Ideal gases

Temperature – Cambridge IGCSE/A‑Level (9702)

Learning Objective

Develop a solid understanding of how temperature is defined, measured and related to molecular energy. By the end of this section you should be able to:

• Explain thermal equilibrium and describe a simple experimental demonstration.
• Convert between the Celsius and Kelvin scales (Fahrenheit is optional).
• Apply the concepts of specific heat capacity and specific latent heat in quantitative problems.
• State why the Kelvin scale is required for any gas‑law calculation (A‑Level extension).

1. Thermal Equilibrium

Definition: Two or more bodies are in thermal equilibrium when they are at the same temperature and no net heat flows between them over a measurable period of time.

• Heat always flows from the hotter body (higher temperature) to the colder body (lower temperature) until equilibrium is reached.
• Temperature determines the direction of heat flow; it is not the amount of heat itself.
• Experimental demonstration (calorimeter): Place two metal blocks of different initial temperatures in an insulated container and insert a thermometer. After a few minutes the thermometer reads a single temperature – the system has reached thermal equilibrium.
• Zero °C = freezing point of water at 1 atm; 0 K = absolute zero, the theoretical point at which molecular motion ceases.

2. Temperature Scales

• Celsius (°C) – based on the freezing (0 °C) and boiling (100 °C) points of water at 1 atm.
• Kelvin (K) – the absolute (thermodynamic) scale; 0 K = absolute zero. One kelvin has exactly the same magnitude as one degree Celsius (ΔK = Δ°C).
• Fahrenheit (°F) – optional, mainly used in the United States. It is not required for the IGCSE syllabus and is therefore placed in a side box.

Optional – Fahrenheit (extra)

Water freezes at 32 °F and boils at 212 °F. Conversion formulas are given in the next table for completeness.

3. Conversion Between Scales

General conversion formulas (required)

• From Celsius to Kelvin:
  \$T(\mathrm{K}) = T(^{\circ}\mathrm{C}) + 273.15\$
• From Kelvin to Celsius:
  \$T(^{\circ}\mathrm{C}) = T(\mathrm{K}) - 273.15\$

Optional Fahrenheit formulas

• °C → °F: \$T(^{\circ}\mathrm{F}) = \frac{9}{5}\,T(^{\circ}\mathrm{C}) + 32\$
• °F → °C: \$T(^{\circ}\mathrm{C}) = \frac{5}{9}\,\bigl[T(^{\circ}\mathrm{F}) - 32\bigr]\$

4. Thermodynamic Temperature & Kinetic Energy

Kelvin is defined so that temperature is directly proportional to the average translational kinetic energy of the particles in an ideal gas:

\$\langle E{\text{kin}}\rangle = \frac{3}{2}\,k{\!B}\,T\$

where \(k_{\!B}=1.38\times10^{-23}\ \mathrm{J\,K^{-1}}\) is Boltzmann’s constant. Because this relationship holds only when the zero point is absolute zero, Kelvin must be used in any equation that links temperature to molecular energy, such as the ideal‑gas law.

5. Specific Heat Capacity

Definition: The amount of heat required to raise the temperature of 1 kg of a substance by 1 K (or 1 °C). Symbol: \(c\). Units: J kg⁻¹ K⁻¹.

Heat‑energy equation:

\$Q = mc\Delta T\$

• \(Q\) – heat added or removed (J)
• \(m\) – mass of the substance (kg)
• \(c\) – specific heat capacity (J kg⁻¹ K⁻¹)
• \(\Delta T = T{\text{final}}-T{\text{initial}}\) (K or °C)

Reference Table – Specific Heat Capacities (selected)

Worked Example – Heating Water

How much heat is needed to raise 50 g of water from 20 °C to 80 °C?

1. Convert mass: \(m = 50\ \text{g} = 0.050\ \text{kg}\).
2. Temperature change: \(\Delta T = 80 - 20 = 60\ \text{K}\).
3. Apply \(Q = mc\Delta T\):
   

   \$Q = (0.050)(4180)(60) = 1.25\times10^{4}\ \text{J}\$

4. Result: \(Q \approx 12.5\ \text{kJ}\) of heat must be supplied.

6. Specific Latent Heat

Latent heat is the heat required to change the phase of a substance without changing its temperature.

• Fusion (melting) – \(L_{\!f}\): heat needed to melt 1 kg of solid.
• Vapourisation – \(L_{\!v}\): heat needed to vaporise 1 kg of liquid.

Heat‑energy equation for a phase change:

\$Q = mL\$

Units: J kg⁻¹.

Reference Table – Specific Latent Heats (water)

Worked Example – Melting Ice

How much heat is required to melt 200 g of ice at 0 °C?

1. Convert mass: \(m = 0.200\ \text{kg}\).
2. Apply \(Q = mL_{\!f}\):
   

   \$Q = (0.200)(3.34\times10^{5}) = 6.68\times10^{4}\ \text{J}\$

3. Result: \(Q \approx 66.8\ \text{kJ}\).

Worked Example – Boiling Water (Vapourisation)

How much heat is needed to convert 150 g of water at 100 °C into steam at the same temperature?

1. Convert mass: \(m = 0.150\ \text{kg}\).
2. Use the vapourisation value \(L_{\!v}=2.26\times10^{6}\ \text{J kg}^{-1}\).
   

   \$Q = mL_{\!v}= (0.150)(2.26\times10^{6}) = 3.39\times10^{5}\ \text{J}\$

3. Result: \(Q \approx 339\ \text{kJ}\).

7. Link to the Ideal‑Gas Law (A‑Level Extension)

Because the ideal‑gas equation uses absolute temperature, Kelvin must be used – this is the direct link to Syllabus 15 (Ideal gases).

The ideal‑gas equation is

\$pV = nRT\$

• \(p\) – pressure (Pa or atm)
• \(V\) – volume (m³ or L)
• \(n\) – amount of substance (mol)
• \(R\) – universal gas constant
• \(T\) – absolute temperature (K)

Two convenient values of \(R\):

• \(R = 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}\) (Pa·m³ units)
• \(R = 0.0821\ \mathrm{L\,atm\,mol^{-1}\,K^{-1}}\) (L·atm units)

Worked Example – A‑Level

Calculate the pressure exerted by 2.00 mol of an ideal gas confined in a 5.00 L container at 25 °C.

1. Convert temperature: \(T = 25 + 273.15 = 298.15\ \text{K}\).
2. Use \(R = 0.0821\ \mathrm{L\,atm\,mol^{-1}\,K^{-1}}\) (because \(V\) is in litres).
3. Apply the equation (solve for \(p\)):
   

   \$\$p = \frac{nRT}{V}

   = \frac{(2.00)(0.0821)(298.15)}{5.00}

   = 9.8\ \text{atm}\$\$

4. Convert to pascals if required: \(1\ \text{atm}=1.013\times10^{5}\ \text{Pa}\) → \(p \approx 9.9\times10^{5}\ \text{Pa}\).

8. Common Mistakes to Avoid

• Using °C or °F directly in equations that require absolute temperature – always convert to kelvin first.
• Mixing units (e.g., litres with the SI value of \(R\)). Choose the form of \(R\) that matches the volume units you are using.
• Confusing heat (energy) with temperature change – remember \(Q = mc\Delta T\) or \(Q = mL\) involve energy, not temperature.
• Assuming the numerical value of the gas constant is the same in all unit systems.
• Omitting the “no net heat flow” condition when describing thermal equilibrium.