Question

The kinetic energy of rotation of diatomic gas at ${27}^{\circ }$ will be $(k=1.38\times {10}^{-23}J/K)$

• A. $3.07\times {10}^{-21}\text{Joule/molecule}$
• B. $4.14\times {10}^{-21}\text{Joule/molecule}$
• C. $2.07\times {10}^{-27}\text{Joule/molecule}$
• D. $4.14\times {10}^{-21}\text{Joule/molecule}$

Answer 1

The correct option is B $4.14\times {10}^{-21}\text{Joule/molecule}$

We know that: $E=\frac{f}{2}kT$
Given, temperature of the gas,
$T={27}^{\circ }=(27+273)K=300K$
A diatomic gas molecule possesses two rotational degrees of freedom,
$f=2$
$\because$ total rotational kinetic energy
$=2\times \frac{1}{2}KT=kT$
$1.38\times {10}^{-23}\times 300$
$4.14\times {10}^{-23}\text{J/molecule}$
$4.14\times {10}^{-21}\text{J/molecule}$