# Combining the first and second laws

Pulling together some characteristics of entropy

• $S$ is a function of the state of the system. So we can write it as a function of any two state variables / functions, for example, $S=S(T,P)$, or $=S(T,V)$, or $=S(P,V)$, or sometimes even $S=S(U,P)$.
• Entropy $S$ is extensive. Specific entropy $s=S/m$ is intensive.
• Units for $S$ are Joules / Kelvin.
• In a reversible system, $dS = \delta Q_r / T$.

Spakovszky's definition of reversibility:

A process is called completely reversible if, after the process has occurred, both the system and its surroundings can be wholly restored by any means to their respective initial states.

### The first law and the second law, together

Always true.... $$dU = \delta Q - \delta W.$$

Sometimes true: for reversible work: $$\delta W = \delta W_r= P\,dV.$$ $$dU = \delta Q - P\,dV.$$

for reversible heat: $$\delta Q=\delta Q_r = T\,dS.$$ $$dU = T\,dS - \delta W.$$

Clearly, it is at least sometimes true that (substituting the reversible forms for $\delta Q$ and $\delta W$): $$dU = T\,dS - P\,dV.$$

But wait a sec'... actually, *all* the quantities in this equation are state variables or functions of the state, so we conclude that this statement, which is Gibbs Law is always true:

$$dU = T\,dS - P\,dV.$$

[Gibbs Law] is by far the most important relation in classical thermodynamics. -Ashley Carter

Now do the first two examples in 6.8 before doing these...

### Entropy changes for some reversible processes

We figured out expressions for the work done in isochoric, isobaric, etc.... processes. Now let's do the same thing for the entropy change in different processes.

Gibbs law holds for all kinds of processes, but for reversible processes can be written: $$dU = T\,dS-P\,dV = \delta Q_r - P\,dV.$$

Adiabatic process: $$\delta Q_r = 0 \Rightarrow dS =\frac{\delta Q_r}{T}= 0.$$ So, this is an isentropic process.

(However, there are also irreversible, adiabatic processes.)

Isothermal process: $$\Delta s =s_2-s_1= \int \frac{dq_r}{T} = \frac{q_r}{T}.$$

Carnot cycle:
Consists of reversible adiabatic compressions/expansions and isothermal compressions/expansions. [Graph a Carnot cycle on a $T$ vs $s$ plot.]

The adiabatic parts of a Carnot cycle are reversible, so from the above considerations we conclude these are isentropic (constant entropy). The graph of a Carnot cycle on a $T-s$ plot is then a very simple figure...

The area under a process line on a $T-s$ plot corresponds to...what??

$$\int T\,ds = \int T$\frac{\delta q_r}{T}$=\int \delta q_r = q_r.$$

So, apparently the shaded area on this $T-s$ Carnot plot is... $$q_2+q_1 = w.$$

Phase change:
If a phase change (e.g. ice, melting to water) takes place in an open system, it is both isobaric and isothermal.

The heat absorbed (or given off) per kilomole is the latent heat $l$, so we conclude that the change of entropy is... $$\Delta s = \int \frac{\delta q}{T} = \frac{l}{T}.$$

Isochoric process:
Normally $u=u(v,T)$. For an isochoric process, $u=u(T)$ and that means $du = c_v\,dT$. Since $dv=0$, the first law reads $du=\delta q$.

$$\Delta s = \int \frac{\delta q}{T} = \int \frac{du}{T} = \int_1^2 \frac{c_v}{T}dT.$$

If $c_v$ is independent of temperature, then $\Delta s = c_v\ln$T_2/T_1$$.

Isobaric process:
For reversible processes, the enthalpy is convenient, since... $$dh = du+Pdv+vdP=\delta q_r -Pdv+Pdv+vdP=\delta q_r+vdP.$$

For an isobaric process, $h=h(T,P) = h(T)$ and $dh = c_P\,dT$. Dividing by $T$: $$\Delta s = \int_1^2 c_P \frac{dT}{T}.$$

When $c_P$ is constant over this temperature range, $$\Delta s=c_P\ln$T_2/T_1$.$$

### Entropy changes for an ideal gas

Making no assumptions of reversibility, the Gibbs relation still holds... $$du = Tds-Pdv.$$

For an ideal gas, the internal energy is only a function of $T$, so $du = c_v\,dT$... $$c_v\,dT = T\,ds-P\,dv \Rightarrow ds= c_v\frac{dT}{T}+\frac{P}{T}dv.$$

Using the ideal gas law to write $P/T = R/v$: $$ds = c_v\frac{dT}{T}+R\frac{dv}{v},$$ which relates the fractional temperature and volume changes to entropy changes.

We can write this in terms of $P$ and $v$ instead of $T$ and $v$ by the following trick: Take the natural log of the ideal gas law:

$$\ln P +\ln v= \ln R + \ln T.$$

Taking differentials... $$\frac{dP}{P}+\frac{dv}{v} = \frac{dT}{T}.$$

Subbing this equation for $dT/T$ in our $ds$ expression: \begin{align}ds = & c_v$\frac{dP}{P}+\frac{dv}{v}$+R\frac{dv}{v} = c_v \frac{dP}{P}+$c_v + R$\frac{dv}{v}\\ = & c_v \frac{dP}{P}+ c_P\frac{dv}{v} \end{align}

Using $c_P/c_v = \gamma$, $$\frac{ds}{c_v} = \frac{dP}{P} + \gamma\frac{dv}{v}.$$

Integrating from state 1 to state 2:

$$\frac{\Delta s}{c_v} =\ln$\frac{P_2}{P_1}$ + \gamma \ln$\frac{v_2}{v_1}$ = \ln$\frac{P_2v_2^\gamma}{P_1v_1^\gamma}$.$$

Taking $e^$...$$ of both sides... $$\frac{P_2v_2^\gamma}{P_1v_1^\gamma} = e^{\Delta s/c_v}.$$

This expression is good for ideal gases in both reversible and irreversible processes. But it shows, in a deeper way, that in general for isentropic processes with an ideal gas the result we previously had found for adiabatic processes, that: $$Pv^{\gamma} = K.$$