Question

The average translational k.5 of air molecules is 0.040 ev $1ev=1.6\times {10}^{-19}J$. Calculate the temperature of the air.

• A. 200 k
• B. 400 k
• C. 310 k
• D. 420 k

Answer 1

The correct option is C 310 k

$K.E.=\frac{3}{2}kT0.040\times 1.6\times {10}^{-19}=\frac{3}{2}\times 1.38\times {10}^{-23}\times T\Rightarrow T=310k$