Why is kinetic energy NOT proportional to degrees F/C

[babacanoosh]

Hello all,
I have a question, that regards a problem I was assigned. The problem is:

Why is kinetic energy NOT proportional to degrees F/C?

-I understand it has something to do with absolute zero Kelivn...not C. Thanks

Any help would be greatly appreciated...I need a response by tomorrow. Thanks again

 

[AbedeuS]

$$PV = \frac{1}{3}Nmv_{rms}^{2}$$
$$PV = nRT$$

Therefore:
$$\frac{1}{3}Nmv_{rms}^{2} = nRT$$

Rearranging:
$$v_{rms} = \sqrt{\frac{3nRT}{Nm}}$$

Since $$N = nN_{a}$$ and $$R = kN_{a}$$ it can be simplified to:

$$v_{rms} = \sqrt{\frac{3nN_{a}kT}{nN_{a}m}} = \sqrt{\frac{3kT}{m}}$$

Therefore:
$$v_{rms}\propto \sqrt{T}$$

EDIT: Similar looking results are found if you doing the same with Boltzmann curves or whatever else tickles your fancy.

 

[Blueshift5]

Hi there,

Kinetic energy is a measure of the energy an object possesses due to its motion. It is dependent on the mass and velocity of the object. Degrees Fahrenheit (F) and degrees Celsius (C) are units of temperature, which is a measure of the average kinetic energy of the particles in a substance.

While it is true that absolute zero, which is 0 Kelvin, represents the absence of any kinetic energy, the relationship between kinetic energy and temperature is not directly proportional. This is because temperature is a measure of the average kinetic energy of all the particles in a substance, while kinetic energy is a measure of the energy of a single object.

Additionally, the conversion between degrees F/C and Kelvin is not a simple linear relationship, as the zero points for each scale are different. F and C both have a zero point of the freezing point of water, while Kelvin has a zero point of absolute zero. This means that the same change in temperature in F/C may not result in the same change in kinetic energy as in Kelvin.

I hope this helps clarify why kinetic energy is not directly proportional to degrees F/C. If you have any further questions, please let me know. Good luck with your assignment!