What I'm trying to say is that $$S=\int\limits_{T_1}^{T_2}\frac{\mathrm dQ}{T}\tag{1}$$ depends only on the initial and final states. Why is that so? Is it like a "law" (like Newton's law of gravity or any other law instated by Nature) or somehow can it be deducted that even though temperature ($T$) and heat supplied ($\mathrm dQ$) vary depending on the path taken, $S=\int\limits_{T_1}^{T_2}\frac{\mathrm dQ}{T}$ will be independent of the path?

  • 1
    $\begingroup$ That $\oint \frac{dQ_{rev}}{T} = 0$ is part of Clausius Theorem, which you should be able to find a proof of in any thermodynamics textbook (unfortunately I don't know a good reference online). From this it follows that entropy is a function of state. Also note that it is reversible heat transfer that appears in the definition of entropy and only along these paths that the integral is path independent. $\endgroup$ – By Symmetry Mar 31 at 10:36
  • 2
    $\begingroup$ The integral of dQ/T to get $\Delta S$ is valid only for reversible paths. So it is not independent of path, since it cannot be used for irreversible paths. And, for reversible paths, the integral is the same for all of them, irrespective of the reversible path chosen. $\endgroup$ – Chet Miller Mar 31 at 11:40
  • $\begingroup$ @ChetMiller Can we comment anything on the Entropy associated with Irreverisible process? $\endgroup$ – Vilvanesh Apr 1 at 15:36
  • 1
    $\begingroup$ Entropy is a physical property of the materials comprising the system, rather than a characteristic of any process that the system is subjected to. For an irreversible process, there are only two ways that the entropy of a system can change: 1. Entropy flow across the boundary of the system with its surroundings as a result of heat transfer at the boundary at the boundary temperature. 2. Entropy generation within the confines of the system as a result of finite viscous heat generation and heat conduction at finite temperature gradients within the system. . $\endgroup$ – Chet Miller Apr 1 at 17:09
  • $\begingroup$ Additional remark regarding @ChetMiller's comment: His remarks are valid as long as the system is closed. For an open system obviously there might be an additional entropy transport due to mass transfer. Have a look at the entropy balance equation (Wikipedia or here) if you are interested in that. $\endgroup$ – 2b-t Apr 2 at 10:09

For the integral

$$ S_T = \int_{T=T_1}^{T_2} \frac{\delta Q_{rev}}{T} $$

to be valid you have to assume that a process, including the heat transfer, is completely reversible. The system is in thermodynamic equilibrium throughout the entire process. For this reason also the index $rev$ is used. This is generally not the case as friction is involved but for modelling some processes you assume such an ideal "comparison" process where the direction of every heat transfer is assumed to be arbitrary as well (continuous quasi-equilibrium), e.g. reversible cycles in thermodynamics where the starting point of your cycle is equal to the endpoint. In this case the thermodynamic entropy is a state variable.

I have already taken some notes on it a while back and I will simply copy and paste them over here. The following section explains what qualifies a state variable and derives why this is the case for the entropy. The basic idea is that for a state variable the order of derivation with respect to the underlying variables does not matter. It is shown that for a perfect gas and a reversible paths the specific heat is no state but instead a process variable but introducing an integrating factor $\frac{1}{T}$ a state variable, the specific thermodynamic entropy $s_T$, can be generated.

State and process variables

In thermodynamics one distinguishes between two main types of quantities: State variables depend only on the current equilibrium state of a system and do not depend on the path it has taken to reach the corresponding condition whereas process variables are an adequate measure for the path taken to reach the corresponding state. In order to differentiate these two types of quantities the latter is generally denoted by small Greek delta $\delta$ instead of exact differentials $d$.

For a state variable $f(u,v)$ (a function of two arbitrary variables $u$ and $v$) the order of derivatives does not matter, the symmetry of second derivatives, the Schwarz's theorem

$$ \frac{\partial}{\partial u} \left( \frac{\partial f}{\partial v} \right) = \frac{\partial}{\partial v} \left( \frac{\partial f}{\partial u} \right), \tag{1}\label{1}$$

must hold: The order of the derivatives has no impact on the change of the state variable.

State variables can be further classified: Extensive variables such as energy $E$ or volume $V$ depend on the size of the system whereas intensive variables such as pressure $p$ or temperature $T$ do not. Dividing an extensive variable by the mass of the corresponding system yields so called specific variables. In the case of the volume this leads to the specific volume $v$ which is the reciprocal of the density $\rho$

$$ v := \frac{V}{m} = \frac{1}{\rho}. $$

Internal energy

One main measure of the state of a thermodynamic system is the so called internal energy, an abstraction for a variety of different forms of energy such as kinetic and rotational energy of the corresponding molecules that reflect the internal state of a fluid regardless of the macroscopic fluid flow. The change of the internal energy can be described using temperature and volume as

$$ dE_{in} = \left( \frac{\partial E_{in}}{\partial T} \right)_V dT+ \left( \frac{\partial E_{in}}{\partial V} \right)_T dV. \tag{2}\label{2}$$

The first law of thermodynamics for a closed system

The change of the internal energy may be described using a basic conservation of energy given by the first law of thermodynamics. A closed system is able to store energy either in macroscopic external energy $E_{ex}$ such as potential or kinetic energy of the fluid flow or the aforementioned internal energy $E_{in}$. These quantities are state variables and are therefore denoted with exact differentials.

First law for a closed system with negligible external forces

For a closed system (no mass transport) the energy can only be changed by the work $W$ and heat $Q$ transferred to the system, which generally depend on the exact process path and therefore are denoted by partial differentials. The most common type of work is the expansion of the corresponding control volume against the surrounding pressure $p$ that can be calculated to $\delta W_V = - p \, dV$. Therefore for a closed system the following energy budget must hold:

$$ \delta Q + \delta W = \underbrace{dE_{in} + dE_{ex}}_{dE} \tag{3}\label{3}$$

Ideal gas

So far no assumption regarding the equation of state was made. However, the equations are simplified significantly when assuming gases characterized by simple interactions. One such simplified model is the ideal gas, an idealised model of a real gas where particles of infinitesimal size interact with each other only in elastic collisions. This allows for a simple description including a very simple equation of state given by

$$ p V = R_m T. \tag{4}\label{4}$$

Additionally it can be found experimentally (Joule expansion) that the internal energy of such an ideal gas is no function of the volume the gas occupies

$$ \left( \frac{\partial E_{in}}{\partial V} \right)_T dV \approx 0. $$

but rather only a function of the temperature

$$ dE_{in} \approx \left( \frac{\partial E_{in}}{\partial T} \right)_V dT. $$

Perfect gas

An even more basic behaviour can be achieved by neglecting the intermolecular forces resulting in constant heat capacities. Such a model gas is referred to as a calorically perfect gas.

Enthalpy and heat capacities

Combining the first law of thermodynamics for a close system (equation \eqref{3}) with the differential of the internal energy (equation \eqref{2}), neglecting changes due to the external energy (see the figure above), yields

$$ \delta Q \underbrace{- p \, dV}_{\delta W_V} = \left( \frac{\partial E_{in}}{\partial T} \right)_V dT+ \left( \frac{\partial E_{in}}{\partial V} \right)_T dV. \tag{5}\label{5}$$

Considering an isochor system, meaning the terms including $dV$ vanish in equation \eqref{5}, and dividing the equation by the mass $m$, results in an equation of the specific heat

$$ \partial q = \left( \frac{\partial e_{in}}{\partial T} \right)_v dT $$

where the term

$$ c_v := \left( \frac{\partial e_{in}}{\partial T} \right)_v $$

is referred to as the heat capacity at constant volume. Analogously the enthalpy $H$ is introduced as

$$ H := E_{in} + p V $$

which takes the place of the internal energy for isobar processes and leads to the heat capacity at constant pressure

$$ c_p := \left( \frac{\partial h}{\partial T} \right)_p. $$

One can find a correlation between the two heat capacities dividing the right-hand side of equation \eqref{5} by the mass $m$

$$ de_{in} = c_v dT + \left( \frac{\partial e_{in}}{\partial v} \right)_T dv $$

which can be rewritten using the first law of thermodynamics (equation \eqref{3}) to

$$ \partial q - c_v dT = \left[ \left( \frac{\partial e_{in}}{\partial v} \right)_T + p \right] dv $$

Dividing by the differential of the temperature $d T$ we yield

$$ \left(\frac{\delta q}{d T} \right)_p - c_v = \underbrace{ \left[ \underbrace{ \left( \frac{\partial e_{in}}{\partial v} \right)_T }_{\approx 0} + p \right] \left(\frac{\partial v}{\partial T} \right)_p }_{R_m} = c_p - c_v. $$

As for ideal gases the internal energy is only a function of temperature but not of the specific volume with equation \eqref{4} the expression above leads to

$$ c_p - c_v = R_m. $$

Additionally the heat capacity ratio is defined as the ratio between the two specific heats

$$ \gamma := \frac{c_p}{c_v}. $$

Entropy on the continuum level

In classical mechanics all laws are reversible: The laws are deterministic in the past as well as in the future, they are essentially symmetric in time - Information about prior states is not lost but instead conserved. This contradicts though our perception of the world as heat always flows from points of higher temperature to points of lower and thus gives the physical laws direction. Nonetheless using the findings from the section above as well as assuming a fictive reversible process, where also the heat flux may be arbitrarily reversed, we are able to formulate the specific reversible heat for an ideal gas as (dividing equation \eqref{3} by the mass $m$ and further considering an ideal gas $ \left( \frac{\partial e_{in}}{\partial \hat v} \right)_T d \hat v \approx 0$)

$$ \delta q_{rev} := de_{in} + p d \hat{v} = c_{\hat v} dT + p d \hat{v}. \tag{6}\label{6}$$

If the heat was a state variable it would have to fulfil the symmetry of second order derivatives. This is though not the case due for a perfect gas as (right-hand side of equation \eqref{6} and the ideal gas law \eqref{4})

$$ \left( \frac{\partial c_{\hat v}}{\partial \hat{v}} \right)_T \neq - \left( \frac{\partial \left( \frac{R_m T}{\hat{v}} \right)}{\partial T} \right)_{\hat{v}} = - \left( \frac{\partial p}{\partial T} \right)_{\hat{v}}. $$

As $c_v$ is by definition independent of the specific volume the left side is equal to zero whereas the right side in the case of an ideal gas yields $-\frac{R_m}{\hat{v}}$. If the temperature on the right side would not appear, the Schwartz's theorem would be fulfilled and we would have found a state variable. This can be achieved introducing the thermodynamic Clausius entropy $s_T$, using the integrating factor $\frac{1}{T}$, which yields

$$ ds_T := \frac{\delta q_{rev}}{T} = \frac{c_{\hat v}}{T} dT + \frac{p}{T} d \hat{v}. \tag{7}\label{7}$$

This is sort of an obscure state variable as its meaning is not directly accessible at first. For reversible processes, where the heat flux may be assumed as reversible, this entity vanishes whereas for every irreversible process, where losses due to dissipation happen, which is the norm for common processes found in nature, but no energy may be discharged (closed system), it steadily increases (second law of thermodynamics). Entropy is responsible for the asymmetry of physical laws in time, it gives processes direction: Differences in temperature drive every process and are the source of this irreversibility. Integrating the differential definition (equation \eqref{7}) assuming a perfect gas yields

$$ \frac{p}{p_1} = \left( \frac{\hat{v}_1}{\hat{v}} \right)^{\gamma} e ^{\frac{s-s_1}{c_v}} = \left( \frac{\rho}{\rho_1} \right)^{\gamma} e ^{\frac{s-s_1}{c_v}} = \left( \frac{T}{T_1} \right)^{\frac{\gamma}{\gamma -1}} e ^{-\frac{s-s_1}{R_m}}, $$

which can be rewritten assuming isentropy $s-s_1 = 0$ to $$ \frac{p}{\rho^{\gamma}} = const. $$

| cite | improve this answer | |

What I'm trying to say is that $$S=\int\limits_{T_1}^{T_2}\frac{\mathrm dQ}{T}\tag{1}$$ depends only on the initial and final states.

First, regarding the equation, it should be

$$dS=\frac{\delta Q_{rev}}{T}$$

$$\Delta S=S_{2}-S_{1}=\int_1^2\frac{\delta Q_{rev}}{T}$$

The integral is not defined between two temperatures $T_2$ and $T_1$, but between two equilibrium states 2 and 1. You can have a change in entropy with no change in temperature, that is, a change in entropy where $T_{2}=T_{1}$. An example is an isothermal process (as discussed below)..

Why is that so? Is it like a "law" (like Newton's law of gravity or any other law instated by Nature)

As a matter of fact it is. Entropy is a property of a system that depends only on the initial and final states. It's in part a consequence of a law of nature that heat transfer has never been observed to occur spontaneously from a low temperature body to a high temperature body. The Clausius statement of the second law is

"It is not possible to construct a device that operates in a cycle and whose sole effect is to transfer heat from a colder body to a hotter body"

Since the spontaneous transfer of heat from a cold to hot body did not violate the first law of thermodynamics (conservation of energy), the second law was developed. The second law had to recognize the fact that heat can only naturally go from a hotter body to a colder body. The law necessitated a new property that not only takes into account conservation of energy, like the first law, but also the direction in which natural processes are possible, including the temperatures that dictate the direction of heat transfer.

The second law says the total change in entropy of the system + surroundings is equal to or greater than zero. This is sometimes referred to as the principle of entropy increase. Therefore, according to the second law:

$$\Delta S_{total}=\Delta S_{system}+\Delta S_{surroundings}≥0$$

where the equality applies to an ideal reversible process and the inequality applies to all irreversible (real) processes.

To illustrate how the second law precludes the possibility of spontaneous heat transfer from a cold body to a hot body, consider heat transfer between two bodies A and B that are thermal reservoirs (constant temperature heat source and heat sink). Either body can be considered the system or surroundings.

Let the temperature of body A be $T_A$ and the temperature of body B be $T_B$. Let a quantity of heat, $\delta Q$, transfer isothermally out of A and into B. Since the heat out of A equals the heat into B, and no work is involved, the first law (conservation of energy) is satisfied. Since the temperatures of the two bodies are constant they come out of the integral for determining the changes in entropy of each body and the total entropy change. Those changes become:

$$\Delta S_{A}=-\frac{\delta Q}{T_{A}}$$

$$\Delta S_{B}=+\frac{\delta Q}{T_{B}}$$

Applying the second law for the total change in entropy

$$\Delta S_{tot}=\Delta S_{A}+\Delta S_{B}=-\frac{\delta Q}{T_{A}}+\frac{\delta Q}{T_{B}}$$

In the limit, when the temperature difference approaches zero ($T_{A}=T_{B}$), $\Delta S$ approaches zero and the process is said to be reversible. However, since a finite temperature difference ($T_{A}>T_{B}$) is always necessary for heat transfer to occur, $\Delta S>0$ and all real processes are necessarily irreversible.

Now let's instead consider the possibility that heat transfer occurs in the reverse direction, that is from the colder body B to the hotter body A. Applying the second law for total entropy change we have

$$\Delta S_{tot}=\frac{+\delta Q}{T_A}+\frac{-\delta Q}{T_B}$$

Which gives us, for all $T_{A}>T_{B}$, $\Delta S_{tot}<0$, in violation of the principle of entropy increase and the second law.

Hope this helps.

| cite | improve this answer | |

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.