0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Thanks for the answer. The only thing is: in this approach where do we consider that the process is adiabatic? We assumed $V$ constant, but not adiabatic process here, right? $\endgroup$ – user1620696 Oct 30 '16 at 13:35
  • $\begingroup$ What exactly is the question? Pressure increased adiabatically from $P_0$ to $P_1$, find final temperature given $\kappa_T$ and $\beta_p$ (at initial temperature?)....the thing inside the brackets...."at initial temperature $T_0$" is explicitly mentioned? $\endgroup$ – Prasad Mani Oct 30 '16 at 13:47
  • $\begingroup$ Yes, the pressure is increased adiabatically from $P_0$ to $P_1$, we know $\kappa_T$ and $\beta_p$ at $T_0$ and we want the final temperature $T_1$. The thing inside the brackets you mean, to use the Gibbs representation? I just used the Gibbs free energy representation because the variables are $P$ and $T$, this is not explicitly mentioned. $\endgroup$ – user1620696 Oct 30 '16 at 13:50
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.