At what temperature the kinetic energy of a gas molecule is half of the value at 27° C?

Question

At what temperature the kinetic energy of a gas molecule is half of the value at 27° C? (A)  13.5° C (B) 150° C (C) 75 K (D) 13.5 K

Tardigrade Admin · Accepted Answer

Kinetic energy of a gas molecule E = (3/2) kT where k is Boltzmann's constant. ∴ E propto T or ( E1 / E2 ) = ( T1 / T2 ) or ( E / (E / 2) ) = ( 300/ T2 ) or T2 = 150 K T2 = 150 - 273 = - 123° C

Q. At what temperature the kinetic energy of a gas molecule is half of the value at $2 7^{\circ}$ C?

$13. 5^{\circ}$ C

$15 0^{\circ}$ C

75 K

13.5 K

- $12 3^{\circ}$ C

Kinetic energy of a gas molecule
E = $\frac{3}{2} k T$
where k is Boltzmann's constant.
$∴ E \propto T$
or $\frac{E _{1}}{E _{2}} = \frac{T _{1}}{T _{2}}$
or $\frac{E}{( E /2 )} = \frac{300}{T _{2}}$
or $T_{2} = 150$ K
$T_{2} = 150 - 273$
= - $12 3^{\circ}$ C

Q. At what temperature the kinetic energy of a gas molecule is half of the value at 27∘ C?

13.5∘ C

150∘ C