Question

Consider a single molecule of ethanol (CH3CH2OH) at T = 256 K. Let it be moving at a speed 372 m/sec, which equals the average speed of a large collection of ethanol molecules at this temperature.What is the kinetic energy of this molecule?

Answer 1

Answer:

The average kinetic energy of the molecule is, $5.299\times 10^{-21}J$.

Explanation:

The formula for average kinetic energy is:

$K.E=\frac{3}{2}kT$

where,

k = Boltzmann’s constant = $1.38\times 10^{-23}J/K$

T = temperature = 372 K

Now put all the given values in the above average kinetic energy formula, we get:

$K.E=\frac{3}{2}\times (1.38\times 10^{-23}J/K)\times (256 K)$

$K.E=7.659\times 10^{-21}J$

The average kinetic energy of the molecule is, $5.299\times 10^{-21}J$.