Thermodynamic Temperature Scale (Kelvin)

1. Definitions (Must Know)

A. Thermodynamic temperature scale (Kelvin scale)

• The thermodynamic temperature scale (the Kelvin scale) is an absolute temperature scale that is independent of the behaviour of any particular thermometric substance.
• The unit is the kelvin (K).
• Absolute zero: the lowest possible temperature on the thermodynamic scale, defined as 0 K (about -273.15°C).

B. Celsius vs kelvin

• Celsius temperature is written θ/°C.
• Thermodynamic temperature is written T/K.

2. Key Ideas (What Earns Marks)

• A thermodynamic temperature scale is absolute (it has an absolute zero) and is not tied to any single material property.
• Convert between Celsius and kelvin:
  • T/K = θ/°C + 273.15
• Temperature intervals are the same size:
  • Δ T/K = Δ θ/°C
• Use kelvin whenever you substitute into thermodynamics/ideal gas equations (e.g. pV = NkT).

3. Detailed Explanations

A. Why “thermodynamic” (and not just “centigrade”)?

A centigrade thermometer assumes the chosen thermometric property varies linearly with temperature. In practice, different thermometric substances (e.g., mercury vs alcohol) do not expand in exactly the same way, so different thermometers can give slightly different readings for the same physical situation.

The thermodynamic scale aims to be substance-independent, so “T” means the same physical temperature regardless of which thermometer you use.

B. Absolute zero (what it means in exams)

• Absolute zero is the zero point of the thermodynamic scale: T = 0 K.
• It corresponds to θ ≈ -273.15°C on the Celsius scale.
• You cannot have a negative thermodynamic temperature in this syllabus context: T starts from 0 K.

C. Conversion and “same size intervals”

The Kelvin scale is just the Celsius scale shifted by 273.15:

T/K = θ/°C + 273.15

Because it is a pure shift, a temperature difference of 1°C is the same size as a difference of 1 K: Δ T = Δ θ

4. Common Mistakes

• Using θ in Celsius directly inside gas/thermodynamics equations (convert to kelvin first).
• Writing the unit as “°K” (incorrect). The unit is K, not degree-K.
• Mixing up temperature with temperature change:
  • you add 273.15 when converting a temperature,
  • you do not add 273.15 to a temperature change.
• Treating Celsius as an absolute scale (it is not: 0°C is not “no thermal energy”).

5. Exam Tips

• Write temperatures in Kelvin when using thermodynamic/ideal gas equations.
• Use the approximate conversion T ≈ θ + 273 if the question only needs an estimate.
• If you later use Δ U = Q + W, keep track of signs and label “work done on” vs “work done by” clearly.

6. Worked Examples

Example 1: Convert Celsius to kelvinCore

Convert θ = 25°C to kelvin.

Show Answer

T = 25 + 273.15 = 298.15 K ≈ 298 K

Example 2: Convert kelvin to CelsiusCore

Convert T = 310 K to Celsius.

Show Answer

θ = 310 - 273.15 = 36.85°C ≈ 36.9°C

Example 3: Temperature changesCore

A gas warms from 18°C to 31°C. Find Δ T in kelvin.

Show Answer

Δ θ = 31 - 18 = 13°C And since Δ T = Δ θ: Δ T = 13 K

Example 4: Why kelvin matters in a gas-law calculationCore

At constant volume, a gas has pressure 100 kPa at 27°C. Estimate its pressure at 127°C.

Show Answer

Convert to kelvin: T₁ = 27 + 273 = 300 K, T₂ = 127 + 273 = 400 K

At constant volume for an ideal gas, p propto T: p₂/p₁ = T₂/T₁ p₂ = 100 × 400/300 ≈ 133 kPa

Example 5: Constant pressure heating (volume change)Core

An ideal gas has volume V₁ = 2.0 L at 20°C. It is heated at constant pressure to 80°C. Find the new volume.

Show Answer

Convert: T₁ = 20 + 273.15 = 293.15 K, T₂ = 80 + 273.15 = 353.15 K

At constant pressure for an ideal gas, Vpropto T: V₂/V₁ = T₂/T₁ V₂ = 2.0 × 353.15/293.15 ≈ 2.41 L

7. Mind Stretchers

Mind stretcher 1: Doubling absolute temperatureExtension

An object is at 20°C. Its thermodynamic temperature is doubled. Find the final temperature in °C.

Show Answer

Initial: T₁ = 20 + 273.15 = 293.15 K

Double it: T₂ = 2T₁ = 586.3 K

Convert back: θ₂ = 586.3 - 273.15 = 313.15°C ≈ 313°C

Mind stretcher 2: Why must T be in kelvin for gas laws?Extension

Explain why you must use thermodynamic temperature (kelvin) in equations like pV = NkT.

Show Answer

The constant k is defined so that pV is proportional to absolute temperature. If you use Celsius, the zero point is arbitrary (0°C is not absolute zero), so the proportionality would fail.

Using kelvin makes the relationship physically meaningful: T starts from 0 K at absolute zero.

Mind stretcher 3: Constant-volume gas thermometerExtension

A real route to an “absolute” scale is to use a gas at low pressure (close to ideal). At constant volume, pressure varies approximately linearly with thermodynamic temperature. Extrapolating the pressure–Celsius graph to zero pressure gives an intercept near -273.15°C (absolute zero).

8. Practice, Quiz and Next Step

Close your notes and use Thermodynamic Temperature Scale (Kelvin) in the supplied context below. This requires a constructed explanation or working, not recognition of an option.

Fresh context: An unfamiliar data set or physical system requires you to apply Thermodynamic Temperature Scale (Kelvin) while stating the model, regime and assumptions.

1. Retrieve: define thermodynamic temperature scale (kelvin) in your own words, including units, sign or conditions where relevant.
2. Represent: Choose and label an appropriate diagram, graph, table or symbolic model; derive or justify the relationship used.
3. Apply: Reach a conclusion, then evaluate it using units, uncertainty, a limiting case and one practical or modelling limitation.

Check the response before looking back

• The model, regime, coordinates and assumptions are explicit.
• The derivation or multi-step reasoning is visible rather than implied.
• The conclusion is tested against units, data quality and a limiting case.
• A practical control, uncertainty or model limitation is evaluated where applicable.

If one check fails, name that exact gap, revisit the matching explanation or worked example, and redo the task with different values or a different situation. Then use theA-Level Physics course hub orpractice browser for an independent re-test.