Question

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

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

Answer 1

Answer: (C)

${27}^{\circ }C$

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

$\frac{E_1}{E_2}=\frac{\frac{3}{2}R{T}_1}{\frac{3}{2}R{T}_2}$

$\frac{1}{1/2}=\frac{600}{T_2}$

$T_2=300K$

$t={27}^{\circ }C$

Hence, the correct option is $\text{C}$