Gases

The Gas Laws

Charles’ law

At constant pressure, the volume of a given mass of an ideal gas is directly proportional to its absolute temperature, in Kelvin. In other words, as the temperature of a gas increases, its volume also increases proportionally, and as the temperature decreases, the volume decreases proportionally, as long as the pressure remains constant.

The law stops working a very low temperatures < $-200\degree C$

Example:
A balloon containing 10 cm³ of air is heated from $200\degree K$ to $300\degree K$ what is the new volume?
$\implies V_f = V_i \times \frac{T_f}{T_i}$
$\implies V_f = 10cm^3 \times \frac{300\degree K}{200\degree K}$
$\implies V_f = 10cm^3 \times \frac32$
$\implies V_f = 15cm^3$

Avogadro’s law

The number of particles or moles in a gas is proportional to the volume of the gas. Don’t overthink it, more gas means more volume.

Example:
A balloon of size 10 cm³ contains 2 moles of oxygen to one more mole is adde what is the new volume?
$\implies V_f = V_i \times \frac{M_f}{M_i}$
$\implies V_f = 10cm^3 \times \frac32$
$\implies V_f = 15cm^3$

Boyle’s Law

The law states that the pressure of a gas is inversely proportional to its volume:
$P ∝ 1/V$
In other words, as the volume of a gas decreases, its pressure increases, and vice versa.

Example:
A syringe containing 50 cm³ starts a 1 atm of pressure and get compressed to 10cm³ what is the new pressure?
$\implies P_f = P_i \times \frac{V_i}{V_f}$
$\implies V_f = 1 \space atm\times \frac{50cm^3}{10cm^3}$
$\implies V_f = 5\space atm$

The Ideal Gas Law

Combination of previous 3 laws:
$V = \frac{nRT}{P}$ or $PV = nRT$, where:

• n = number of moles
• R = universal gas constant = 8.3145 J/mol K
• V = volume in liters
• T = absolute temperature in kelvin
• P = pressure in kpa

Stoichiometry Involving Gases

A mix of limiting reagents, balancing equations and the ideal gas law.

Example:
3 grams of hydrogen and an unknown amount of oxygen produce 16 grams of water at stp what is the volume of the initial gases combined?

To solve this problem, we need to use the limiting reagents and the ideal gas law.

First, let’s calculate the number of moles of water produced:

The molar mass of water (H2O) is 18.015 g/mol, so 16 grams of water is equal to:

16 g / 18 g/mol = 0.8887 mol

According to the balanced chemical equation for the reaction, 1 mole of hydrogen reacts with 1/2 mole of oxygen to produce 1 mole of water. Therefore, the number of moles of oxygen used in the reaction is:

0.8887 mol x 1/2 = 0.4444 mol

The number of moles of hydrogen used in the reaction is:

3 g / 2.016 g/mol = 1.4881 mol

So, the initial number of moles of oxygen and hydrogen must have been:

moles of oxygen = 1.4881 mol
moles of hydrogen = 0.4444 mol

Now, we can use the ideal gas law to calculate the volume of the initial gases combined. At STP (standard temperature and pressure), the temperature is 273.15 K and the pressure is 101.325. The ideal gas law is:

$V = nRT / P$

Plugging in the values, we get:

$V = \frac{(0.4444 mol + 1.4881 mol) \times 8.3145 J/mol\space K \times 273.15 K}{101.325 kpa}$

$V = 43.3$ Litres

Written with StackEdit.