Answer:
It increases but less than double
Explanation:
As the temperature of a gas increase, the average kinetic energy of the gas increases. The kinetic energy of a gas is the thermal energy that the gas contains.
We know, the kinetic energy of an ideal gas is given by :
![$V_{avg} = \sqrt{\frac{8R}{\pi M}}$](https://tex.z-dn.net/?f=%24V_%7Bavg%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B8R%7D%7B%5Cpi%20M%7D%7D%24)
where, R = gas constant
T = absolute temperature
M = molecular mass of the gas
From the above law, we get
![$V_{avg} \propto \sqrt{T}$](https://tex.z-dn.net/?f=%24V_%7Bavg%7D%20%5Cpropto%20%5Csqrt%7BT%7D%24)
Thus, if we increase the temperature then the average kinetic energy of the ideal gas increases.
In the context, if the temperature of the ideal gas increases from 100°C to 200°C, then
![$\frac{(V_{avg)_2}}{(V_{avg)_1}} =\sqrt{\frac{T_2}{T_1}}$](https://tex.z-dn.net/?f=%24%5Cfrac%7B%28V_%7Bavg%29_2%7D%7D%7B%28V_%7Bavg%29_1%7D%7D%20%3D%5Csqrt%7B%5Cfrac%7BT_2%7D%7BT_1%7D%7D%24)
![$\frac{(V_{avg)_2}}{(V_{avg)_1}} =\sqrt{\frac{473.15}{373.15}}$](https://tex.z-dn.net/?f=%24%5Cfrac%7B%28V_%7Bavg%29_2%7D%7D%7B%28V_%7Bavg%29_1%7D%7D%20%3D%5Csqrt%7B%5Cfrac%7B473.15%7D%7B373.15%7D%7D%24)
![$\frac{(V_{avg)_2}}{(V_{avg)_1}} =\sqrt{1.26}$](https://tex.z-dn.net/?f=%24%5Cfrac%7B%28V_%7Bavg%29_2%7D%7D%7B%28V_%7Bavg%29_1%7D%7D%20%3D%5Csqrt%7B1.26%7D%24)
![$\frac{(V_{avg)_2}}{(V_{avg)_1}} =1.12$](https://tex.z-dn.net/?f=%24%5Cfrac%7B%28V_%7Bavg%29_2%7D%7D%7B%28V_%7Bavg%29_1%7D%7D%20%3D1.12%24)
![$(V_{avg})_2 = 1.12\ (V_{avg})_1$](https://tex.z-dn.net/?f=%24%28V_%7Bavg%7D%29_2%20%3D%201.12%5C%20%28V_%7Bavg%7D%29_1%24)
Therefore, ![$(V_{avg})_2 > (V_{avg})_1$](https://tex.z-dn.net/?f=%24%28V_%7Bavg%7D%29_2%20%3E%20%28V_%7Bavg%7D%29_1%24)
Thus the average kinetic energy of the molecule increases but it increases 1.12 times which is less than the double.
Thus, the answer is " It increases but less that double".