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, .

Step-by-step 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, .

Step-by-step explanation:

Answer 2

The average kinetic energy of the molecule is,
`5.299* 10^(-21)J`.

Step-by-step explanation:

The formula for average kinetic energy is:

`K.E=(3)/(2)kT`

where,

k = Boltzmann’s constant =
`1.38* 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=(3)/(2)* (1.38* 10^(-23)J/K)* (256 K)`

`K.E=7.659* 10^(-21)J`

The average kinetic energy of the molecule is,
`5.299* 10^(-21)J`.