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.

Attachments

Chestermiller
Mentor
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
Mentor
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
Mentor
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
Mentor
##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.