Gas laws

Boyle's Law

Boyle's Law describes the relationship between the pressure and volume of a gas at constant temperature and number of moles. The law states that the pressure of a gas is inversely proportional to its volume. Mathematically, it is expressed as: $P1V1 = P2V2$
Where:

$P_1$ = Initial pressure
$V_1$ = Initial volume
$P_2$ = Final pressure
$V_2$ = Final volume

Example:
A gas occupies a volume of 10 L at a pressure of 2 atm. If the pressure is increased to 4 atm while keeping the temperature constant, what is the new volume?
$P1 = 2 atm$ $V1 = 10 L$
$P2 = 4 atm$ $V2 = ?$
Using Boyle's Law: $(2 atm)(10 L) = (4 atm)(V2)$ $V2 = \frac{(2 atm)(10 L)}{4 atm} = 5 L$

Charles's Law

Charles's Law describes the relationship between the volume and temperature of a gas at constant pressure and number of moles. The law states that the volume of a gas is directly proportional to its temperature. Mathematically, it is expressed as:
$\frac{V1}{T1} = \frac{V2}{T2}$
Where:

$V_1$ = Initial volume
$T_1$ = Initial temperature
$V_2$ = Final volume
$T_2$ = Final temperature

Example:
A gas occupies a volume of 5 L at 27°C. If the temperature is increased to 227°C while keeping the pressure constant, what is the new volume?
First, convert Celsius to Kelvin:
$T1 = 27 + 273.15 = 300.15 K$ $T2 = 227 + 273.15 = 500.15 K$
$V1 = 5 L$ $V2 = ?$
Using Charles's Law: $\frac{5 L}{300.15 K} = \frac{V2}{500.15 K}$ $V2 = \frac{(5 L)(500.15 K)}{300.15 K} = 8.33 L$

Temperature Unit

When solving gas law equations, the temperature must be in Kelvin (K). To convert Celsius (°C) to Kelvin (K), use the following formula:
$K = °C + 273.15$

Gay-Lussac's Law

Gay-Lussac's Law describes the relationship between the pressure and temperature of a gas at constant volume and number of moles. The law states that the pressure of a gas is directly proportional to its temperature. Mathematically, it is expressed as:
$\frac{P1}{T1} = \frac{P2}{T2}$
Where:

$P_1$ = Initial pressure
$T_1$ = Initial temperature
$P_2$ = Final pressure
$T_2$ = Final temperature

Example:
The pressure of a gas in a container is 3 atm at 25°C. If the temperature is increased to 50°C, what is the new pressure?
First, convert Celsius to Kelvin:
$T1 = 25 + 273.15 = 298.15 K$ $T2 = 50 + 273.15 = 323.15 K$
$P1 = 3 atm$ $P2 = ?$
Using Gay-Lussac's Law: $\frac{3 atm}{298.15 K} = \frac{P2}{323.15 K}$ $P2 = \frac{(3 atm)(323.15 K)}{298.15 K} = 3.25 atm$

Pressure Units

Different units are used to measure pressure:

Atmosphere (atm)
Pascal (Pa)
Kilopascal (kPa)
Millimeters of mercury (mmHg) or Torr

Conversions:

1 atm = 101.325 kPa
1 atm = 760 mmHg

Standard Temperature and Pressure (STP)

STP is defined as:

Temperature: 273.15 K (0 °C)
Pressure: 1 atm

Avogadro's Law

Avogadro's Law describes the relationship between the volume and the number of moles of a gas at constant temperature and pressure. The law states that the volume of a gas is directly proportional to the number of moles. Mathematically, it is expressed as:
$\frac{V1}{n1} = \frac{V2}{n2}$
Where:

$V_1$ = Initial volume
$n_1$ = Initial number of moles
$V_2$ = Final volume
$n_2$ = Final number of moles

Example:
A gas occupies a volume of 10 L with 2 moles. If the number of moles is increased to 4, what is the new volume, assuming constant temperature and pressure?
$V1 = 10 L$ $n1 = 2 moles$
$V2 = ?$ $n2 = 4 moles$
Using Avogadro's Law: $\frac{10 L}{2 moles} = \frac{V2}{4 moles}$ $V2 = \frac{(10 L)(4 moles)}{2 moles} = 20 L$

Combined Gas Law

The Combined Gas Law combines Boyle's, Charles's, and Gay-Lussac's laws into a single equation that relates pressure, volume, and temperature of a fixed amount of gas:
$\frac{P1V1}{T1} = \frac{P2V2}{T2}$
Where:

$P_1$ = Initial pressure
$V_1$ = Initial volume
$T_1$ = Initial temperature
$P_2$ = Final pressure
$V_2$ = Final volume
$T_2$ = Final temperature

Example:
A gas occupies a volume of 5 L at a pressure of 2 atm and a temperature of 27°C. If the pressure is changed to 4 atm and the temperature is changed to 54°C, what is the new volume?
First, convert Celsius to Kelvin:
$T1 = 27 + 273.15 = 300.15 K$ $T2 = 54 + 273.15 = 327.15 K$
$P1 = 2 atm$ $V1 = 5 L$
$T1 = 300.15 K$ $P2 = 4 atm$
$V2 = ?$ $T2 = 327.15 K$
Using the Combined Gas Law: $\frac{(2 atm)(5 L)}{300.15 K} = \frac{(4 atm)(V2)}{327.15 K}$ $V2 = \frac{(2 atm)(5 L)(327.15 K)}{(4 atm)(300.15 K)} = 2.72 L$

Ideal Gas Law

The Ideal Gas Law relates pressure, volume, temperature, and the number of moles of a gas:
$PV = nRT$
Where:

$P$ = Pressure
$V$ = Volume
$n$ = Number of moles
$R$ = Ideal gas constant
$T$ = Temperature

R values:

$R = 0.0821 \frac{L \cdot atm}{mol \cdot K}$
$R = 8.314 \frac{J}{mol \cdot K}$

Example:
What is the volume of 1 mole of a gas at STP?
At STP:
$P = 1 atm$
$n = 1 mole$
$R = 0.0821 \frac{L \cdot atm}{mol \cdot K}$
$T = 273.15 K$
Using the Ideal Gas Law: $(1 atm)(V) = (1 mole)(0.0821 \frac{L \cdot atm}{mol \cdot K})(273.15 K)$
$V = \frac{(1 mole)(0.0821 \frac{L \cdot atm}{mol \cdot K})(273.15 K)}{1 atm} = 22.4 L$

Dalton's Law

Dalton's Law of Partial Pressures states that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of the individual gases:
$P{total} = P1 + P2 + P*3 + …$

Example:
A container contains nitrogen gas at a pressure of 2 atm, oxygen gas at a pressure of 3 atm, and carbon dioxide at a pressure of 1 atm. What is the total pressure in the container?
$P{N2} = 2 atm$ $P{O2} = 3 atm$
$P{CO_2} = 1 atm$
Using Dalton's Law: $P{total} = 2 atm + 3 atm + 1 atm = 6 atm$

Graham's Law

Graham's Law describes the relationship between the rate of effusion or diffusion of a gas and its molar mass. The law states that the rate of effusion or diffusion is inversely proportional to the square root of its molar mass:
$\frac{Rate1}{Rate2} = \sqrt{\frac{M2}{M1}}$
Where: