1. Introduction to the Gas Laws

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
These physical similarities allow us to establish mathematical laws to explain the behavior of gases. These gas laws all derive from the kinetic theory of gases

3. Kinetic Theory of Gases
Gas molecules are very spread out. Because of this, a volume of gas molecules is mostly empty space. Based on this, it is “assumed” that gas molecules do not interact with each other (e.g. no intermolecular forces of attraction). Therefore, they act as “independent” bodies of mass, which is why the physical behaviors are similar.
Gas molecules are in constant motion and collide frequently with each other and with the walls of the container. These collisions with the container are the source of the pressure. The collisions between molecules are elastic.
The kinetic energy of a gas is directly proportional to the temperature according to the equation below, where 𝑉 is the root mean square speed (square root of the average speed) and R is the gas constant:
1 2 𝑚 𝑉 2 = 3 2 𝑅𝑇

4. Gas Constant
Kinetic energy is directly proportional to temperature by the gas constant, R, which can be expressed as a molar quantity in terms of Joules or L•atm
𝐑=𝟖.𝟑𝟏𝟒 𝐉 𝐦𝐨𝐥 𝐊 𝐨𝐫 𝟎.𝟎𝟖𝟐𝟏 𝐋 𝐚𝐭𝐦 𝐦𝐨𝐥 𝐊

5. 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.

6. 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.

7. 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).

8. Atmospheric Pressure
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.
Closed, under vacuum h
Open end

9. Units of Pressure

10. 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 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:
𝐏 𝟏 𝐕 𝟏 = 𝐏 𝟐 𝐕 𝟐 (𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 𝐓)

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

12. 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?
* 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

13. 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 both the molarity and pressure of air in the lungs to decrease (LP).

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

15. 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.
Air flows out LP LP LP LP HP HP

16. 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

17. Charles’ Law
HOT
COLD

18. 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

19. 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

20. 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

21. Ideal Gases
Any gas that follows the ideal gas law is considered an ideal gas. These gases behave according to kinetic theory (assumes no intermolecular forces of attraction)
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.

22. 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, because it actuality, there are intermolecular attractions, but those deviations are reasonably small.

23. 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.

24. Example, continued. PV = nRT = 1.72 moles N2(g)
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 = oC = 293 oK
R = 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 𝐾
Rearrange the equation to solve for n.
𝐧= 𝐏𝐕 𝐑𝐓 = (𝟒.𝟏𝟓 𝒂𝒕𝒎)(𝟏𝟎.𝟎 𝑳) (.𝟎𝟖𝟐𝟏 𝑳∙𝒂𝒕𝒎∙𝒎𝒐 𝒍 −𝟏 𝑲 −𝟏 )(𝟐𝟗𝟑 𝑲)
= moles N2(g)

25. Example 2KClO3(s)  2KCl(s) + 3O2(g)
The reaction above is carried out at 2.1 atm and 70oC, and is known to proceed with 91% yield. If 250 mL of oxygen are produced, how many grams of potassium chlorate were consumed?
Find actual moles of oxygen! Use this to find mass of potassium chlorate.
𝑛= 𝑃𝑉 𝑅𝑇 → 𝑛= 2.1 𝑎𝑡𝑚 𝐿 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 𝐾 (343 𝐾) → 𝑚𝑜𝑙
𝑥 𝑔 𝐾𝐶𝑙 𝑂 3 𝑥 1 𝑚𝑜𝑙 𝐾 𝐶𝑙𝑂 𝑔 𝐾𝐶𝑙𝑂 3 𝑥 3 𝑚𝑜𝑙 𝑂 2 2 𝑚𝑜𝑙 𝐾𝐶𝑙𝑂 3 𝑥 0.91= 𝑚𝑜𝑙 𝑂 2
𝐱=𝟏.𝟔𝟕 𝐠

26. Example for you to work on at home. . . .
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. Given that the reaction generates 1.85L of chlorine gas, calculate the % yield.