NVAEasyJEE 2025Pressure & Temperature Relation

JEE Physics 2025 Question with Solution

For a particular ideal gas, which of the following graphs represents the variation of mean squared velocity of the gas molecules with temperature?

Four graphs of mean squared velocity versus temperature labeled 1 to 4, where graph 3 is a straight line with positive slope through the origin.

Answer

Correct answer:3

Step-by-step solution

Standard Method

Given: We need the graph showing variation of mean squared velocity with temperature for an ideal gas.

Find: Which graph matches the relation between mean squared velocity and temperature.

For an ideal gas,

v2=3kTm\langle v^2 \rangle = \frac{3kT}{m}

where kk is the Boltzmann constant, TT is the temperature, and mm is the mass of a gas molecule.

This shows that v2\langle v^2 \rangle is directly proportional to TT.

So the graph of mean squared velocity versus temperature must be a straight line with positive slope passing through the origin.

Therefore, the correct graph is Graph 3.

Common mistakes

  • Confusing mean squared velocity with mean velocity or rms velocity. Here the relation is for v2\langle v^2 \rangle, which is directly proportional to TT, so the graph is linear.

  • Choosing a curved graph by assuming speed varies as T\sqrt{T}. That applies to rms speed vrmsv_{\text{rms}}, not to mean squared velocity. For v2\langle v^2 \rangle, use direct proportionality with TT.

Practice more Pressure & Temperature Relation questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Related questions