B3-1. Gas Pressure and Amount of Substance

1. Gas Pressure

Pressure is defined as the normal (perpendicular) force applied per unit area by a fluid on a surface. In a gas, this pressure arises from the macroscopic sum of billions of microscopic particle collisions with the walls of the container.

Pressure Formula:

$$p = \dfrac{F}{A}$$
  • $p$: Pressure in Pascals (Pa) or Newtons per square metre (N m⁻²)
  • $F$: Normal force (N)
  • $A$: Cross-sectional area (m²)

Example 1: Calculating Gas Pressure

Problem: A gas trapped in a cylinder pushes against a movable circular piston with a normal force of $450$ N. If the piston has a cross-sectional area of $0.015$ m², calculate the pressure exerted by the gas.

Solution:

  • Step 1: Identify the known variables.
    $F = 450$ N
    $A = 0.015$ m²
  • Step 2: Apply the Pressure formula.
    $$p = \dfrac{F}{A} = \dfrac{450}{0.015} = \mathbf{30,000 \text{ Pa}}$$ (or $30$ kPa)

2. Amount of Substance (Moles)

Because atoms and molecules are extraordinarily small, physicists and chemists use the "mole" to group them into manageable, measurable quantities.

The Mole ($n$):

One mole of any substance contains exactly $6.02 \times 10^{23}$ elementary entities (atoms, molecules, etc.). This number is known as Avogadro's Constant ($N_A$).

$$n = \dfrac{N}{N_A}$$
  • $n$: Number of moles (mol)
  • $N$: Total number of individual molecules/atoms
  • $N_A$: Avogadro's constant ($6.02 \times 10^{23}$ mol⁻¹)

Molar Mass ($M$ or $m_r$):

The mass of exactly one mole of a specific substance. It allows us to relate the physical mass of a gas sample to the number of moles.

$$n = \dfrac{m}{M}$$
  • $m$: Total mass of the substance (g or kg depending on units of M)
  • $M$: Molar mass (typically g mol⁻¹ or kg mol⁻¹)

Example 2: Moles from the Number of Molecules

Problem: A sealed glass bulb contains $1.505 \times 10^{24}$ molecules of nitrogen gas. How many moles of nitrogen are inside the bulb?

Solution:

  • Step 1: Identify the known variables.
    $N = 1.505 \times 10^{24}$ molecules
    $N_A = 6.02 \times 10^{23}$ mol⁻¹
  • Step 2: Apply the Avogadro formula.
    $$n = \dfrac{N}{N_A} = \dfrac{1.505 \times 10^{24}}{6.02 \times 10^{23}} = \mathbf{2.5 \text{ mol}}$$

Example 3: Moles from Total Mass

Problem: A party balloon is filled with $12.0$ g of helium gas. Given that the molar mass of helium is $4.0$ g mol⁻¹, calculate the number of moles of helium inside the balloon.

Solution:

  • Step 1: Identify the known variables.
    $m = 12.0$ g
    $M = 4.0$ g mol⁻¹
  • Step 2: Apply the Molar Mass formula.
    $$n = \dfrac{m}{M} = \dfrac{12.0}{4.0} = \mathbf{3.0 \text{ mol}}$$
  • Note: Because both the mass and molar mass were given in grams, there was no need to convert to kilograms—the ratio remains the same!