Enthelpy change and Entropy change; given the equation for the substance

[sac.lalit]

Homework Statement

Equation for the substance given for a certain range of pressure and temperature:

v=(RT/p) - C/T³

Where C=Constant

Show that
(i) The change of enthalpy for this substance in an isothermal process is

(h₂-h₁)_T=4C/T³(p₁-p₂)_T

(ii) The change of entropy for this substance in an isothermal process is

(S₂-S₁)_T=Rln(p₁/p₂)+3C/T⁴(p₁-p₂)_T

The Attempt at a Solution

taking h=U+pv

dh=du+d(pv) {du=0 for isothermal process}

dh=d(pv)=-Cdp/T³

but result which i need to show has factor 4 also; which i have no idea how to get.

 

[physicsworks]

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
$$dU = C_V dT - \frac{3C}{T^3}dp$$
________________________________________________________________

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:
$$\Delta H_{12} = - \frac{4C}{T^3} \Delta p_{12}$$