Q20 / 46 · NTA Abhyas (NEET) 2020 · Physics · Kinetic Theory of Gases

Temperature at which the kinetic energy of gas molecule is half of the value of kinetic energy at $27 ° C is$

A
$13.5 ° C$
B
$150 ° C$
C
$75 K$
Correct answer
D
$- 123 ° C$

Solution

Kinetic energy of a gas molecule

E = \frac{3}{2} k T

where

k

is Boltzmann's constant.

∴ E \propto T

or

\frac{E_{1}}{E_{2}} = \frac{T_{1}}{T_{2}} o r \frac{E}{(E / 2)} = \frac{300}{T_{2}}

or

T_{2} = 150 K

T_{2} = 150 - 273 = - 123 ℃

Hence, the correct answer is option: 'D'.

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