Change in temperature during adiabatic increase of pressure

Question

Let us consider a system with a certain volume $V$ and at a particular temperature $T_0$. Suppose further that the pressure is increased adiabatically from $P_0$ to $P_1$. I want to be able to determine the change in temperature $\Delta T$.

My initial idea was: the change in entropy can be written (using the Gibbs representation)

$$dS = \dfrac{\partial S}{\partial T}dT + \dfrac{\partial S}{\partial P}dP.$$

Now since the process is adiabatic $dS = 0$. We also know that $\partial S/\partial T = C_p/T$, where $C_p$ is the specific heat at constant pressure, so that

$$-\dfrac{C_p}{T}dT=\dfrac{\partial S}{\partial P}dP.$$

Now one can reduce this derivative. It is quite simple to see that

$$\dfrac{\partial S}{\partial P}=-\dfrac{\partial^2 G}{\partial T\partial P}=-\dfrac{\partial^2 G}{\partial P\partial T}=-\dfrac{\partial V}{\partial T}=-V\beta_p.$$

Thus we end up with

$$\dfrac{C_p}{T}dT=V\beta_p dP\Longrightarrow \dfrac{1}{T}\dfrac{C_p}{\beta_p}dT=VdP$$

Now, this requires knowlege of $C_p$ and $\beta_p$ so that it seems no good.

I derived this to try to solve the particular problem where the system is a cylinder filled with a volume $V$ of water that starts at $T_0$. The only data I have is $\beta_p$ and $\kappa_T$ at $T_0$.

Since this is water I can assume $V$ constant, but $C_p$ and $\beta_p$ remain free. I only know $\beta_p$, for instance, for a particular $T = T_0$, not as a function of $T$.

All I know is that

$$V\Delta P=\int_{T_0}^{T_1}\dfrac{1}{T}\dfrac{C_p}{\beta_p}dT,$$

which is an equation for $T_1$, but depends on $C_p$ and $\beta_p$.

How can I use all of this to find $T_1$? Is my approach correct? In this adiabatic increase of pressure of a system of water, how do I find the change in temperature?

Answer 1

Dont know if this is helpful but writing $V$ as $V(T,P)$ and then differentiating $V$, $$dV= \frac{\partial V}{\partial T} dT \space+ \frac{\partial V}{\partial P}dP = 0$$ which gives $$\frac{1}{V}\frac{\partial V}{\partial T}dT = \frac{-1}{V}\frac{\partial V}{\partial P}dP $$ then $$\beta_P dT= \kappa_T dP$$, assuming that $\kappa_T$ is the isothermal compressibility. Both $\beta_p$ and $\kappa_T$ are mostly used as constants, not as a varying function of temperature or any other thermodynamic variable

Answer 2

You have roughly the right idea. One way to do this is to create a table of entropy as a function of temperature and pressure (and interpolate in the table at constant entropy). Assuming that $P_0$ and/or $P_1$ are large, you can take as a datum S = 0 at say 1 bar and 298 K. You only need to know $C_p$ as a function of temperature at P = 1 bar. The first thing you do is integrate $dS=C_pdT/T$ at P = 1 bar from 298 to T. This gives you the entropy at 1 Bar and temperature T. Then, you hold T constant, and use $dS=-V\beta(T) dP$ from P = 1 bar to pressure P. So, for this step $\Delta S=-V\beta (T)\Delta P$. This assumes that you can neglect the dependence of V and $\beta$ on P. So, $$S(T,P)=\int_{298}^T{C_p(T',1bar)}\frac{dT'}{T'}-V\beta(T)(P-1bar)$$where T' is a dummy variable of integration.