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Question

(D) The mean kinetic energy $(K)$ of an ideal gas molecule is given by the formula $K = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $

Vedclass · Accepted Answer

$63$

Vedclass · Answer

(D) The mean kinetic energy $(K)$ of an ideal gas molecule is given by the formula $K = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature in Kelvin.From this relation,we can see that $K \propto T$.Given,at $T_1 = 399^{\circ} C = (399 + 273) K = 672 K$,the kinetic energy is $E_1 = E$.We need to find the temperature $T_2$ at which the kinetic energy $E_2 = E/2$.Using the proportionality $E_1 / E_2 = T_1 / T_2$,we get:$E / (E/2) = 672 / T_2$$2 = 672 / T_2$$T_2 = 672 / 2 = 336 K$.To convert this back to Celsius: $T_2(^{\circ} C) = 336 - 273 = 63^{\circ} C$.Therefore,the correct option is $D$.

The mean kinetic energy of the molecules of an ideal gas at $399^{\circ} C$ is $E$. The temperature at which the mean kinetic energy of its molecules will be $E/2$ is: (in $^{\circ} C$)

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