Thermodynamics: adiabatic compression

[Another]

Homework Statement

Question
If changed isothermal compression process to adiabatic compression process. find the final temperature of process.

Homework Equations

$\alpha = \frac{1}{v} (\frac{∂v}{∂T})_{P}$ expansivity
$\beta = -\frac{1}{v} (\frac{∂v}{∂P})_{T}$ compressibility
$P_1^{1-γ}T_1^{γ}=P_2^{1-γ}T_2^{γ}$

$P_1 = 0.1 Mpa$
$P_2 = 100 Mpa$
$T_1 =15+273.15 K$

$γ= \frac{C_p}{C_v}$

The Attempt at a Solution

I don't know γ . So I must find the γ value.
$Tds = \frac{\beta C_v}{\alpha} dP+\frac{C_p}{\alpha v} d$
In adiabatic process ds = 0 So.
$0= \frac{\beta C_v}{\alpha} dP_s+\frac{C_p}{\alpha v} dv_s$
$-\frac{\beta C_v}{\alpha} dP_s = \frac{C_p}{\alpha v} dv_s$
$-\frac{\beta \alpha v}{\alpha} dP_s = \frac{C_p}{C_v} dv_s$
So
$γ dv_s =- \beta v dP_s$
$γ =-\beta v (\frac{dP}{dv})_s$
I don't have a idea to find γ value . Please help me to find final temperature in a Question.

 

[Chestermiller]

You are trying to do the adiabatic reversible compression, correct?

What is the general equation for dS in terms of dT and dP?

 

[Another]

You are trying to do the adiabatic reversible compression, correct?

Yes. sir

What is the general equation for dS in terms of dT and dP?

let s = f(T,P)
$ds = (\frac{∂s}{∂T})_PdT + (\frac{∂s}{∂P})_TdP$

 

[Chestermiller]

Yes. sir
let s = f(T,P)
$ds = (\frac{∂s}{∂T})_PdT + (\frac{∂s}{∂P})_TdP$

In terms of $C_P$, what is $\frac{∂s}{∂T})_P$?
In terms of $\left(\frac{\partial V}{\partial T}\right)_P$, what is $(\frac{∂s}{∂P})_T$?

 

[Another]

In terms of $C_P$, what is $\frac{∂s}{∂T})_P$?
In terms of $\left(\frac{\partial V}{\partial T}\right)_P$, what is $(\frac{∂s}{∂P})_T$?

let $s = f(P,T)$ , $u = f(P,Y)$ and $Tds = du + pdv$

$ds = (\frac{∂s}{∂P})_T dP + (\frac{∂s}{∂T})_P dT$
$du = (\frac{∂u}{∂P})_T dP + (\frac{∂u}{∂T})_P dT$
$Tds = (\frac{∂u}{∂P})_T dP + (\frac{∂u}{∂T})_P dT + Pdv$
$dp = 0$
$Tds = (\frac{∂u}{∂T})_P dT + Pdv$
$T(\frac{∂s}{∂T})_P = (\frac{∂u}{∂T})_P + P (\frac{∂v}{∂T})_P$
$(\frac{∂s}{∂T})_P = \frac{C_P}{T}$

In maxwell relation
$(\frac{∂s}{∂P})_T = - \left(\frac{\partial V}{\partial T}\right)_P$

 

[Chestermiller]

let $s = f(P,T)$ , $u = f(P,Y)$ and $Tds = du + pdv$

$ds = (\frac{∂s}{∂P})_T dP + (\frac{∂s}{∂T})_P dT$
$du = (\frac{∂u}{∂P})_T dP + (\frac{∂u}{∂T})_P dT$
$Tds = (\frac{∂u}{∂P})_T dP + (\frac{∂u}{∂T})_P dT + Pdv$
$dp = 0$
$Tds = (\frac{∂u}{∂T})_P dT + Pdv$
$T(\frac{∂s}{∂T})_P = (\frac{∂u}{∂T})_P + P (\frac{∂v}{∂T})_P$
$(\frac{∂s}{∂T})_P = \frac{C_P}{T}$

In maxwell relation
$(\frac{∂s}{∂P})_T = - \left(\frac{\partial V}{\partial T}\right)_P$

Good. So, if ds=0, what is dT/dP?

 

[Another]

Good. So, if ds=0, what is dT/dP?

$ds = (\frac{∂s}{∂P})_T dP + (\frac{∂s}{∂T})_P dT$
$0 = (\frac{∂s}{∂P})_T dP + (\frac{∂s}{∂T})_P dT$
$- (\frac{∂s}{∂P})_T dP = (\frac{∂s}{∂T})_P dT$
$(\frac{∂v}{∂T})_P = (\frac{∂s}{∂T})_P (\frac{∂T}{∂P})_s$ and $(\frac{∂v}{∂T})_P = v \alpha$
$(\frac{∂T}{∂P})_s = \frac{Tv \alpha}{C_p}$

I am not sure

 

[Chestermiller]

$ds = (\frac{∂s}{∂P})_T dP + (\frac{∂s}{∂T})_P dT$
$0 = (\frac{∂s}{∂P})_T dP + (\frac{∂s}{∂T})_P dT$
$- (\frac{∂s}{∂P})_T dP = (\frac{∂s}{∂T})_P dT$
$(\frac{∂v}{∂T})_P = (\frac{∂s}{∂T})_P (\frac{∂T}{∂P})_s$ and $(\frac{∂v}{∂T})_P = v \alpha$
$(\frac{∂T}{∂P})_s = \frac{Tv \alpha}{C_p}$

I am not sure

This is correct. So substitute the numbers in and let's see what you get for the temperature rise (don't forget to use absolute temperature).

 

[Another]

This is correct. So substitute the numbers in and let's see what you get for the temperature rise (don't forget to use absolute temperature).

I'm thankful for that.