States of Matter and Gaseous StatehardNUMERICAL

At what temperature (in Kelvin) will the total translational kinetic energy of of Helium () be exactStates 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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