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I'm new to the field of thermodynamics, but have been thinking about it recently, and wondered whether this statement has any truth in it. I doubt it is true, however I need to see whereabouts the mistake is.

A definition of entropy says that the change in entropy is equal to a change in heat over temperature.

$dS=dQ/T$

For an ideal gas, I make the assumption that a change in heat is proportional to a change in temperature, (by a value I think is the product of the number of particles and the Boltzmann constant).

$dQ=Nk_B dT$

Substituting this into the first equation, gives:

$dS=(Nk_BdT)/T$

If we integrate this, then we find:

$S=Nk_B. ln(T)+C$

This can be manipulated to give:

$S=k_B. ln(mT^N)$

Where $k_Bln(m)=C$

Another definition of entropy says that the entropy is proportional (by the Boltzmann constant) to the natural log of the number of microscopic configurations of the particles.

$S=k_B. ln(\Omega)$

Equating the two definitions, I find that:

$k_Bln(mT^N)=k_Bln(\Omega)$

This leads me to believe that the number of configurations can be expressed as:

$\Omega=mT^N$

I find this hard to believe, however, and so I would quite like to ask where I have gone wrong. Thank you for reading, and I give my sincerest apologies for any errors I have made along the way.

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    $\begingroup$ This is actually a well-known result! you would have obtained the proper scaling law in three dimensions, $\Omega \sim T^{\frac{3}{2}N}$, using the proper proportional constant $dQ=\frac{3}{2}Nk_B dT$ $\endgroup$ – Javi Feb 22 at 17:04
  • $\begingroup$ Heat is not a function of state. There is nothing like a change of heat of a body. $\endgroup$ – GiorgioP Feb 22 at 18:23

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