Question

Calculate the kinetic energy of a molecule of a perfect gas at ${227}^{\circ }$C.

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

K.E. of one molecule = $\frac{3}{2}\frac{RT}{N}$
$\therefore K.E=\frac{3}{2}\times 1.38\times {10}^{-23}\times 500$
$K.E=1.035\times {10}^{-20}$ Joule