Question

At what temperature the kinetic energy of a gas molecule is half of the value at $ 27^\circ$ C?

• A. $ 13.5^\circ$ C
• B. $150^\circ$ C
• C. 75 K
• D. 13.5 K
• E. - $123^\circ$ C

Answer 1

Answer: (E)

Kinetic energy of a gas molecule
E = $ \frac{3}{2} kT $
where k is Boltzmann's constant.
$\therefore E \propto T $
or $ \frac{ E_1 }{ E_2 } = \frac{ T_1 }{ T_2 } $
or $ \frac{ E }{ (E / 2) } = \frac{ 300}{ T_2 } $
or $ T_2 = 150 $ K
$ T_2 = 150 - 273 $
= - $123^\circ$ C