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(C) The average kinetic energy of a gas molecule is given by $K = \frac{f}{2} kT$,where $f$ is the degrees of freedom,$k$ is the Boltzmann constant,an

Vedclass · Accepted Answer

$327$

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(C) The average kinetic energy of a gas molecule is given by $K = \frac{f}{2} kT$,where $f$ is the degrees of freedom,$k$ is the Boltzmann constant,and $T$ is the absolute temperature in Kelvin.Since $f$ and $k$ are constants,the kinetic energy is directly proportional to the absolute temperature: $K \propto T$.Given the initial temperature $T_1 = 27^{\circ}\,C = 27 + 273 = 300\,K$.Let $K_1$ be the kinetic energy at $T_1$ and $K_2$ be the kinetic energy at temperature $T_2$.According to the problem,$K_2 = 2K_1$.Using the proportionality $K_1 / K_2 = T_1 / T_2$,we get:$1 / 2 = 300 / T_2$$T_2 = 600\,K$.Converting the temperature back to Celsius: $T_2 = 600 - 273 = 327^{\circ}\,C$.

The temperature at which the kinetic energy of oxygen molecules becomes double its value at $27^{\circ}\,C$ is $............^{\circ}\,C$.

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