1. Introduction to Gases
Lecture 6
CH 13

2. 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.

3. 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.

4. 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.

5. 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.

6. Atmospheric Pressure
Closed, no air
In the 17th century, scientists believed that the atmosphere had no weight.
To test this, Torricelli, a student of Galileo, invented the barometer, a glass tube that is closed on one end and filled with mercury. He then inverted the tube into a dish of mercury.
He observed that only some of the mercury flowed out of the tube. He also noticed that the amount was variable. Why?

7. Atmospheric Pressure h
Mercury in the dish experiences the downward force of earth’s atmosphere, which forces mercury up into the tube.
The flow of mercury out of the barometer stops when the atmospheric pressure outside is equal to the pressure caused by the weight of the mercury inside.
The height of the mercury column, h, must then be directly related to the atmospheric pressure.
As Torricelli walked the barometer up a mountain, the mercury level dropped. Why?
Open

8. Units of Pressure
Standard atmospheric pressure at sea level is the pressure sufficient to support a column of mercury 760 mm high. We define a unit of standard pressure called an atmosphere.
1 atm=760 mm Hg
1 atm= x Pa
1 mm Hg=1 torr

9. Gas Laws: Boyle’s Law
Robert Boyle was able to show that, at constant temperature, volume and pressure are inversely proportional.
𝑃 ∝ 1 𝑉
In other words, if we compress 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:
𝐏 𝟏 𝐕 𝟏 = 𝐏 𝟐 𝐕 𝟐 (𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 𝐓)

10. Boyle’s Law Low Pressure (P1) High Pressure (P2) Constant Temperature
𝑉 1
𝑉 2
Constant Temperature

11. Example
GP 1
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?
* 𝑽 𝟐 =𝟎.𝟔𝟓 𝑽 𝟏
𝑃 2 = 𝑃 1 𝑉 1 𝑉 𝑃 2 = 𝑃 1 𝑉 𝑉 𝑃 2 = 𝑃 1 𝑃 2 =1.54 𝑃 1
b. If V1 = 10L and P1 = 3.5 atm, solve.
𝑃 2 =5.39 atm

12. 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).

13. C.I.R.L .: Boyle’s Law
The pressure outside of the lung remains high (HP). Now, there is a pressure gradient. LP HP
Immediately, air flows from the HP region (atmosphere) to the LP region (lungs) until the pressures are equal.
Inhalation

14. C.I.R.L .: Boyle’s Law
When you exhale, the lungs shrink, increasing the air pressure (and air molarity). Now, the pressure inside the lung is higher than the pressure outside. The gradient is reversed.
Air flows out LP LP LP LP HP HP

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

16. Relationship between Volume and Temperature: Charles’s Law
The average kinetic energy of a gas is given by:
𝐸 𝐾 = 3 2 RT
where R is a value known as the gas constant, J mol-1 K-1
Since kinetic energy is equal to 𝑚 𝑉 2 , where m is the mass of a gas molecule in kg/mol, and 𝑉 is the velocity in meters/sec, we can rewrite the expression above as:
Increasing T causes the molecules to move faster and cause higher energy collisions. This causes the gas to expand, increasing the volume.
1 2 𝑚 𝑉 2 = 3 2 𝑅𝑇
GP2

17. 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.
𝑉 ∝𝑇
We can express this as:
𝑽 𝟏 𝑻 𝟏 = 𝑽 𝟐 𝑻 𝟐
Constant Pressure

18. Charles’ Law
𝑇 1
2 𝑇 1
Constant Pressure

19. 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
𝟏.𝟑 𝐋 𝟐𝟗𝟖 𝐨 𝐊 = 𝐕 𝐅 𝟐𝟎𝟎 𝐨 𝐊
𝐕 𝐅 =𝟎. 𝟖𝟕 𝐋
GP3

20. 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

21. The Ideal Gas Law PV = nRT
If we take the 3 previously discussed gas laws:
V α n (n = moles)  Avogadro’s Law
P α 1 𝑉  Boyle’s Law
V α T  Charles’s Law
We can combine these laws to obtain the IDEAL GAS LAW:
PV = nRT
In this equation, R is converted to 𝟎.𝟎𝟖𝟐𝟏 𝑳 𝒂𝒕𝒎 𝒎𝒐𝒍 𝑲 . Therefore, P is in atmospheres, and V is in liters.

22. 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.

23. Comparisons of Real Gases to the Ideal Gas Law Approximation at STP
Molar Volume at STP (L)
Ideal Gas
Cl2
CO2 He H2
Deviations from the ideal gas law exist, but those deviations are reasonably small.

24. 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.

25. Example, continued. PV = nRT
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
We are solving for moles.
V = 10.0 L
P = 4.15 atm.
T = 293 K
R = L•atm•mol-1• K-1
Rearrange the equation to solve for n.
GP 4
𝐧= 𝐏𝐕 𝐑𝐓 = (𝟒.𝟏𝟓 𝒂𝒕𝒎)(𝟏𝟎.𝟎 𝑳) (.𝟎𝟖𝟐𝟏 𝑳∙𝒂𝒕𝒎∙𝒎𝒐 𝒍 −𝟏 𝑲 −𝟏 )(𝟐𝟗𝟑 𝑲)
= moles N2(g)

26. Example 3 PV = nRT 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 L
T = 20 oC  293 oK
n = ?
R = L•atm•mol-1• K-1

27. Before we can find P, we must find n
2KClO3(s)  2KCl(s) + 3O2(g)
1.34 g
.0164 mol
.0109 mol
.0109 𝑚𝑜𝑙 𝐾𝐶𝑙 𝑂 3 𝑥 3 𝑚𝑜𝑙 𝑂 2 (𝑔) 2 𝑚𝑜𝑙 𝐾𝐶𝑙 𝑂 3 =.𝟎𝟏𝟔𝟒 𝒎𝒐𝒍 𝑶 𝟐 (𝒈)
𝑃= 𝑛𝑅𝑇 𝑉 = (.𝟎𝟏𝟔𝟒 𝒎𝒐𝒍)(𝟎𝟖𝟐𝟏 𝑳∙𝒂𝒕𝒎∙𝒎𝒐 𝒍 −𝟏 𝑲 −𝟏 )(𝟐𝟗𝟑 𝑲) (.𝟐𝟓𝟎 𝑳) =1.57 atm
GP 5