Question

Singly charged gas ions are accelerated from rest through a voltage of 10.3 V. At what temperature (in K) will the average kinetic energy of gas molecules be the same as that given these ions?

Answer 1

Answer:

Temperature of the gas molecules is 7.96 x 10⁴ K

Explanation:

Given :

Ions accelerated through voltage, V = 10.3 volts

The work done to change the position of singly charged gas ions is given by the relation :

W = q x V

Here q is charge of the ions and its value is 1.6 x 10⁻¹⁹ C.

Average kinetic energy of gas molecules is given by the relation:

K.E. = $\frac{3}{2}kT$

Here T is temperature and k is Boltzmann constant and its value is 1.38 x 10⁻²³ J/K.

According to the problem, the average kinetic energy of gas is equal to the work done to move the singly charged ions, i.e. ,

K.E. = W

$\frac{3}{2}kT = qV$

Rearrange the above equation in terms of T :

$T= \frac{2qV}{3k}$

Substitute the suitable values in the above equation.

$T=\frac{2\times1.6\times10^{-19}\times10.3 }{3\times1.38\times10^{-23} }$

T = 7.96 x 10⁴ K