MixedJEE Physics · Original learning card5 original chapter questions

Temperature and Translational Kinetic Energy

Equipartition assigns (1/2)k_BT to each independent quadratic translational coordinate, so every ideal-gas molecule has mean translational energy (3/2)k_BT.

Why this shows up in the exam

Comparing gases at the same temperature · Converting energy per molecule to temperature · Testing kinetic-theory statements

Learn the idea

Absolute temperature measures average translational kinetic energy, independent of molecular identity. At equilibrium, light and heavy molecules have the same average translational energy, although the lighter ones move faster.

🧠 Memory hook: Same T means same average translational energy, not same speed.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • <K_trans> = (3/2) k_B T — mean translational energy per molecule at thermal equilibrium
  • K_trans,total = (3/2) nRT = (3/2)PV — total translational energy of n moles of an ideal gas

How to approach it

  1. 1Convert temperature to kelvin
  2. 2Decide per molecule or whole sample
  3. 3Use k_B or R consistently

Common slip-ups that cost marks

  • •Making average kinetic energy depend on molar mass
  • •Using Celsius as proportional temperature
  • •Counting rotational energy as wall-pressure energy

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Original chapter practice

Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.

Question 1 of 5

A gas has rms molecular speed 300 m/s at 300 K. What is its rms speed at 1200 K, assuming ideal behavior?

Take a timed JEE Physics sectional mock