Answer
See the detailed answer below.
Work Step by Step
a) We know that the average translational kinetic energy of a molecule is given by
$$K_{avg}=\frac{3}{2}k_BT$$
So it is obvious that $K_{avg}\propto T$, and hence when the temperature is doubled, the average translational kinetic energy will be doubled as well.
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b) We know that the rms speed of a molecule is given by
$$v_{\rm rms}=\sqrt{\dfrac{3k_BT}{m}}$$
So it is obvious that $v_{\rm rms}\propto \sqrt{T}$, and hence when the temperature is doubled, the rms speed will be increased by a factor of $\sqrt{2}$.
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c) We know that the mean free path is given by
$$\lambda=\dfrac{1}{4\sqrt{2}\pi (N/V)r^2}$$
We can see that non of the variables here as $N$, $V$, or $r$ depends on temperature.
Thus, when the temperature is doubled, the mean free path remains constant.
