Lets say we have a tank with a fixed mass of liquid at atmospheric pressure and room temperature. How do we influence the temperature when we exert pressure (e. g., with a piston) on the liquid? Are pressure-enthalpy diagrams the key for that question?

If it is incompressible, temperature would not change, since the applied force does no work, right?

Somehow this collides with my understanding of the microscopic realm, since I associate higher pressure with more microscopic movement and more energy per particle.


This can be computed for small changes in the pressure by considering the partial derivative of the temperature w.r.t. pressure at constant entropy. If we suddenly raise the pressure a bit, then this is to a good approximation an isentropic process as no heat is exchanged and it is not a violent process causing large irreversible effects. So, we want to evaluate:

$$\left(\frac{\partial T}{\partial P}\right)_S \text{ (1)}$$

To express this in terms of the readily measurable quantities (such has heat capcity, thermal expansion coefficient, compressibility etc.) we need to get rid of the entropy in this expression. A direct application of a Maxwell relation will not work because then you end up getting the entropy inside the parial derivative. Instead, we consider dS expressed in terms of dP and dT:

$$dS = \left(\frac{\partial S}{\partial T}\right)_P dT + \left(\frac{\partial S}{\partial P}\right)_T dP$$

The derivative (1) is the ratio of dT and dP at constant S, so we can obtain this by putting dS = 0 and solving for that ratio. This yields:

$$\left(\frac{\partial T}{\partial P}\right)_S = -\frac{\left(\frac{\partial S}{\partial P }\right)_T}{\left(\frac{\partial S}{\partial T }\right)_P}$$

In the denominator we recognize the heat capacity at constant pressure, T dS at constant P is the heat supplied to the system at constant pressure, therefore

$$\left(\frac{\partial S}{\partial T }\right)_P = \frac{C_P}{T}$$

The numerator can be simplified by using an appropriate Maxwell relation. This works as follows, you start with the fundametal thermodynamic relation:

$$dE = T dS - P dV$$

We can read off from this that the temperature is the partial derivative of the internal energy w.r.t. S at constant volume. Also the pressure is minus the partial derivative of the internal energy w.r.t. volume at constant entropy. Now, the second derivative of the internal energy w.r.t. S and V can then be computed by differentiating the temperature w.r.t. the volume at constant S or by differentiating minus the pressure w.r.t. S at constant V. The symmetry of partial derivatives implies that the two quantities are equal, such an equality is called a Maxwell relation.

To rewrite the partial derivative in the numerator, we manipulate the fundamental thermodynamic identity using Leibnitz rule for the differential of a product until we get an expression involving S and dP:

$$\begin{split} &dE = T dS - P dV = d(TS) - S dT - d(PV) + V dP\Longrightarrow\\ &dG = -S dT + V dP \end{split}$$

where $G = E - T S + P V$ is the Gibbs energy. The Maxwell relation derived from G is then:

$$\left(\frac{\partial S}{\partial P}\right)_T=-\left(\frac{\partial V}{\partial T}\right)_P$$

The thermal expansion coefficient $\alpha$ is defined as:

$$\alpha = \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_P $$

Therefore the result is:

$$\left(\frac{\partial T}{\partial P}\right)_S = \frac{T V\alpha}{C_P}$$

In terms of the specific heat capacity (per unit mass) $c_P$ this is:

$$\left(\frac{\partial T}{\partial P}\right)_S = \frac{T\alpha}{\rho c_P}$$

where $\rho$ is the density. For water at 20 C this is about $1.45\times 10^{-8}\frac{K}{\text{Pa}}$, so 1000 bar pressure will raise the temperature by about 1.45 C.

  • $\begingroup$ Thanks for pointing out the typo. I use the lower case c for the specific heat capacity, the upper case C is the total heat capacity. $\endgroup$ – Count Iblis Jul 23 '14 at 2:03

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