The average translational kinetic energy of $$N_2$$ gas molecules at _________ °C becomes equal to the K.E. of an electron accelerated from rest through a potential difference of 0.1 volt.
(Given $$k_B = 1.38 \times 10^{-23}$$ J K$$^{-1}$$) (Fill the nearest integer).

For any ideal gas, the average translational kinetic energy per molecule is given first:

$$\text{Average K.E. per molecule} = \dfrac{3}{2}\,k_B\,T,$$

where $$k_B$$ is the Boltzmann constant and $$T$$ is the absolute temperature in kelvin.

An electron that starts from rest and is accelerated through a potential difference $$V$$ acquires kinetic energy according to the work-energy principle. The work done on a charge $$q$$ in moving through a potential difference $$V$$ is

$$\text{K.E. gained} = q\,V.$$

For an electron, the magnitude of the charge is $$e = 1.6 \times 10^{-19}\ \text{C}$$. Given that the accelerating potential difference is $$V = 0.1\ \text{V}$$, we have

$$\text{K.E. of the electron} = eV = (1.6 \times 10^{-19}\ \text{C})(0.1\ \text{V}) = 1.6 \times 10^{-20}\ \text{J}.$$

According to the statement of the question, this electron kinetic energy must equal the average translational kinetic energy of one $$\mathrm{N_2}$$ molecule. Hence we equate the two expressions:

$$\dfrac{3}{2}\,k_B\,T = 1.6 \times 10^{-20}\ \text{J}.$$

Now we substitute the given value $$k_B = 1.38 \times 10^{-23}\ \text{J K}^{-1}$$ and solve algebraically for $$T$$:

$$T = \dfrac{2}{3}\,\dfrac{1.6 \times 10^{-20}}{1.38 \times 10^{-23}}\ \text{K}.$$

Simplifying step by step, we first divide the numerical factors:

$$\dfrac{1.6}{1.38} \approx 1.1594,$$

and we handle the powers of ten separately:

$$10^{-20}\, /\,10^{-23} = 10^{3}.$$

So the fraction becomes

$$\dfrac{1.6 \times 10^{-20}}{1.38 \times 10^{-23}} \approx 1.1594 \times 10^{3} = 1.1594 \times 10^{3}.$$

Multiplying by the factor $$\dfrac{2}{3}$$ gives

$$T = \dfrac{2}{3} \times 1.1594 \times 10^{3}\ \text{K}.$$

Carrying out the multiplication,

$$\dfrac{2}{3} \times 1.1594 \approx 0.7730,$$

and finally,

$$T \approx 0.7730 \times 10^{3}\ \text{K} = 773\ \text{K}.$$

To express this temperature in degrees Celsius, we use the relation

$$T\;({}^{\circ}\text{C}) = T\;(\text{K}) - 273,$$

so

$$T\;({}^{\circ}\text{C}) = 773 - 273 = 500^{\circ}\text{C}.$$

The question asks for the nearest integer value, which is already an integer. Hence, the correct answer is Option 500.