1
$\begingroup$

I'm learning entropy change for incompressible substances and on the way I saw this equality of $C_p=C_v$. I can't seem to understand this equality and I'm not able to visualize any such example. My professor used T-ds relationships to derive the entropy change for incompressible substances wherein he says, "For incompressible substances, there is no distinction between Cp & Cv, and there is no change in volume and pressure (dV=0 & dP=0)". I'm feeling uncomfortable with this statement as I cant understand and I can't visualize an example that conveys this message. Please help.

$\endgroup$

2 Answers 2

3
$\begingroup$

he says, "For incompressible substances, there is no distinction between Cp & Cv, and there is no change in volume and pressure (dV=0 & dP=0)". I'm feeling uncomfortable with this statement as I cant understand and I can't visualize an example that conveys this message.

In order to help visualize what your professor is saying, consider that a substance, whether compressible or incompressible, obeys the first law for a closed system (no change in mass of the system) which is

$$\Delta U=Q-W=Q-\int pdV$$

Since the volume of the solid or a liquid does not increase appreciably when heated we can say that, for all practical purposes the change in volume is negligible and the $pdV$ work done by the solid or liquid is approximately zero. Therefore

$$\Delta U=Q=mC\Delta T$$

Where $C$ is the specific heat of the solid and $m$ is its mass. This expression applies for a wide range of external pressures so that there is no distinction between between $C_{p}$ and $C_{v}$ for incompressible substances, although sometimes the specific heat is given as $C_{p}$.

So we can say that, for an incompressible solid or liquid, essentially all the heat added to the substance results in an increase in the temperature of the substance, assuming no phase change, regardless of the pressure or volume, so that the substance can be considered to have a single value of specific heat.

This is not the case for a compressible substance, like a gas. A gas when heated can expand unless constrained. An example is a gas in a cylinder fitted with a movable piston. When it does expand against some external pressure the gas does work. So for a gas that does work expanding against constant external pressure we have

$$Q=nC_{p}\Delta T=\Delta U +p\Delta V$$

where $C_{p}$ is the specific heat of the gas at constant pressure and $n$ is the number of moles of the gas. Some of the heat transfer does work and only some of the heat increases the internal energy and temperature of the gas.

On the other hand if the piston is fixed so the volume can't change, $W=0$ and

$$Q=nC_{v}\Delta T = \Delta U$$

Where $C_{v}$ is the specific heat of the gas at constant volume. In this case all the heat results in an increase in internal energy and temperature.

So for the two cases involving the same heat transfer $Q$ the one involving work (constant pressure process) results in less temperature rise than the one involving no work (constant volume process). This means the specific heat at constant pressure is greater than the specific heat at constant volume.

For the special case of an ideal gas, where $\Delta U=nC_{v}\Delta T$ for any process, the relationship between the specific heats is

$$c_{p}-c_{v}=R$$

where $R$ is the specific gas constant.

Hope this helps.

$\endgroup$
1
$\begingroup$

Here's another way to think about it that pulls in the relation between the two heat capacities:

$$C_P-C_V=VT\frac{\alpha^2}{\beta_T}=VT\left[\left(\frac{\partial P}{\partial T}\right)_V\right]^2\beta_T, \tag{1}$$

as derived here, where $C_P$ and $C_V$ are the constant-pressure and constant-volume heat capacities, respectively, $V$ is the volume, $T$ is the temperature, $\alpha\equiv\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_P$ is the thermal expansion coefficient, and $\beta_T\equiv-\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T$ is the compressibility. Note that I applied the triple product rule above to obtain $\beta_T$ in the numerator.

From this we obtain the notable result that $C_P > C_V$ always holds, as the volume, temperature, and compressibility must be positive for a real stable material.

(It's a common misconception, propagated here, for example, that $C_P>C_V$ holds solely because heated materials expand against the atmosphere and therefore do work on it, so we must pump in more energy to obtain a certain temperature rise. It is true that most materials expand and do work against the atmosphere—or whatever is applying the external pressure $P$—when heated. However, the presence of the squared $\alpha^2$ in equation (1) tells us that the inequality holds even when a material contracts with increasing temperature. Thus, internal energy must be a playing a part, and this is clear from the definition of $C_P\equiv\left(\frac{\partial U}{\partial T}\right)_P+P\left(\frac{\partial V}{\partial T}\right)_P=\left(\frac{\partial U}{\partial T}\right)_P+\alpha P$. The misconception arises from forgetting aspects of the $\left(\frac{\partial U}{\partial T}\right)_P$ term.)

Anyway, when we idealize a material as incompressible, we're saying that we'll approximate the compressibility $\beta_T$ as zero (and crucially, also take the thermal expansion coefficient $\alpha$ to be zero, to avoid an indefinite result in equation (1)), or equivalently that the bulk modulus $1/\beta_T$ is very high (usually relative to gases). Thus, applying this idealization to equation (1), $C_P-C_V=0$, QED.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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