Question

At room temperature, an oxygen molecule, with mass of 5.31x10^-26kg , typically has a kinetic energy of about 6.21x10^-21J. How fast is it moving?

Answer 1

Answer:

V=483.63 m/s

Explanation:

Given that

mass ,m= 5.31 x 10⁻²⁶ kg

Kinetic energy KE= 6.21 x 10⁻²¹ J

As we know that the kinetic energy of the mass m with moving velocity V given as

$KE=\dfrac{1}{2}mV^2$  

Now by putting the values

$6.21\times 10^{-21}=\dfrac{1}{2}\times 5.31\times 10^{-26}\times V^2$

$V^2=\dfrac{2\times 6.21\times 10^{-21}}{5.31\times 10^{-26}}$

$V^2=233898.30$

$V=\sqrt{233898.30}\ m/s$

V=483.63 m/s

Therefore the velocity of the molecule at the room temperature will be 483.63 m/s.