Question

At what temperature is the kinetic energy of a gas molecule half of its value at ${327}^{\circ }\text{C}$?

• A. ${13.5}^{\circ }\text{C}$
• B. ${150}^{\circ }\text{C}$
• C. ${27}^{\circ }\text{C}$
• D. $-{123}^{\circ }\text{C}$

Answer 1

The correct option is C ${27}^{\circ }\text{C}$

Average kinetic energy of gas particles $(K.E)$ $=\frac{3}{2}RT$
$\frac{(K.E{)}_1}{(K.E{)}_2}=\frac{\frac{3}{2}R{T}_1}{\frac{3}{2}R{T}_2}$, where $(K.E.{)}_1$ and $(K.E.{)}_1$ are the kinetic energies at $T_1$ and $T_2$.
According to the question:
$\frac{1}{0.5}=\frac{600}{T_2}$
$T_2=300\text{K}$
$\Rightarrow t={27}^{\circ }\text{C}$