At what temperature (in Kelvin) will the total translational kinetic energy of of Helium () be exact — States of Matter and Gaseous State Chemistry Question
Question
At what temperature (in Kelvin) will the total translational kinetic energy of $0.30\text{ mole}$ of Helium ($He$) be exactly the same as the total translational kinetic energy of $0.40\text{ mole}$ of Argon ($Ar$) at $300\text{ K}$?
Answer: 400
💡 Solution & Explanation
Total kinetic energy $KE = (3/2) nRT$. We equate the two: $KE_{He} = KE_{Ar} \implies (3/2) \times 0.30 \times R \times T = (3/2) \times 0.40 \times R \times 300$. Simplifying: $0.30 \times T = 0.40 \times 300 = 120$. Therefore, $T = 120 / 0.30 = 400\text{ K}$.
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