Now in order for d u ( v , T ) to be an exact differential, namely that u ( v , T ) be continuous with continuous partial derivatives, its second mixed partial derivatives must also be equal, ∂ ∂ v T 2 u = ∂ ∂ T v 2 u . Then with c v = ∂ ∂ T u this condition can be written simply as ∂ ∂ v c ( v , T ) = ∂ ∂ T [ T 2 ∂ ∂ T ( p / T ) ] . Differentiating p / T for the vdW equation gives T 2 ∂ ∂ T ( p / T ) ] = a / v 2 , so ∂ ∂ v c v = 0 . Consequently c v = c v ( T ) for a vdW fluid exactly as it is for an ideal gas. To keep things simple it is regarded as a constant here, c v = c R with c a number. Then both integrals can be easily evaluated and the result is u − − C u = c R T − − a / v This is the energetic equation of state for a perfect vdW fluid. By making a dimensional analysis (what might be called extending the principle of corresponding states to other thermodynamic properties) it can be written simply in reduced form as u r − − C u = c T r − − 9 / ( 8 v r ) where u r = u / ( R T c ) and C u is a dimensionless constant.