL. Then, the water is raised to its boiling temperature and the air expands. What is the new volume of air?

𝑽 𝒊

𝑻 𝒊=𝑽 𝒇

𝑻 𝒇

• Here, we are looking for the final volume. We will use Charles’s Law and plug in the given information.

.𝟏𝟖𝟎 𝒙𝟏𝟎−𝟑𝑳𝟐𝟕𝟑𝒐𝑲

=𝑽 𝑭

𝟑𝟕𝟑𝒐𝑲

USE KELVIN SCALE !!!

𝑽 𝑭=.𝟐𝟒𝟔𝒎𝑳

Avogadro’s Law

• Avogadro determined that equal volumes of gases at a given pressure and temperature must contain equal numbers of molecules, and thus, equal moles

Volume is directly related to moles, and both are related to the stoichiometry

1L H2(g) 1L Cl2(g) 2L HCl(g)

2L H2(g) 1L O2(g) 2L H2O(g)

The Ideal Gas Law

• If we take the 3 previously discussed gas laws:

• V α n (n = moles) Avogadro’s Law• V α Boyle’s Law• V α T Charles’s Law

• We can combine these laws to obtain the IDEAL GAS LAW:

PV = nRT

Depending on the units of Pressure, we can use multiple values of R

Ideal Gases

• Any gas that follows the ideal gas law is considered an ideal gas.

• One mole of an ideal gas at 0 oC and 1 atmosphere of pressure occupies 22.4 L of space. The value of R is based on these values of n, T, P, and V.

• The ideal gas law is valid only at low pressures

• The conditions listed above (0 oC, 1 atm) are referred to as standard temperature and pressure (STP)

• Note: The Ideal gas law is a theoretical approximation, and no gas follows this law exactly, but most gases are within a few percent of this approximation, so it is very useful.

Comparisons of Real Gases to the Ideal Gas Law Approximation at STP

Ideal Gas

22.4 L22.06 L 22.31 L

22.41 L 22.42 L

Cl2 CO2 He H2

Mol

ar V

olum

e at

STP

(L)

Deviations from the ideal gas law exist, but those deviations are reasonably small.

Example

• The pressure in a 10.0 L gas cylinder containing N2(g) is 4.15 atmospheres at 20.0 oC. How many moles of N2(g) are there in the cylinder?

*When dealing with ideal gas law questions, follow these steps:

1) Determine what it is you are solving for.

2) List the given information. Pay attention to the units of each parameter. Convert as needed, and MAKE SURE THAT THE TEMPERATURE IS IN KELVIN !

3) Rearrange the ideal gas law equation accordingly to solve for the desired parameter.

Example, continued.

We are solving for moles.V = 10.0 LP = 4.15 atm.T = 20. 0 oC 293 oK R = .0821 L•atm•mol-1• K-1

Rearrange the equation to solve for n.

The pressure in a 10.0 L gas cylinder containing N2(g) is 4.15 atmospheres at 20.0 oC. How many moles of N2(g) are there in the cylinder?

PV = nRT

𝐧=𝐏𝐕𝐑𝐓

=(𝟒 .𝟏𝟓𝒂𝒕𝒎)(𝟏𝟎 .𝟎𝑳)

(.𝟎𝟖𝟐𝟏 𝑳 ∙𝒂𝒕𝒎 ∙𝒎𝒐𝒍−𝟏𝑲 −𝟏)(𝟐𝟗𝟑𝑲 )= 1.72 moles N2(g)

Example 2

2KClO3(s) 2KCl(s) + 3O2(g)

• In the reaction above, 1.34 g of potassium chlorate is heated inside a container to yield oxygen gas and potassium chloride. The oxygen occupies 250 mL at 20.0 oC. What will the pressure of the gas be, in atmospheres?

We are solving for pressure in atm.V = .250 LT = 20 oC 293 oKn = ?R = 0.0821 L•atm•mol-1• K-1

PV = nRT

2KClO3(s) 2KCl(s) + 3O2(g)1.34 g

.0109 mol

.0109𝑚𝑜𝑙 𝐾𝐶𝑙𝑂3𝑥3𝑚𝑜𝑙𝑂2(𝑔)2𝑚𝑜𝑙𝐾𝐶𝑙𝑂3

=.𝟎𝟏𝟔𝟒𝒎𝒐𝒍𝑶𝟐(𝒈 )

• Before we can find P, we must find n

𝑃=𝑛𝑅𝑇𝑉

=( .𝟎𝟏𝟔𝟒𝒎𝒐𝒍 )( .𝟎𝟖𝟐𝟏 𝑳 ∙𝒂𝒕𝒎 ∙𝒎𝒐𝒍−𝟏𝑲−𝟏)(𝟐𝟗𝟑𝑲 )

(.𝟐𝟓𝟎 𝑳)=1.57atm

Partial Pressures in Gas Mixtures

• To this point, we have not discusses mixtures of gases, yet gaseous mixtures are an important part of everyday life

• For example, the air we breathe is a gas mixture 78% N2 (g)

21% O2 (g)

1 % Ar and other gases

• In a gas mixture, each gas exerts its own pressure. If we had a sample of air in a container, the total pressure would be equal to the sum of the partial pressures (Dalton’s Law)…

Ptotal = PN2 + PO2

+ PAr …..

Partial Pressures

• The total pressure can further be defined using the Ideal gas law

• From this expression, we can determine the mole fraction (xi) of each gas

tot

N

tot

N

tot

N

totArONtot

ArONtot

n

n

nV

RT

nV

RT

P

P

nV

RTnnn

V

RTP

V

RTn

V

RTn

V

RTnP

2

2

2

22

22

𝑃𝑂2

𝑃 𝑡𝑜𝑡

=𝑛𝑂2

𝑛𝑡𝑜𝑡

𝑃 𝐴𝑟

𝑃 𝑡𝑜𝑡

=𝑛𝐴𝑟

𝑛𝑡𝑜𝑡

XN2XO2

XAr

Partial Pressures

• Therefore, we see that the partial pressure of a gas is dictated by its mole fraction.

• So, if we know the moles of a given gas in a mixture, and the total moles of gas, we can easily calculate the partial pressure:

𝑃𝑁2=𝑥𝑁2

𝑃 𝑡𝑜𝑡

𝑃𝑂2=𝑥𝑂2

𝑃𝑡𝑜𝑡

𝑃 𝐴𝑟=𝑥𝐴𝑟 𝑃 𝑡𝑜𝑡

The sum of the mole fractions must equal 1. + + = 1𝒙𝑵 𝟐

𝒙𝑶𝟐𝒙 𝑨𝒓

Example• A 0.428 g sample of air at 2 atmospheres is contained with a vessel.

Using the mass percentages previously mentioned, what are the partial pressures of N2, O2, and Ar ?

• Find moles of each gas78% N2 .012 mol N2

21% O2 .0028 mol O2

1 % Ar .0001 mol Ar

• Calculate mole fractions, then partial pressures

.805

.188

𝑥𝐴𝑟=1−𝑥𝑁2−𝑥𝑂2

=.007

𝑃𝑁2=𝑥𝑁2

𝑃 𝑡𝑜𝑡= ( .805 ) (2𝑎𝑡𝑚 )=1.6 𝑎𝑡𝑚

𝑃𝑂2=𝑥𝑂2

𝑃𝑡𝑜𝑡=.376𝑎𝑡𝑚

𝑃 𝐴𝑟=𝑥𝐴𝑟 𝑃 𝑡𝑜𝑡=.014𝑎𝑡𝑚