# 2.4: Thermodynamics Second Law

There are several definitions of the second law. No matter which definition is used to describe the second law it will end in a mathematical form. The most common mathematical form is Clausius inequality which state that

$\oint \dfrac {\delta Q} { T} \ge 0 \label{thermo:eq:clausius}$

The integration symbol with the circle represent integral of cycle (therefor circle) in with system return to the same condition. If there is no lost, it is referred as a reversible process and the inequality change to equality.

$\oint \dfrac {\delta Q} { T} = 0 \label{thermo:eq:clausiusE}$

The last integral can go though several states. These states are independent of the path the system goes through. Hence, the integral is independent of the path. This observation leads to the definition of entropy and designated as $$S$$ and the derivative of entropy is

$ds \equiv \left( \dfrac{ \delta Q}{T} \right)_{\hbox{rev}} \label{thermo:eq:engropy}$

Performing integration between two states results in

$S_2 -S_1 = \int^2_1 \left( \dfrac{ \delta Q}{T} \right)_{\hbox{rev}} = \int^2_1 dS \label{thremo:eq:deltaS}$

One of the conclusions that can be drawn from this analysis is for reversible and adiabatic process $$dS=0$$. Thus, the process in which it is reversible and adiabatic, the entropy remains constant and referred to as isentropic process. It can be noted that there is a possibility that a process can be irreversible and the right amount of heat transfer to have zero change entropy change. Thus, the reverse conclusion that zero change of entropy leads to reversible process, isn't correct. For reversible process equation (12) can be written as

$\delta Q = T\, dS \label{thermo:eq:dQ}$

and the work that the system is doing on the surroundings is

$\delta W = P\,dV \label{thermo:eq:dW}$

Substituting equations (15) (16) into (10) results in

$T\,dS = d\,E_U + P\, dV \label{thermo:eq:Tds}$

Even though the derivation of the above equations were done assuming that there is no change of kinetic or potential energy, it still remain valid for all situations. Furthermore, it can be shown that it is valid for reversible and irreversible processes.

## Enthalpy

It is a common practice to define a new property, which is the combination of already defined properties, the enthalpy of the system.

$H = E_U + P\,V \label{thermo:eq:enthalpy}$

The specific enthalpy is enthalpy per unit mass and denoted as, $$h$$. Or in a differential form as

$dH = dE_U + dP\,V + P\,dV \label{thermo:eq:dEnthalpy}$

Combining equations (18) the (17) yields

(one form of) Gibbs Equation

$\label{thermo:eq:TdSH}T\,dS = dH -V\,dP$

For isentropic process, equation (17) is reduced to $$dH=VdP$$. The equation (17) in mass unit is

$T\,ds = du + P\,dv = dh - \dfrac{dP}{\rho}\label{thermo:eq:Tdsh}$

when the density enters through the relationship of $$\rho=1/v$$.

## Specific Heats

The change of internal energy and enthalpy requires new definitions. The first change of the internal energy and it is defined as the following

Specific Volume Heat

$\label{thermo:eq:cv}C_v \equiv \left( \dfrac {\partial E_u }{\partial T} \right)$

And since the change of the enthalpy involve some kind of boundary work is defined as

Specific Pressure Heat

$\label{thermo:eq:cp}C_p \equiv \left( \dfrac {\partial h }{\partial T} \right)$

The ratio between the specific pressure heat and the specific volume heat is called the ratio of the specific heat and it is denoted as, $$k$$.

Specific Heats Ratio

$\label{thermo:eq:k}k \equiv \dfrac {C_p}{C_v}$

For solid, the ratio of the specific heats is almost 1 and therefore the difference between them is almost zero. Commonly the difference for solid is ignored and both are assumed to be the same and therefore referred as $$C$$. This approximation less strong for liquid but not by that much and in most cases it applied to the calculations. The ratio the specific heat of gases is larger than one.

## Equation of state

Equation of state is a relation between state variables. Normally the relationship of temperature, pressure, and specific volume define the equation of state for gases. The simplest equation of state referred to as ideal gas. And it is defined as

$P = \rho\, R\, T\label{thermo:eq:idealGas}$

Application of Avogadro's law, that 'all gases at the same pressures and temperatures have the same number of molecules per unit of volume,' allows the calculation of a "universal gas constant.'' This constant to match the standard units results in

$\bar{R} = 8.3145 \dfrac{kj} {kmol\; K } \label{thermo:eq:Rbar}$

Thus, the specific gas can be calculate as

$R = \dfrac{\bar{R}} {M} \label{thermo:eq:R}$

The specific constants for select gas at 300K is provided in table 2.1.

Table 2.1 Properties of Various Ideal Gases [300K]
Gas Chemical Formula Molecular Weight $$R\, \left[\dfrac{kj}{Kg K}\right]$$ $$C_p\, \left[\dfrac{kj}{Kg K}\right]$$ $$C_V\, \left[\dfrac{kj}{Kg K}\right]$$ $$k$$
Air - 28.970 0.28700 1.0035 0.7165 1.400
Argon $$Ar$$ 39.948 0.20813 0.5203 0.3122 1.400
Butane $$C_4H_{10}$$ 58.124 0.14304 1.7164 1.5734 1.091
Carbon Dioxide $$CO_2$$ 44.01 0.18892 0.8418 0.6529 1.289
Carbon Monoxide $$CO$$ 28.01 0.29683 1.0413 0.7445 1.400
Ethane $$C_2H_6$$ 30.07 0.27650 1.7662 1.4897 1.186
Ethylene $$C_2H_4$$ 28.054 0.29637 1.5482 1.2518 1.237
Helium $$He$$ 4.003 2.07703 5.1926 3.1156 1.667
Hydrogen $$H_2$$ 2.016 4.12418 14.2091 10.0849 1.409
Methane $$CH_4$$ 16.04 0.51835 2.2537 1.7354 1.299
Neon $$Ne$$ 20.183 0.41195 1.0299 0.6179 1.667
Nitrogen $$N_2$$ 28.013 0.29680 1.0416 0.7448 1.400
Octane $$C_8H_{18}$$ 114.230 0.07279 1.7113 1.6385 1.044
Oxygen $$O_2$$ 31.999 0.25983 0.9216 0.6618 1.393
Propane $$C_3H_8$$ 44.097 0.18855 1.6794 1.4909 1.327
Steam $$H_2O$$ 18.015 0.48152 1.8723 1.4108 1.327

From equation (25) of state for perfect gas it follows

$d\left(P\,v\right) = R\,dT \label{thermo:eq:stateD}$

For perfect gas

$dh = dE_u + d(Pv) = dE_u + d(R\,T) = f(T)\hbox{ (only)} \label{thermo:eq:dhIdeal}$

From the definition of enthalpy it follows that

$d(Pv) = dh - dE_u \label{thermo:eq:defHd}$

Utilizing equation (28) and subsisting into equation (29) and dividing by $$dT$$ yields

$C_p - C_v = R \label{thermo:eq:CpCvR}$

This relationship is valid only for ideal/perfect gases. The ratio of the specific heats can be expressed in several forms as

$$C_v$$ to Specific Heats Ratio

$\label{thermo:eq:Cv} C_v = \dfrac{R}{k-1}$

$$C_p$$ to Specific Heats Ratio

$\label{thermo:eq:Cp} C_p = \dfrac{k\,R}{k-1}$

The specific heat ratio, $$k$$ value ranges from unity to about 1.667. These values depend on the molecular degrees of freedom (more explanation can be obtained in Van Wylen "F. of Classical thermodynamics.'' The values of several gases can be approximated as ideal gas and are provided in Table 2.1. The entropy for ideal gas can be simplified as the following

$s_2 - s_1 = \int_1^2 \left(\dfrac{dh}{T}- \dfrac{dP}{\rho\, T}\right) \label{thermo:eq:deltaSidealI}$

Using the identities developed so far one can find that

$s_2 - s_1 = \int_1^2 C_p \dfrac{dT}{T} - \int_1^2 \dfrac{R\,dP}{P} = C_p \, \ln \dfrac{T_2}{T_1} - R\, \ln \dfrac{P_2}{P_1} \label{thermo:eq:deltaSideal}$

Or using specific heat ratio equation (35) transformed into

$\dfrac{s_2 - s_1} {R} = \dfrac{k}{ k -1} \,\ln \dfrac{T_2}{T_1} - \ln \dfrac{P_2}{P_1} \label{thermo:eq:deltaSidealK}$

For isentropic process, $$\Delta s=0$$, the following is obtained

$\ln \dfrac{T_2}{T_1} = \ln \left(\dfrac{P_2}{P_1} \right) ^ {\dfrac{k -1 }{k}} \label{thermo:eq:sZero}$

There are several famous identities that results from equation (37) as

Ideal Gas Isentropic Relationships

$\label{thermo:eq:famousIdeal} \dfrac{T_2}{T_1} = \left(\dfrac{P_2}{P_1} \right) ^ {\dfrac{k -1 }{k}} = \left(\dfrac{V_1}{V_2} \right) ^ {k -1 }$

The ideal gas model is a simplified version of the real behavior of real gas. The real gas has a correction factor to account for the deviations from the ideal gas model. This correction factor referred as the compressibility factor and defined as

Z deviation from the Ideal Gas Model

$\label{thermo:eq:Z}Z = \dfrac{P\,V}{R\,T}$