(p+a/v^2)v=RT,derive the cyclic rule.i.e. show the products of the partial derivatives of p,v&T taken in a order is equal to -1?

DISCLAIMER: LONG AND MATH-HEAVY ANSWER!

I am pretty sure that you mean to write #barV = V/n# in place of #V#. Then, it seems that your equation of state is:

#(P + a/(barV^2))barV = RT#

or

#PbarV + a/(barV) = RT#

THE CYCLIC RULE OF PARTIAL DERIVATIVES

The cyclic rule in general says:

#((delx)/(dely))_z((dely)/(delz))_x((delz)/(delx))_y = -1#.

Since #P# is a function of #barV# and #T#, then if we let #x = barV#, #y = T#, and #z = P#:

#((delbarV)/(delbarT))_P((delT)/(delP))_(barV)((delP)/(delbarV))_(T) = ?#

where #P# is implicitly a function of #T# and #barV#.

PROVING THE CYCLIC RULE USING P, V/n, T

To prove the cyclic rule, write the total derivative of #P = P(T,barV)#:

#bb(dP = ((delP)/(delT))_(barV)dT + ((delP)/(delbarV))_TdbarV)#

Now if we divide by #dbarV_P# (the differential change in molar volume at a constant pressure), we get:

#cancel(((delP)/(delbarV))_P)^(0) = ((delP)/(delT))_(barV)((delT)/(delbarV))_P + ((delP)/(delbarV))_Tcancel(((delbarV)/(delbarV))_P)^(1)#

#0 = ((delP)/(delT))_(barV)((delT)/(delbarV))_P + ((delP)/(delbarV))_T#

#-((delP)/(delbarV))_T = ((delP)/(delT))_(barV)((delT)/(delbarV))_P#

Now if you recall that #((delx)/(dely))_z = 1/(((dely)/(delx))_z)#, then multiply by the respective reciprocal partial derivatives to get:

#-((delbarV)/(delT))_P*((delT)/(delP))_(barV)*((delP)/(delbarV))_T = cancel(((delT)/(delP))_(barV)*((delP)/(delT))_(barV)((delbarV)/(delT))_P*((delT)/(delbarV))_P)^(1)#

Thus:

#color(blue)(((delbarV)/(delT))_P((delT)/(delP))_(barV)((delP)/(delbarV))_T = -1)#

APPLYING THE CYCLIC RULE OF PARTIAL DERIVATIVES

To use the cyclic rule for your equation of state, try rewriting it in terms of #P#.

P-form of the equation of state

#bb(P = (RT)/(barV) - a/(barV^2))#

#=> color(green)(((delP)/(delbarV))_T = -(RT)/(barV^2) + (2a)/(barV^3))#

We can use the reciprocal property of partial derivatives to not have to rewrite this in terms of #barV# as the dependent variable, which would have proven quite difficult.

#=> color(green)(((delT)/(delP))_(barV)) = 1/((delP)/(delT))_(barV)#

#= 1/(R/(barV)) = color(green)((barV)/R)#

T-form of the equation of state

Now to rewrite in terms of #T# and evaluate the #((delbarV)/(delT))_P# derivative:

#T = 1/R[PbarV + a/(barV)]#

#=> color(green)(((delbarV)/(delT))_P) = 1/(((delT)/(delbarV))_P)#

#= 1/[P/R - a/(RbarV^2)]#

#= 1/[(PbarV^2 - a)/(RbarV^2)]#

#= color(green)((RbarV^2)/[PbarV^2 - a])#

Showing that these derivatives explicitly reduce to -1

Now we combine these to see if we get #-1#:

#((delbarV)/(delT))_P((delT)/(delP))_(barV)((delP)/(delbarV))_T stackrel(?)(=) -1#

#= (RbarV^2)/[PbarV^2 - a] (barV)/R(-(RT)/(barV^2) + (2a)/(barV^3))#

#= (RbarV^2)/[PbarV^2 - a] (-(T)/(barV) + (2a)/(RbarV^2))#

#= -(TRbarV^2)/(barV(PbarV^2 - a)) + (2aRbarV^2)/(RbarV^2(PbarV^2 - a))#

#= -(TR^2barV^3)/(RbarV^2(PbarV^2 - a)) + (2aRbarV^2)/(RbarV^2(PbarV^2 - a))#

Now make sure that you recall that

#-x/(a + b) + y/(a + b) = (-(x - y))/(a + b) = -(x - y)/(a + b)#.

Thus:

#=> (-(TR^2barV^3 - 2aRbarV^2))/(RbarV^2(PbarV^2 - a))#

#= (-cancel(RbarV^2)(TRbarV - 2a))/(cancel(RbarV^2)(PbarV^2 - a))#

#= (-(RTbarV - 2a))/(PbarV^2 - a)#

Now we substitute #RT = PbarV + a/(barV)# from the original equation to get:

#=> (-((PbarV + a/(barV))barV - 2a))/(PbarV^2 - a)#

#= -(PbarV^2 + a - 2a)/(PbarV^2 - a)#

#= -cancel((PbarV^2 - a)/(PbarV^2 - a))#

#=> color(blue)(-1 = ((delbarV)/(delT))_P((delT)/(delP))_(barV)((delP)/(delbarV))_T)#

as expected!