Question

The average translational kinetic energy of air molecules is 0.040 eV (1 eV = 1.6 × 10−19 J). Calculate the temperature of the air. Boltzmann constant k = 1.38 × 10−23 J K−1.

Answer 1

Answer:

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Answer 2

The average translational kinetic energy of air molecules is 0.040 eV (1 eV = 1.6 × 10−19 J. Then the temperature of the air is  308.8 K  

Explanation:

Step 1:

Given data in the question :-  

Molecules air  is 0.040 eV (1 eV = 1.6 × 10−19 J).

Boltzmann's  constant i.e k = 1.38 × 10−23 J K−1.

Step 2:

We learn from the cinematic gas theory that the average translational energy per molecule is   $\frac{3}{2} K T$

$\begin{aligned}&\mathrm{E}_{\mathrm{a} \mathrm{Wg}}=0.040 \mathrm{eV}=0.040 \times 1.6 \times 10^{-19}=6.4 \times 10^{-21} \mathrm{J}\\&6.40 \times 10^{-21}=32 \times 1.38 \times 10^{-23} \times T\end{aligned}$

Step 3:

To find T

$\begin{aligned}&I=\frac{2}{3} \times \frac{6.40 \times 10^{-21}}{1.38 \times 10^{-23}}\\&T=0.666 \times \frac{6.40 \times 10^{2}}{1.38}\\&T=0.666 \times 4.63 \times 10^{2}\\&T=3.088 \times 10^{2}\\&T=308.8 K\end{aligned}$  

Thus the temperature is 308.8 K