JEE Main 2025 · 28 January · morning · Single correct · +4 / −1

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

  1. (a)
    Option (a)
  2. (b)
    Option (b)
  3. (c)
    Option (c)
  4. (d)
    Option (d)
Answer and solution

Answer (c)

Solution

Physics sir · 4 steps

Mean squared velocity is just kinetic energy in disguise, and kinetic energy goes straight with T.

  1. Start from kinetic theory. The average kinetic energy of a gas molecule is 1/2mv̅²̅ = 3/2kT. Rearranging gives the mean squared velocity directly.

    v̅²̅ = 3kT/m = 3RT/M

  2. Read the shape. For a particular gas, M is fixed and R is a constant. So v̅²̅ ∝ T, which is the form y = cx: a straight line through the origin.
  3. Check the options. Graph (a) has a peak, graph (b) bends downward, and graph (d) bends upward. Only graph (c) is a straight line starting from the origin.
  4. So. The variation of v̅²̅ with temperature is shown by graph (c).

The idea to keep. Write each quantity as a power of T: mean square speed goes as T, while rms, average and most probable speeds go as √T. The power tells you the shape of the graph.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Kinetic Theory of Gases

Asked about 130 times in JEE Main since 2020. This question is from 2025.