Introduction to Gases

Ch 13Suggested HW: 1, 3, 15, 16, 31, 39, 40, 49, 53, 55

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. A volume of gas consists mostly of empty space. Thus, each gas atom/molecule behaves as if the others are not there.

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

• The mass of a 1m2 column of air extending through the entire atmosphere would be approximately 104 kg.

• The force exerted on the surface would be:

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

• Then, P = F/A = 105 N/m2 = 105 Pa

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

Atmospheric Pressure

• A barometer is shown to the left. It is composed of a glass tube that is open on one end and closed on the other. The tube is under vacuum (no air in tube).

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

• The flow stops when the atmospheric pressure is equal to the pressure inside the tube. Pressure inside the tube is caused by the weight of mercury. The height h is proportional to the atmospheric pressure.

h

Closed, under vacuum

Open

Units of Pressure

1 atm = 760 mm Hg

Gas Laws: Boyle’s Law

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

• 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 against the container walls, there will be more collisions per second in a smaller volume if the molecules are moving at the same speed.

• We can express this relationship as:

𝐏𝟏𝐕𝟏=𝐏𝟐𝐕𝟐(𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 𝐓)

High Pressure (P2)Low Pressure (P1)

Constant Temperature

Boyle’s Law

𝑉 1

𝑉 2

Example

a. A container sealed with a movable plug holds a gas that is initially at a volume V1 and pressure P1. Then, the volume of the system is decreased by 35%. What is P2 in terms of P1?

What do we know?*

b. If V1 = 10L and P1 = 3.5 atm, solve.

C.I.R.L .: Boyle’s Law

• Everyday, without thinking about it, you move nearly 8500 L of air in and out of your lungs, equating to about 25 lbs.

• The ability of the lungs to create pressure gradients (differences in pressure between two regions) is what allows us to breathe.

• When you inhale, your lungs expand (increased volume). This expansion causes the molarity of air in the lungs to decrease, yielding low pressure (LP).

• The pressure outside of the lung remains high (HP). Now, there is a pressure gradient.

LP LP

HP

HP

HP

HP• Immediately, air flows from the HP

region (atmosphere) to the LP region (lungs) until the pressures are equal.

Inhalation

C.I.R.L .: Boyle’s Law

• When you exhale, the lungs shrink, increasing the air molarity, and the air pressure. Now, the pressure inside the lung is higher than the pressure outside. The gradient is reversed.

HP HP

LP

LP

LP

LP• Air flows out

C.I.R.L .: Boyle’s Law

C.I.R.L. Boyle’s Law Applied to the Rules of Deep Sea Diving

1. Never hold your breath!• If you ascend even a few feet with air

in your lungs, the decrease in pressure will cause the air to expand in your lungs! POP!!

2. Ascend slowly!• The solubility of gas in fluid increases

with pressure. Air is 79% N2, and N2 is quite soluble in water. As you ascend, all of that N2 comes out of your blood. If this happens too fast…..POP!!

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)

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.

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.

Constant Pressure𝑽𝟏

𝑻 𝟏=𝑽 𝟐

𝑻 𝟐

𝑉 ∝𝑇

• We can express this as:

Charles’ Law

𝑇 1 2𝑇1Constant Pressure

Example

• At 25oC, the volume of the air in a balloon is 1.3L. Then, the balloon is cooled to -73 oC using N2 (L). What is the new volume of air?

𝑽 𝒊

𝑻 𝒊=𝑽 𝒇

𝑻 𝒇

𝟏 .𝟑𝑳𝟐𝟗𝟖𝒐𝑲

=𝑽 𝑭

𝟐𝟎𝟎𝒐𝑲

𝑽 𝑭=𝟎 .𝟖𝟕𝑳

C.I.R.L.: Charles’ Law, Boyle’s Law, and The Internal Combustion Engine

Gas Laws: Avogadro’s Law

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

𝑉 ∝𝑛 constant T, P

The Ideal Gas Law

• If we take the 3 previously discussed gas laws:

• V α n (n = moles) Avogadro’s Law• P 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.

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.

Example, continued.

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 3

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 = 0.821 L•atm•mol-1• K-1

PV = nRT

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

.0109 mol

.0109𝑚𝑜𝑙 𝐾𝐶𝑙𝑂3𝑥3𝑚𝑜𝑙𝑂2(𝑔)2𝑚𝑜𝑙 𝐾𝐶𝑙𝑂3

=.𝟎𝟏𝟔𝟒𝒎𝒐𝒍𝑶𝟐(𝒈 )

• Before we can find P, we must find n

.0164 mol

𝑃=𝑛𝑅𝑇𝑉 =

(.𝟎𝟏𝟔𝟒𝒎𝒐𝒍)(𝟎𝟖𝟐𝟏 𝑳∙𝒂𝒕𝒎 ∙𝒎𝒐𝒍−𝟏𝑲−𝟏)(𝟐𝟗𝟑𝑲 )( .𝟐𝟓𝟎 𝑳)

=1.57 atm

Group Example (KNOW HOW TO DO THIS!!)

• 18.7g of solid manganese (IV) oxide is added to 600 mL of an aqueous solution of 1.33M hydrochloric acid in a sealed container at standard temperature and pressure. The reaction produces aqueous manganese (II) chloride, water, and chlorine gas.

• a. Write a balanced reaction• b. Given that reaction generates 1.85L of chlorine gas, calculate the % yield.