Ideal gases

Celsius (°C) 📏

The Celsius scale is based on the freezing point of water at 0 °C and the boiling point at 100 °C under standard atmospheric pressure. It’s the scale most people use daily.

• 0 °C = 32 °F (freezing point of water)
• 100 °C = 212 °F (boiling point of water)
• Used in most scientific contexts outside the USA.

Kelvin (K) 🌡️

Kelvin is the SI unit of temperature. It starts at absolute zero, the point where molecular motion stops.

• Absolute zero = 0 K = –273.15 °C
• 1 K ≈ 1 °C (difference only)
• Used in physics and chemistry because it eliminates negative numbers.

Conversion: $K = °C + 273.15$

Fahrenheit (°F) ❄️

The Fahrenheit scale is common in the United States. It sets the freezing point of water at 32 °F and the boiling point at 212 °F.

• 0 °F = –17.78 °C
• 100 °F = 37.78 °C (average human body temperature)

Conversion: $°F = (°C × 9/5) + 32$

Quick Conversion Table 🔄

The Ideal Gas Law

The behaviour of an ideal gas is described by the equation:

$$PV = nRT$$

• $P$ = pressure (Pa)
• $V$ = volume (m³)
• $n$ = amount of gas (mol)
• $R$ = ideal gas constant = 8.314 J mol⁻¹ K⁻¹
• $T$ = temperature (K)

Analogy: Think of gas molecules as people in a room. If you increase the room size ($V$) or add more people ($n$), the pressure ($P$) goes up unless you also increase the temperature ($T$) to keep them moving faster.

Assumptions of an Ideal Gas

1. Gas molecules are point particles with no volume.
2. There are no intermolecular forces (except perfectly elastic collisions).
3. All collisions are perfectly elastic.
4. Average kinetic energy depends only on temperature.

These assumptions hold well at high temperatures and low pressures.

Real Gas Deviations

At high pressures or low temperatures, real gases deviate from ideal behaviour.

• At high pressure, molecules occupy volume → reduces available space.
• At low temperature, attractive forces become significant → reduces pressure.

The Van der Waals equation corrects for these effects:

$$\left(P + \frac{a n^2}{V^2}\right)(V - nb) = nRT$$

Where $a$ accounts for attraction and $b$ for volume.

Example Problem 🧪

A sample of 2.0 mol of an ideal gas occupies 0.050 m³ at 300 K. What is its pressure?

1. Write the Ideal Gas Law: $PV = nRT$.
2. Insert known values: $P = \frac{nRT}{V} = \frac{(2.0)(8.314)(300)}{0.050}$.
3. Calculate: $P \approx \frac{4988.4}{0.050} = 99,768\ \text{Pa}$.
4. Convert to bar (1 bar = 10⁵ Pa): $P \approx 0.997\ \text{bar}$.

Answer: $P \approx 1.0 \times 10^5\ \text{Pa}$ or 1.0 bar.

Exam Tips 📚

• Always keep units consistent – Kelvin for temperature, m³ for volume, Pa for pressure.
• Remember $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ or $0.0821\ \text{L atm mol}^{-1}\text{K}^{-1}$ if using litres and atmospheres.
• Check if the problem asks for a real gas correction; if so, use Van der Waals parameters.
• For conversion questions, write the formula first, then plug in numbers.
• Use the “check your answer” step: does the magnitude make sense?