Question

Derive the third Tds equation

Tds = cv \(\left(\cfrac{∂T}{∂p}\right)_v\) dp + c_p \(\left(\cfrac{∂T}{∂v}\right)_p\) dv

and also show that this may be written as

Tds = \(\cfrac{c_v}{\beta}\) Kdp + \(\cfrac{v_p}{\beta v}\) dv.

Answer 1

s = f(p, v)

Substituting these values in the above Tds equation, we get

Tds = \(\cfrac{c_vK}\beta dp\) + \(\cfrac{c_p}{\beta v }dv\)