as the gas molecules gain velocity from the rotating fan. Measurements of pressure is taken after the fan has stopped completely.
What will happen? Does the contained gas retains its pressure given constant temperature and volume?
Guess law?
It is likely that all kinetic energy imparted by the fan is lost through collisions with the containment walls in time with temperature held constant.
In that case, we postulate that the total kinetic energy of the gas is made up of,
\(KE_{total}=KE+KE_{T}\)
where \(KE_{T}\) is Thermal Kinetic Energy.
Thermal kinetic energy is the kinetic energy associated with temperature particles \(T^{+}\) caught in the orbit of electrons around the nucleus.
\(KE\) is the difference of \(KE_{T}\) from \(KE_{total}\). The origin of this part of kinetic energy is left opened for the time being.
A plot of \(P\) over time after the fan is switched off would indicate the effect of collisions on kinetic energy.
Why would the gas retain any kinetic energy? Is has always been assumed that a gas left to stand will have only kinetic energy associated with its temperature. Is this true?
A plot of \(P\) vs \(v^2\) with the fan at different speed settings might allow the pressure due solely to \(KE_{T}\) to be found when the graph y-intercept is extrapolated.
If a solid can exert pressure on the containment then pressure is not due to \(KE\) and collisions.
How much of the pressure is due to collisions and how much of the pressure is due to the distribution of temperature particles on the inner surface of the containment? If temperature particles exist.
My guess is, collisions are not involved in pressure without agitations. Pressure is due solely to the distribution of temperature particles in the rested state.
Good night...zzz..
