ated from rest through a potential difference of 'V' volts, we start by comparing their respective kinetic energies.

The kinetic energy of an electron accelerated through a potential difference V is given by:

$$ E_{\text{electron}} = eV $$

where:

$$e$$ is the electronic charge.
$$V$$ is the potential difference in volts.

For a molecule in an ideal gas, the average translational kinetic energy is given by:

$$ E_{\text{translational}} = \frac{3}{2} k_B T $$

where:

$$k_B$$ is the Boltzmann constant.
$$T$$ is the absolute temperature in Kelvin.

Now, we equate the kinetic energy of the electron to the average translational kinetic energy of a molecule in the gas:

$$ eV = \frac{3}{2} k_B T $$

Solving for temperature T, we get:

$$ T = \frac{2 eV}{3 k_B} $$

For a gas with number of molecules $$N$$ and using the relationship between Boltzmann constant $$k_B$$ and gas constant $$R$$ (i.e., $$k_B = \frac{R}{N}$$), we substitute this value into the equation:

$$ T = \frac{2 eV}{3} \cdot \frac{N}{R} $$

Simplifying, we obtain:

$$ T = \frac{2 eVN}{3R} $$

Thus, the correct option is:

Option A

$$\frac{2 \mathrm{eVN}}{3 \mathrm{R}}$$

Heat And Thermodynamics

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