At what temperature will the total kinetic energy of 0.30 moles of $\mathrm{He}$ be same as the total…
At what temperature will the total kinetic energy of 0.30 moles of $\mathrm{He}$ be same as the total kinetic energy of 0.40 moles of Ar at $400 \mathrm{~K}$ ?
$400 \mathrm{~K}$
$300 \mathrm{~K}$
$273 \mathrm{~K}$
$533 \mathrm{~K}$
Solution
Kinetic energy (KE) $=\frac{3 n R T}{2}$
Here, $n=$ number of moles
$T=$ temperature
$R=$ rate constant
Now, kinetic energy of $\mathrm{He}=$ total $\mathrm{KE}$ of $\mathrm{Ar}$
Now, $\frac{\mathrm{KE} \text { of } \mathrm{He}}{\mathrm{KE} \text { of } \mathrm{Ar}}=\frac{(\text { mole of } \mathrm{He}) \times(\text { temp. of He })}{(\text { mole of Ar) } \times(\text { temp. of Ar })}$
$
\begin{aligned}
1 & =\frac{0.30 \times T_1}{0.40 \times 400} \Rightarrow T_1=\frac{0.40 \times 400}{0.30} \\
T_1 & =533 \mathrm{~K}
\end{aligned}
$
Hence, at temperature $533 \mathrm{~K}, \mathrm{KE}$ of $\mathrm{He}$ and Ar will be same