Gibbs Paradox
Storyboard 
If one has two identical systems and joins them together it has twice the volume and twice the number of particles. In this context, the internal energy of both systems must and is equal to the sum of that of each system separately. However, if the entropy is calculated, it turns out that the sum of the added system is different from the sum of the entropies of each system separately, which does not make sense. This contradiction is the so-called Gibbs paradox and its resolution has profound implications for how nature behaves. Their solution makes it necessary to accept that the particles of the systems that are being studied are indistinguishable, that is, they do not have something that makes them distinguishable.
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Gibbs Paradox
Definition 
Si se tiene un volumen de gas
| $ S = k_B N \left(\ln V + \displaystyle\frac{3}{2}\ln k_B T + \displaystyle\frac{3}{2}\ln\left(\displaystyle\frac{2 \pi m }{ h ^2}\right)+\displaystyle\frac{3}{2}\right)$ |
\\n\\nSi ahora consideramos un volumen del doble de tamaño, o sea de
$S=k_B2N\left(\ln 2V + \displaystyle\frac{3}{2}\ln k_BT + \displaystyle\frac{3}{2}\ln\left(\displaystyle\frac{2\pi m}{h^2}\right)+\displaystyle\frac{3}{2}\right)$
\\n\\nlo que no es igual a el doble de la entropía. El problema esta en que\\n\\n
$2k_BN\ln V \neq k(2N)\ln(2V)$
El problema de que la entropía no resulte extensible se denomina la paradoja de Gibbs y apunta a que en el calculo de la función partición se omitió un termino.
ID:(653, 0)
Gibbs Paradox
Storyboard 
If one has two identical systems and joins them together it has twice the volume and twice the number of particles. In this context, the internal energy of both systems must and is equal to the sum of that of each system separately. However, if the entropy is calculated, it turns out that the sum of the added system is different from the sum of the entropies of each system separately, which does not make sense. This contradiction is the so-called Gibbs paradox and its resolution has profound implications for how nature behaves. Their solution makes it necessary to accept that the particles of the systems that are being studied are indistinguishable, that is, they do not have something that makes them distinguishable.
Variables
Calculations
Calculations
Equations
Examples
La entrop a se defin a en base a el numero de estados con
$Z=\displaystyle\frac{1}{h^{3N}}\left(\displaystyle\frac{2\pi m}{\beta}\right)^{3N/2}V^N$
\\n\\ny como la energ a interna resulto\\n\\n
$U=\displaystyle\frac{3}{2}k_BNT$
se tiene que la entrop a de un gas ideal es con
Si se tiene un volumen de gas
$S=k_B2N\left(\ln 2V + \displaystyle\frac{3}{2}\ln k_BT + \displaystyle\frac{3}{2}\ln\left(\displaystyle\frac{2\pi m}{h^2}\right)+\displaystyle\frac{3}{2}\right)$
\\n\\nlo que no es igual a el doble de la entrop a. El problema esta en que\\n\\n
$2k_BN\ln V \neq k(2N)\ln(2V)$
El problema de que la entrop a no resulte extensible se denomina la paradoja de Gibbs y apunta a que en el calculo de la funci n partici n se omiti un termino.
Para resolver la paradoja de Gibbs se debe modificar el termino del volumen de modo de que en vez de ser un logaritmo del volumen sea un logaritmo del volumen dividido por el numero de part culas:\\n\\n
$\ln V\rightarrow \ln\displaystyle\frac{V}{N}$
\\n\\nya que en ese caso la duplicaci n del volumen y numero de part culas significar a que la entrop a se duplica del mismo modo. Si se introduce el factor de correcci n, la entrop a tendr a que tener un factor adicional del tipo
$S=kN\left(\ln V + \displaystyle\frac{3}{2}\ln kT + \displaystyle\frac{3}{2}\ln\left(\displaystyle\frac{2\pi m}{h^2}\right)+\displaystyle\frac{3}{2} - \ln N\right)$
\\n\\nlo que significar a que a la funci n partici n le faltar a un termino de la forma
$\ln N!=N\ln N-N$
se ve que una funci n partici n que incluya un factor
Por ello la funci n partici n es finalmente con
donde
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