# Can the number of microscopic configurations of a gas be expressed in terms of its temperature?

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.

• 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 $dQ=\frac{3}{2}Nk_B dT$ – Javi Feb 22 at 17:04
• Heat is not a function of state. There is nothing like a change of heat of a body. – GiorgioP Feb 22 at 18:23