Question

(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?

Answer 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!