How Does the TDS Equation Relate CP/CV to Partial Derivatives?

[foranlogan2]

Explaining your reasoning, show that the ratio of the specifc heat is

CP/CV = (DP/DV)s / (DP/DV)t

hint :use TDS EQUATIONS

I have used tds equation 3 and tds equation 1 and equated them,but i am stuck because i get to an equation like


(CP/CV-1)(DT/DV)p = T(DP/DT)v and just do not know how to get to the ratio

how i used the wrong equations or is it my maths skills,please help

 

[msohl]

that is same as to right in the expression below:
\gamma = k_s/ k_t
which \gamma is the ratio of C_P and C_V heat capacities and k_s and k_t are the isentropic and isothermal compressibility, respectively.
k can be expressed as below:
k_t= -(1/V)(DV/DP)_t
and,
k_s= -(1/V)(DV/DP)_s

once we know these, all you should worry about is to have the ratio of heat capacities ( C_P and C_V ) out of Tds Equations, and that is by using the first and second ones and the third one is of no use here, there we have:

Tds= C_VdT + T(DP/DT)_VdV (the first Tds equation)
and,

Tds= C_PdT - T(DV/DT)_PdP (the second Tds equation)
by putting dS=0 in both equation we wil have,

C_V = - [T(DP/DT)_VdV]/dT
and,

C_P = [T(DV/DT)_PdP]/dT
by dividing the two new equations we will have,

C_P/C_V = - [(DV/DT)_PdP]/[(DP/DT)_VdV]
more explicitly,

C_P/C_V = - [(DV/DT)_P/(DP/DT)_V](DP/DV)_?
we should put "S" instead of "?" for started with the assumption ds=0 and that is how got here. That is,

C_P/C_V = - [(DV/DT)_P/(DP/DT)_V](DP/DV)_S
next time we put dT=0 in the Tds equations (the first and second), then we should have,

ds= (DP/DT)_VdV
and,

ds= - (DV/DT)_PdP
left sides are equal, so we try putting right sides equal,

(DP/DT)_VdV = - (DV/DT)_PdP
more explicitly,

(DP/DT)_V/(DV/DT)_P= - (DP/DV)_?
this time we should put "T" instead of "?" for we started with the assumption dT=0. That is,

[(DP/DT)_V/(DV/DT)_P]= - (DP/DV)_T
with a small change we will have,

[(DV/DT)_P / (DP/DT)_V] = - [ 1/ (DV/DP)_T]
from there we will have:

C_P/C_V = - [- 1 / (DV/DP)_T](DP/DV)_S
no different from,

C_P/C_V = (DP/DV)_S / (DV/DP)_T
================================
one question students may get up to, after working out this problem, is:
Why using TdS equations to solve the problem?
and the answer is not that simple!
you may or maynot find it out later.

One may say it is of the two equal interpretations on thermodynamics, which say:
1- thermodynamics is the science dealing with work, heat and internal energy
2- thermodynamics is the science dealing with entropy and energy.