Question

At what temperature will the total kinetic energy of $0.3$ mole of ${}^{\prime }H{e}^{\prime }$ be the same as that total kinetic energy of $0.4mole$ of argon at $400K$?

• A. $700K$
• B. $500K$
• C. $800K$
• D. $533.33K$

Answer 1

The correct option is D $533.33K$

$KE=\frac{3}{2}nRT$

According to question,
${(\frac{3}{2}\times 0.3\times R\times T)}_{Helium}={(\frac{3}{2}\times 0.4\times R\times 400)}_{Argon}$
$T=533.33K$