Thermal physics: gases, systems and energy transfer
Key idea: Connect thermodynamic temperature and molecular motion to ideal-gas behaviour, internal energy, equilibrium, work, the thermodynamic laws and thermal-property energy balances.
Before you start: Energy & Fields objective chainQuantities & Measurement objective chain
By the end, you can
- Use the absolute thermodynamic scale and ideal-gas equations with particle and mole quantities.
- Apply the kinetic model to derive gas pressure and relate temperature to mean translational kinetic energy.
- Distinguish internal energy, temperature and heating, and explain thermal equilibrium.
- Apply work sign conventions and the zeroth and first laws without changing convention mid-solution.
- Use specific heat capacity and specific latent heat in thermal energy balances.
Starting-point self-check
1. Check your starting point
Attempt all six groups without notes and mark the first scale, particle-count, collision, energy-store or sign decision you cannot justify. Use the recorded topic diagnostic above when you want scoring and a personalised repair plan.
Thermodynamic temperature and Celsius conversion 12(a)–(b)
Question 1
A sample is at 27.0 °C. Convert it to kelvin and explain why doubling 27.0 °C does not double thermodynamic temperature.
Check the model response
T = 27.0 + 273.15 = 300.15 K. Thermodynamic temperature has its zero at absolute zero and is independent of a particular thermometric substance; doubling must therefore use kelvin, not the offset Celsius reading.
repair
2. Repair the common breaks
Use only the correction matching an error, then retry the corresponding diagnostic.
Thermodynamic temperature and Celsius conversion 12(a)–(b)
Check this idea
Misconception: Zero degrees Celsius is absolute zero.
Repair: Absolute zero is 0 K or −273.15 °C; 0 °C is 273.15 K.
worked example
3. Follow six worked models
Follow how each solution fixes the scale, gas amount, collision axis, system boundary or work convention before calculating.
Thermodynamic temperature and Celsius conversion 12(a)–(b)
Model 1
A resistance thermometer suggests a temperature of −40.0 °C. Convert it to kelvin and compare a 10 °C interval with a 10 K interval.
Check the model response
T = −40.0 + 273.15 = 233.15 K. Celsius and kelvin have equal-sized intervals, so a temperature change of 10 °C is exactly 10 K even though their zero points differ.
guided practice
4. Guided practice
Use each hint only to choose the governing definition, equation or sign convention.
Thermodynamic temperature and Celsius conversion 12(a)–(b)
Question 1
A gas changes from 250 K to 310 K. Give both Celsius temperatures and the temperature change.
Hint: Subtract 273.15 for readings; offsets cancel in a difference.
Check the model response
The readings are −23.15 °C and 36.85 °C. The change is 60 K or 60 °C; do not add 273.15 to an interval.
independent practice
5. Independent practice
Solve without repair notes and state every idealisation, system boundary and sign convention used.
Thermodynamic temperature and Celsius conversion 12(a)–(b)
Question 1
Explain the two defining features of the thermodynamic scale and convert 68.0 °C to kelvin.
Check the model response
It is absolute, with zero at absolute zero, and substance-independent. T = 68.0 + 273.15 = 341.15 K.
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
Thermodynamic temperature and Celsius conversion 12(a)–(b)
Question 1
Define thermodynamic temperature, locate absolute zero and convert −18.0 °C to kelvin.
Check the model response
It is an absolute, substance-independent temperature scale; absolute zero is 0 K. T = 255.15 K.
Re-test practice
7. Delayed re-test practice
Return after at least three days and solve these fresh contexts without reopening earlier responses. The recorded plan enforces the delay and uses a separate re-test family for selected-response skill-group evidence.
Thermodynamic temperature and Celsius conversion 12(a)–(b)
Question 1
Convert 315.0 K to Celsius and explain why kelvin has no degree sign.
Check the model response
θ = 315.0 − 273.15 = 41.85 °C. Kelvin is the SI thermodynamic-temperature unit on an absolute scale, written K rather than °K.