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?

Respuesta :

Answer:

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

Explanation:

The formula for average kinetic energy is:

[tex]K.E=\frac{3}{2}kT[/tex]

where,

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

T = temperature = 372 K

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

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

[tex]K.E=7.659\times 10^{-21}J[/tex]

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

Answer:

Answer:

The average kinetic energy of the molecule is, .

Explanation:

The formula for average kinetic energy is:

where,

k = Boltzmann’s constant =

T = temperature = 372 K

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

The average kinetic energy of the molecule is, .

Explanation: