Question 709: Thermodynamics & KTG

Question

Two gases A and B are at absolute temperatures 350 K and 420 K respectively. The ratio of average kinetic energy of the molecules of gas $B$ to that of gas A is

Accepted Answer

The average kinetic energy of a gas molecule is given by:
$$ 	ext{Average kinetic energy} = \frac{3}{2}kT $$
where $k$ is the Boltzmann constant and $T$ is the absolute temperature.
Let $T_A = 350\,	ext{K}$ (for gas A) and $T_B = 420\,	ext{K}$ (for gas B).
The ratio of average kinetic energies:
$ \frac{	ext{Average KE of B}}{	ext{Average KE of A}} = \frac{\frac{3}{2}kT_B}{\frac{3}{2}kT_A} = \frac{T_B}{T_A} $
Substitute the values:
$ \frac{T_B}{T_A} = \frac{420}{350} = \frac{42}{35} = \frac{6}{5} $
Therefore, the ratio is $6:5$.
Correct answer: Option A

Answer

$6: 5$

Answer

$\sqrt{6}: \sqrt{5}$

Answer

$36: 25$

Answer

$5: 6$

Thermodynamics & KTG

Two gases A and B are at absolute temperatures 350 K and 420 K respectively. The ratio of average kinetic energy of the molecules of gas B to that of gas A is

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