Welcome to PF, sac.lalit!
Your mistake is to treat dU=0 in isothermal process. If it were ideal gas with the usual gas law
p(V,T)=\frac{RT}{V}
you would be right, but this is not the case.
In fact, you have
V(p,T) = \frac{RT}{p} - \frac{C}{T^3}~~ \text{or}
p(V,T) = \frac{RT^4}{(VT^3+C)} \left(*\right)
and that's why the usual expression
dU = C_V dT
need to be changed (Note also that if C=0 in the above equation (*), you get the usual form of equation of state).
To change U in appropriate way, you must rederive it from the basical first law of thermodynamics in its form:
dS = \frac{dU}{T} + \frac{p}{T}dV \left( ** \right)
Now, with U=U(T,V) (caloric equation) you have:
dU = \left(\frac{\partial U}{\partial T}\right)_V dT + \left(\frac{\partial U}{\partial V}\right)_T dV
and after substituting this into (**):
dS = \frac{1}{T} \left(\frac{\partial U}{\partial T}\right)_V dT + \frac{1}{T} \left[ p + \left(\frac{\partial U}{\partial V}\right)_T \right] dV
From this equation:
\left(\frac{\partial S}{\partial T}\right)_V = \frac{1}{T} \left(\frac{\partial U}{\partial T}\right)_V
\left(\frac{\partial S}{\partial V}\right)_T = \frac{1}{T} \left[ p + \left(\frac{\partial U}{\partial V}\right)_T \right] dV
and remembering that
\left(\frac{\partial^2 S}{\partial V \partial T}\right) = \left(\frac{\partial^2 S}{\partial T \partial V}\right)
we finally get:
T \left(\frac{\partial p}{\partial T}\right)_V = \left(\frac{\partial U}{\partial V}\right)_T + p \left(***\right)From this equation we can derive U (caloric equation) if we know thermal equation of state p=p(V,T)
In fact, from (*):
\left(\frac{\partial p}{\partial T}\right)_V = \frac{RT^3 (VT^4 + C)}{(VT^3+C)^2}
and substituting this into (***)
\left(\frac{\partial U}{\partial V}\right)_T = \frac{3RCT^4}{(VT^3+C)^2}
Then
dU = \left(\frac{\partial U}{\partial T}\right)_V dT + \left(\frac{\partial U}{\partial V}\right)_T dV = C_V dT + \frac{3RCT^4}{(VT^3+C)^2}dV
or, using (*) for V(p,T) from which
\frac{dV}{dp} = -\frac{RT}{p^2}
and substituting the expression for p=p(V,T) we can rewrite
dU = C_V dT - \frac{3C}{T^3}dp
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Now, as you wrote, but using the new expression for dU:
dH = dU + pdV + Vdp = C_V dT - \frac{3C}{T^3}dp -\frac{RT}{p}dp + \frac{RT}{p}dp - \frac{C}{T^3}dp = C_V dT - \frac{4C}{T^3} dp
For isothermal process
dH = - \frac{4C}{T^3}dp
Now the factor 4 pops up
AND:
\Delta H_{12} = - \frac{4C}{T^3} \Delta p_{12}
