Question

At sufficiently high temperatures, the thermal speeds of gas molecules may be high enough that collisions may ionize a molecule (that is, remove an outer electron). An ionized gas in which each molecule has lost an electron is called a “plasma.” Determine approximately the temperature at which air becomes a plasma.

Short answer

The temperature at which the air becomes plasma is $1.7\times {10}^5\mathrm{K}$.

Step-by-step solution

Defination of plasma.

The fourth state of the matter after solid liquid and Gas is the plasma state, in this state the atoms of the gas becomes Ionized and becomes an electrically conducting medium in which there are roughly equal numbers of positively and negatively chargedparticles.

Determination of the temperature at which air becomes plasma.

To determine the temperature at which the air becomes plasma, we have to use the ratio of energy and absolute temperature, therefore,

$\mathrm{E}=\mathrm{kT}$ ….(i)

Here, K is the Boltzman constant and T is the temperature, and E is the energy,

It is known that the constituents of the air are mainly Nitrogen and Oxygen. Therefore, from the periodic table, we get the ionization energy of Nitrogen and Oxygen as,

Nitrogen has $\mathrm{E}=14.53\mathrm{eV}$.

Oxygen has $\mathrm{E}=13.62\mathrm{eV}$.

By converting the eV unit to an SI unit, solve as:

For Nitrogen:

$\begin{array}{rcl}\mathrm{E}& =& 14.53\mathrm{eV}\\ & =& 14.53\times 1.6\times {10}^{-19}\mathrm{J}\\ & =& 23.246\times {10}^{-19}\mathrm{J}\end{array}$

For Oxygen:

$\begin{array}{rcl}\mathrm{E}& =& 13.62\times 1.6\times {10}^{-19}\\ & =& 21.648\times {10}^{-19}\mathrm{J}\end{array}$

From equation (i), write the expression for:

$\mathrm{T}=\frac{\mathrm{E}}{\mathrm{k}}$

Using the above equation solve as:

The temperature of nitrogen is as follows:

$\begin{array}{rcl}\mathrm{T}& =& \frac{23.246\times {10}^{-19}}{1.38\times {10}^{-23}}\\ & =& 16.8\times {10}^4\mathrm{K}\end{array}$

The temperature of Oxygen is as follows:

$\begin{array}{rcl}\mathrm{T}& =& \frac{21.648\times {10}^{-19}}{1.38\times {10}^{-23}}\\ & =& 15.7\times {10}^4\mathrm{K}\end{array}$

Therefore, to ionize the atmosphere, the temperature must reach a minimum $1.7\times {10}^5\mathrm{K}$.