Question

Find the expression for ds in terms of dT and dp.

Answer 1

Let s = f(T, p)

As per Maxwell relation 

\(\left(\cfrac{\partial s}{\partial p}\right)_T\) = - \(\left(\cfrac{\partial v}{\partial T}\right)_p\)

Substituting this in the above equation, we get

ds = \(\left(\cfrac{\partial s}{\partial T}\right)_p\) dT - \(\left(\cfrac{\partial v}{\partial T}\right)_p\) . dp ....(i)

The enthalpy is given by 

dh = c_pdT = Tds + vdp

Dividing by dT at constant pressure

\(\left(\cfrac{\partial h}{\partial T}\right)_p\) = c_p = T \(\left(\cfrac{\partial s}{\partial T}\right)_p\) + 0

(as dp = 0 when pressure is constant)

Now substituting this in eqn. (i), we get

Substituting this in eqn. (ii), we get

ds = c_p \(\cfrac{dT}T\) - βvdp