For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Dominic Chukwuemeka

# Welcome to The Laws of Ideal Gases

I greet you this day,
Second: view the videos.
Third: solve the questions/solved examples.
Fourth: check your solutions with my thoroughly-explained examples.
Comments, ideas, areas of improvement, questions, and constructive criticisms are welcome. You may contact me.
If you are my student, please do not contact me here. Contact me via the school's system.
Thank you for visiting!!!

Samuel Dominic Chukwuemeka (Samdom For Peace) B.Eng., A.A.T, M.Ed., M.S

## The Laws and Equations of Ideal Gases

### Boyle's Law (Volume - Pressure Relationship @ Constant Temperature)

$\rightarrow$ Attributed to Robert Boyle
$\rightarrow$ Describes the relationship between the volume and the pressure of a gas at constant temperature
$\rightarrow$ States that at a constant temperature, the volume of a given mass of gas is inversely proportional to the pressure of the gas.

$V \:\:\:\alpha\:\:\: \dfrac{1}{P} \:\:\:@constant\:\:temperature...Inverse\:\:Proportion \\[5ex] V = k * \dfrac{1}{P} ... Equation \\[5ex] V = \dfrac{k}{P} \\[5ex] k = V * P \\[3ex] \rightarrow k = V_1 * P_1 = V_2 * P_2 \\[3ex] \therefore V_1P_1 = V_2P_2$

### Charles's Law (Volume - Absolute Temperature Relationship @ Constant Pressure)

$\rightarrow$ Attributed to Jacques Charles
$\rightarrow$ Describes the relationship between the volume and the absolute temperature of a gas at constant pressure
$\rightarrow$ States that at a constant pressure, the volume of a given mass of gas is directly proportional to the absolute temperature of the gas.

$V \:\:\:\alpha\:\:\: T \:\:\:@constant\:\:pressure...Direct\:\:Proportion \\[3ex] V = k * T ... Equation \\[3ex] k = \dfrac{V}{T} \\[5ex] \rightarrow k = \dfrac{V_1}{T_1} = \dfrac{V_2}{T_2} \\[5ex] \therefore \dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}$

The Absolute Temperature or Absolute Zero Temperature
According to Charles' Law, the volume of a gas becomes zero at $0K$ or $-273^oC$

### GayLussac's Law (Pressure - Absolute Temperature Relationship @ Constant Volume)

$\rightarrow$ Attributed to Joseph Gay-Lussac
$\rightarrow$ Also known as Pressure Law
$\rightarrow$ Describes the relationship between the pressure and the absolute temperature of a gas at constant volume
$\rightarrow$ States that at a constant volume, the pressure of a given mass of gas is directly proportional to the absolute temperature of the gas.

$P \:\:\:\alpha\:\:\: T \:\:\:@constant\:\:pressure...Direct\:\:Proportion \\[3ex] P = k * T ... Equation \\[3ex] k = \dfrac{P}{T} \\[5ex] \rightarrow k = \dfrac{P_1}{T_1} = \dfrac{P_2}{T_2} \\[5ex] \therefore \dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}$

### General Gas Equation

Combining the three laws:
Boyle's Law
Charles' Law and
Gay Lussac's Law/Pressure Law

We write the General Gas Equation as:

$V_1P_1 = V_2P_2 ...Boyle's\:\:Law \\[3ex] \dfrac{V_1}{T_1} = \dfrac{V_2}{T_2} ...Charles'\:\:Law \\[5ex] \dfrac{P_1}{T_1} = \dfrac{P_2}{T_2} ...Pressure\:\:Law \\[5ex] Combine\:\:them \\[3ex] \therefore \dfrac{V_1P_1}{T_1} = \dfrac{V_2P_2}{T_2} ...General\:\:Gas\:\:Equation$

### Standard Temperature and Pressure (S.T.P)

Because of Boyle's Law and Charles' Law
In other words; because the volume of a given mass of gas varies with the pressure and with the temperature,
Chemists had to establish a standard temperature and standard pressure (s.t.p) for the volumes of gases

The standard temperature is: $0^\circ C$ or $273K$

The standard pressure is: $1\:atm$ or $760 \:mmHg$ or $760\:torr$ or $101325.0164 \:N/m^2$ or $101325.0164 \:Pa$
The standard pressure is also: $76 \:cmHg$ or $7.6 \:dmHg$ or $0.76 \:mHg$ or $101.3250164 \:KPa$

### Avogadro's Law (Volume - Amount Relationship @ Constant Temperature and Pressure)

$\rightarrow$ Attributed to Amedeo Avogadro
$\rightarrow$ Describes the relationship between the volume and the amount of a given mass of gas in moles at constant temperature and pressure
$\rightarrow$ States that at a constant temperature and pressure, equal volumes of all gases contain the same number of moles (same number of molecules).

$V \:\:\:\alpha\:\:\: n \:\:\:@constant\:\:temperature\:\:and\:\:pressure...Direct\:\:Proportion \\[3ex] V = k * n ... Equation \\[3ex] k = \dfrac{V}{n} \\[5ex] \rightarrow k = \dfrac{V_1}{n_1} = \dfrac{V_2}{n_2} \\[5ex] \therefore \dfrac{V_1}{n_1} = \dfrac{V_2}{n_2} \\[5ex]$ NOTE: Based on Avogadro's Law
(1.) $Avogadro\:\:number = 6.02 * 10^{23} \:\:particles$
(2.) Particles could be: molecules, atoms, ions
(3.) $1 \:mole$ of any substance contains $1 \:Avogadro$ number
(4.) Substance could be: element, compound, solid, liquid, gas
(5.) $1 \:mole$ of any gas at s.t.p has a volume of $22.4 \:dm^3$ or $22.4 \:liters$.
This is also known as the $molar\:volume\:of\:gas\:at\:s.t.p$
$molar\:volume\:of\:gas\:at\:s.t.p = 22.4\:dm^3$

### Ideal Gas Equation

Combining the four laws:
Boyle's Law
Charles' Law
Gay Lussac's Law/Pressure Law and

We write the Ideal Gas Equation as:

$V \:\:\:\alpha\:\:\: \dfrac{1}{P} \:\:\:@constant\:\:temperature ...Boyle's\:\:Law \\[5ex] V \:\:\:\alpha\:\:\: T \:\:\:@constant\:\:pressure ...Charles'\:\:Law \\[3ex] P \:\:\:\alpha\:\:\: T \:\:\:@constant\:\:pressure ...Pressure\:\:Law \\[3ex] V \:\:\:\alpha\:\:\: n \:\:\:@constant\:\:pressure\:\:and\:\:temperature ...Avogadro's\:\:Law \\[3ex] Combine\:\:them \\[3ex] PV \:\:\:\alpha\:\:\: nT ...Proportion \\[3ex] PV = RnT ... Equation ...R\:\:is\:\:the\:\:constant\:\:of\:\:proportionality \\[3ex] R\:\:is\:\:known\:\:as\:\:the\:\:molar\:\:gas\:\:constant \\[3ex]$

$PV = RnT \\[3ex] R = \dfrac{PV}{nT} \\[5ex] From\:\:s.t.p \\[3ex] P = 101325.0164Pa \\[3ex] T = 273K \\[3ex] From\:\:Avogadro \\[3ex] n = 1mol \\[3ex] V = 22.4dm^3 = 0.0224m^3 \\[3ex] \rightarrow R = \dfrac{101325.0164 * 0.0224}{1 * 273} \\[5ex] R = \dfrac{2269.680367}{273} \\[5ex] R = 8.313847499 J/Kmol \\[3ex]$ Some students may ask you to explain how you arrived at that unit, and why you used those values to calculate the value of $R$

Explanation of the Units and Values used in finding the Molar Gas Constant
$1Pa \:(Pascal) = 1N/m^2 \\[3ex] 1J \:(Joule) = 1Nm \\[3ex] 10dm = 1m \\[3ex] \rightarrow 1000dm^3 = 1m^3 \\[3ex]$ We need to convert the $22.4dm^3$ to $m^3$ so we can use it...units have to be consistent

$22.4dm^3 = 0.0224m^3 \\[3ex] \therefore R = \dfrac{101325.0164 \dfrac{N}{m^2} * 0.224m^3}{1mol * 273K} \\[5ex] R = \dfrac{101325.0164Nm * 0.224}{1mol * 273K} \\[5ex] R = \dfrac{22696.80367Nm}{273Kmol} \\[5ex] 1Nm = 1J \\[3ex] R = 8.313847499 J/Kmol \\[3ex]$ Ask students to find the value of $R$ if the standard pressure is $1$ atmosphere
They should write the correct unit for $R$ in that case.
Compare with my result.

$PV = RnT \\[3ex] R = \dfrac{PV}{nT} \\[5ex] From\:\:s.t.p \\[3ex] P = 1atm \:(atmosphere) \\[3ex] T = 273K \\[3ex] From\:\:Avogadro \\[3ex] n = 1mol \\[3ex] V = 22.4dm^3 \\[3ex] \rightarrow R = \dfrac{1 * 22.4}{1 * 273} \\[5ex] R = \dfrac{22.4}{273} \\[5ex] R = 0.08205128205atmdm^3/Kmol \\[3ex]$ NOTE: Normally, you would be given the value of $R$.
Make sure that the units of the other quantities are consistent with that given value of $R$

### Dalton's Law of Partial Pressures

$\rightarrow$ Attributed to John Dalton
$\rightarrow$ Describes the relationship between the total pressure of a mixture of non-reactive gases and the pressures of the individual gases in the mixture
$\rightarrow$ States that the total pressure exerted by a mixture of non-reactive gases in a closed container is the sum of the partial pressures exerted by the individual gases in the mixture.

Assume there are three individual non-reactive gases: $A, B, C$ in a mixture of gases in a closed container
Based on Dalton's Law of Partial Pressures

$P_{total} = P_A + P_B + P_C \\[3ex] P_{total} = total\:\:pressure\:\:exerted\:\:by\:\:gases\:\:A,B,C \\[3ex] P_A = partial\:\:pressure\:\:of\:\:gas\:\:A \\[3ex] P_B = partial\:\:pressure\:\:of\:\:gas\:\:B \\[3ex] P_C = partial\:\:pressure\:\:of\:\:gas\:\:C \\[3ex] n_A = number\:\:of\:\:moles\:\:of\:\:gas\:\:A \\[3ex] n_B = number\:\:of\:\:moles\:\:of\:\:gas\:\:B \\[3ex] n_C = number\:\:of\:\:moles\:\:of\:\:gas\:\:C \\[3ex]$ Assume an individual gas is collected over water, then the gas is probably saturated with water vapor.
In that case; based on Dalton's Law of Partial Pressures
$P_{total}$ = pressure of the gas saturated with water vapour

$P_{total} = P_{dry\:gas} + P_{water\:vapour} \\[3ex] \rightarrow P_{dry\:gas} = P_{total} - P_{water\:vapour}$ For a mixture of unreactive gases in a container;
(1.) The volume occupied by one gas when it is alone in the container is the same volume occupied by the gas when is part of a mixture of the unreactive gases in the container
(2.) The temperature of one gas when it is alone in the container is the same temperature of that gas in a mixture of unreactive gases in the container
(3.) This means that the volume and the temperature of one gas in the mixture is the same volume and temperature of each of the gases in the mixture
(4.) However, the molar mass of each gas is different...because of the different relative atomic mass of the each gas
(5.) Because the molar mass of each gas is different, the number of moles of each gas is also different

For a mixture of unreactive gases: $A, B, C$ in a closed container
From the Ideal Gas Equation;

$PV = RnT \\[3ex] PV = nRT \\[3ex] P = \dfrac{nRT}{V} \\[5ex] R = constant \\[3ex] T_A = T_B = T_C = T \\[3ex] V_A = V_B = V_C = V \\[3ex] P_A = \dfrac{n_ART}{V} ...eqn.(1) \\[5ex] P_B = \dfrac{n_BRT}{V} ...eqn.(2) \\[5ex] P_C = \dfrac{n_CRT}{V} ...eqn.(3) \\[5ex] P_{total} = P_A + P_B + P_C ...Dalton's \:\:Law \\[3ex] \rightarrow P_{total} = P_A + P_B + P_C \\[3ex] P_{total} = \dfrac{n_ART}{V} + \dfrac{n_BRT}{V} + \dfrac{n_CRT}{V} \\[5ex] P_{total} = \dfrac{RT}{V}(n_A + n_B + n_C) ...eqn.(4) \\[5ex] n_A + n_B + n_C = n_{total} \\[3ex] P_{total} = \dfrac{RT}{V} * n_{total} ...eqn.(4) \\[5ex] From \:\:eqn.(1) \\[3ex] P_A = \dfrac{n_ART}{V} ...eqn.(1) \\[5ex] \dfrac{RT}{V} = \dfrac{P_A}{n_A} \\[5ex] Substitute \:\: \dfrac{P_A}{n_A} \:\:for\:\: \dfrac{RT}{V} \:\:in\:\: eqn.(4) \\[5ex] P_{total} = \dfrac{RT}{V} * n_{total} \\[5ex] P_{total} = \dfrac{P_A}{n_A} * n_{total} \\[5ex] Swap...Make \:\:LHS = RHS \:\:and\:\: RHS = LHS \\[3ex] \dfrac{P_A}{n_A} * n_{total} = P_{total} \\[5ex] \dfrac{P_A}{n_A} = \dfrac{P_{total}}{n_{total}} \\[5ex] \rightarrow P_A = n_A * \dfrac{P_{total}}{n_{total}} \\[5ex] \therefore P_A = \dfrac{n_A}{n_{total}} * P_{total} \\[5ex] Similarly \\[3ex] P_B = \dfrac{n_B}{n_{total}} * P_{total} \\[5ex] P_C = \dfrac{n_C}{n_{total}} * P_{total} \\[5ex]$ This means that the partial pressure of any individual gas is the product of the mole ratio of the gas (ratio of the number of moles of the individual gas to the total number of moles of the gaseous mixture) and the total pressure of the gaseous mixture.
This knowledge is useful for solving certain kinds of questions faster.

### Gay Lussac's Law of Combining Volumes

$\rightarrow$ Attributed to Joseph Gay-Lussac
$\rightarrow$ Describes the combining volumes of reacting gases (gaseous reactants) and gaseous products
$\rightarrow$ States that at a constant temperature and pressure, the volumes of gaseous reactants and gaseous products occur in simple whole number ratios.

### References

Chukwuemeka, S.D (2019, April 30). Samuel Chukwuemeka Tutorials - Math, Science, and Technology. Retrieved from https://www.samuelchukwuemeka.com

Ababio, O. Y. (2013). New School Chemistry (6th ed.). Onitsha, Nigeria: African First Publishers Plc.

Chemistry 1983-2004 JAMB Questions. (n.d.). Retrieved June 7, 2020, from https://www.vastlearners.com/wp-content/uploads/2018/12/chemistry.pdf ‌

Chemistry. (n.d.). Waeconline.Org.Ng. Retrieved June 7, 2020, from https://waeconline.org.ng/e-learning/Chemistry/chemmain.html