Question 684: Thermodynamics & KTG

Question

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

Accepted Answer

Step 1: Recall the formula
For an ideal gas, the
mean kinetic energy per molecule
is proportional to absolute temperature:
$ \langle KE angle = \frac{3}{2} k_B T $
So:
$ \langle KE angle \propto T \quad (	ext{in Kelvin}) $
Step 2: Initial conditions
Given that at $T_1 = 399^\circ 	ext{C}$:
$ T_1 = 399 + 273 = 672 \, 	ext{K} $
The mean kinetic energy $E \propto T_1$.
Step 3: Required condition
We want $E/2$. Since KE is proportional to $T$:
$ \frac{E}{2} \propto \frac{T_1}{2} $
So the required temperature:
$ T_2 = \frac{T_1}{2} = \frac{672}{2} = 336 \, 	ext{K} $
Step 4: Convert back to Celsius
$ T_2(\,^\circ 	ext{C}) = 336 - 273 = 63^\circ 	ext{C} $
Ã¢ÂÂ Final Answer:
$ \boxed{63^\circ 	ext{C}} $
Correct option: D.

Answer

$336^{\circ} \mathrm{C}$

Answer

$276^{\circ} \mathrm{C}$

Answer

$123^{\circ} \mathrm{C}$

Answer

$63^{\circ} \mathrm{C}$

Thermodynamics & KTG

The mean kinetic energy of the molecules of an ideal gas at 399^{\circ} \mathrm{C} is ' E '. The temperature at which the mean kinetic energy of its molecules will be ' \mathrm{E} / 2 ', is

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