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Physics · Kinetic Theory of Gases

JEE Main 2025 — 7 April, Evening Shift — Question 53

The helium and argon are put in the flask at the same room temperature ( 300 K ). The ratio of average kinetic energies (per molecule) of helium and argon is (Give : Molar mass of helium =4 g/mol=4 \mathrm{~g} / \mathrm{mol}, Molar mass of argon =40 g/mol=40 \mathrm{~g} / \mathrm{mol} )

  1. Option A:

    1:101: 10

  2. Option B:

    10:110: 1

  3. Option C:

    1:101: \sqrt{10}

  4. Option D:

    1:11: 1

    Correct

Answer: D

Step-by-step solution

Average K.E. =32RT=\frac{3}{2} R T, is not depends on molar/molecular mass. so, (K.E.)He(K.E.)Ar=1\frac{(\mathrm { K.E. })_{\mathrm{He}}}{(\mathrm { K.E. })_{\mathrm{Ar}}}=1

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
7 April, Evening Shift
Subject
Physics
Chapter
Kinetic Theory of Gases
Topic
Equipartition Law of Energy and Degrees of Freedom
The helium and argon are put in the flask at the same room… | JEE Main 2025 PYQ with Solution · DhiX AI