Does Equipartition of Energy Mean K is Proportional to D and T?

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Equipartition of energy states that each degree of freedom contributes (1/2)(k_b)T to the energy of a molecule. The equation K = (D/2)(k_b)T suggests that the total energy K is proportional to both the number of degrees of freedom D and the temperature T. This implies that two objects at the same temperature can have different energy values K due to differing degrees of freedom. For example, molecules with six degrees of freedom will have twice the energy of those with three degrees of freedom at the same temperature. This highlights the relationship between energy, degrees of freedom, and temperature in thermodynamics.
Kenny Lee
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Equipartition of energy states that each degree of freedom contributes an amount of K to each molecule equal to (1/2)(k_b)T.

I wrote:
K = (D/2)(k_b) T, where D is the number of degrees of freedom.

Is this correct? Because if it is, then does it mean that K is proportional to both D and T?

In other words, for two cases where Temperature is the same, the value of D dictates its temperature.

Thank you.

//edit
Well actually, I'm more concerned with the fact that according to that equation, two objects can have the same temperature and yet, a different K value; because of a different D.

Lemme know if you don't understand; I'll try rewording.
 
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Kenny,

I think that you are exactly right. For a given temperature, molecules with 6 degrees of freedom would have twice as much energy as those with just 3 degrees of freedom, on average.

Best Regards,
Walter
 
thats strange.
Thanks.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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