# How do I prove that the cyclic rule of partial derivatives applies to #P#, #T#, and #V#?

##### 1 Answer

The main "tricks" are to write the total derivative of

To derive the **cyclic rule** of partial derivatives of pressure, volume, and temperature, try writing the **total derivative** of the pressure as a function of temperature

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

Now, we can divide through by

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

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

Now all we have to do is subtract the lone derivative to the other side, and cancel it out. Recall that

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

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

In the end, we get these partial derivatives in a row:

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

Notice how if you read each column of partial derivatives, it is in a sequence:

#P, T, T, V, V, P# where the remaining variable

notinvolved in the derivative is heldconstant. It is in that sense cyclic (we went back to#P# ), and that is how you can remember this.

Now, we can use the properties of partial derivatives to evaluate these derivatives (i.e. whatever is held constant can be factored out).

The first one is a straightforward partial derivative:

#P = (RT)/(barV)#

#=> ((delP)/(delT))_barV = (del)/(delT)[(RT)/(barV)]_(barV)#

#= 1/(barV)(d)/(dT)[RT]#

#= R/(barV)# (or#(nR)/V# .)

The second one involves solving for

#T = (PbarV)/R#

#=> ((delT)/(delbarV))_P = (del)/(delbarV)[(PbarV)/R]_P#

#= P(d)/(dbarV)[(barV)/R]#

#= P/R#

And lastly, the third partial derivative.

#barV = (RT)/P#

#=> ((delbarV)/(delP))_T = (del)/(delP)[(RT)/P]_(T)#

#= T(d)/(dP)[(R)/P]#

#= -(RT)/P^2#

Lastly, we multiply these three derivatives together to see if they equal

#((delP)/(delT))_barV ((delT)/(delbarV))_P((delbarV)/(delP))_T#

#= cancel(R)/(barV) cdot cancel(P)/cancel(R) cdot -(RT)/(P^cancel(2))#

#= -(RT)/(PbarV)#

And recalling that *we inherently assumed that the gas was ideal.*

Therefore...

#color(blue)(((delP)/(delT))_barV ((delT)/(delbarV))_P((delbarV)/(delP))_T)#

#= -(RT)/(PbarV) = -1/Z = -1/1 = color(blue)(-1)#