Question

At what temperature KE of a gas molecules is 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

The correct option is

C

${27}^{\circ }c$

Kinetic energy of a gas molecule

E = 3/2kT

where k is Boltzmann's constant.

∴E∝T

$\frac{E_1}{E_2}=\frac{T_1}{T_2}$

$\frac{E}{\frac{E}{2}}=\frac{600}{T_2}\Rightarrow T_2=300k$ or ${27}^{\circ }c$