If the temperature is doubled, the average velocity of a gaseous molecule increases by

The correct option is 1.4.

To understand why, let us look at the relationship between the average velocity of a gaseous molecule and its temperature.
The average velocity ($v_{\text{av}}$) of a gas molecule is given by the formula:
$v_{\text{av}}=\sqrt{\frac{8RT}{\pi M}}$
where:
- $R$ is the universal gas constant,
- $T$ is the absolute temperature in Kelvin,
- $M$ is the molar mass of the gas.

From this formula, we can see that for a given gas (where $R$, $M$, and $\pi$ are constants), the average velocity is directly proportional to the square root of its absolute temperature:
$v_{\text{av}}\propto \sqrt{T}$

Let the initial temperature be $T_1$ and the initial average velocity be $v_1$.
According to the problem, the temperature is doubled. Therefore, the new temperature $T_2$ is:
$T_2=2T_1$

Let the new average velocity be $v_2$. We can set up the ratio of the new velocity to the initial velocity:
$\frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1}}$

Substituting $T_2=2T_1$ into the equation:
$\frac{v_2}{v_1}=\sqrt{\frac{2T_1}{T_1}}=\sqrt{2}$

Thus, the new average velocity is:
$v_2=\sqrt{2}·v_1$

Since the value of $\sqrt{2}$ is approximately 1.414 (or 1.4 when rounded to one decimal place):
$v_2\approx 1.4·v_1$

Therefore, when the absolute temperature is doubled, the average velocity of a gaseous molecule increases by a factor of 1.4.