Question

Using relation for accessible state,(omega)=B(v-b)^Nf(E) here f(E) is function of energy only. Obtain equation of state p(v-b)=NkT

Answer 1

As per the given question,

Average potential energy of the molecules

$u=\dfrac{-a}{NV}$

Free volume per molecule, $V_f=\dfrac{(V-b)}{N}$

The molecular partition function,

$z=(\dfrac{2\pi mkT}{h^2})^{\frac{3}{2}}\int_o^{v_f}e^{\frac{-u}{kT}}dv$

$\Rightarrow z=(\dfrac{2\pi mkT}{h^2})^{\frac{3}{2}}(v_f)e^{\frac{-u}{kT}}$

Now, as per the definition, substituting the values,

$\ln z=\dfrac{3}{2}\ln(\dfrac{2\pi mkT}{h^2})^{\frac{3}{2}}+\ln (V-b)+\dfrac{a}{VNkT}$

Now, free energy equation can be written as,

$F=-NkT(\dfrac{3}{2}\ln(\dfrac{2\pi mkT}{h^2})^{\frac{3}{2}}+\ln (V-b))-\dfrac{a}{V}$

The pressure of such gas containing N molecule in volume V,

$P=NkT(\dfrac{\partial \ln z}{\partial V})_T=\dfrac{NkT}{V-b}-\dfrac{a}{V^2}$

if a=0, then

$P(V-b)=NkT$