Question

Calculate the temperature when total $K.E.$ of $0.3\text{mol}$ of $He$ be the same as the total $K.E.$ of $0.4\text{mol}$ of $Ar$ at $400\text{K}$?

• A. $346\text{K}$
• B. $500\text{K}$
• C. $600\text{K}$
• D. $533\text{K}$

Answer 1

The correct option is D $533\text{K}$

We know total $K.E.$ is given as:
$K.E{.}_{\text{Total}}=\frac{3}{2}nRT$
According to question
${\left[\frac{3}{2}nRT\right]}_{He}={\left[\frac{3}{2}nRT\right]}_{Ar}$

$\Rightarrow 0.3T=0.4\times 400$
$\Rightarrow T=533\text{K}$