Thermodynamics/The Second Law of Thermodynamics

Introduction

The 1st Law of Thermodynamics tells us that an increase in one form of energy, E, must be accompanied by a decrease in another form of energy, E. Likewise the 2nd Law of Thermodynamics tells us which other kinds of processes in nature may or may not occur. For instance, with two objects in thermal contact, heat will spontaneously flow from a warmer to cooler object, but this effect does not spontaneously occur in the opposite direction.

• Irreversible process = processes that occur naturally in only one direction.
• Reversible process = process that can occur in the reverse direction in infinitesimal steps

Why a second law is needed

Suppose you have two separate containers each containing 1 kg of water. One container is at 20°C and the other is at 10°C. What temperature will the water be if you mix it? Can you prove it with the first law of thermodynamics?

Impossible thermodynamic process

As you can see, with just the use of the first law of thermodynamics, both of these scenarios are possible. However, intuitively, for this simple example, we know that this will not happen. The process will only end up at 15°C.

“ It is impossible for a self-acting machine, unaided by any external agency, to convey heat from one body to another at a higher temperature. ” — Kelvin (based on Clausius statement).

In addition, if there is only reservoir a fluid, and there is no other reservoir which is at a separate temperature, no work or exchange of energy will naturally occur.

Impossible thermodynamic process of only one source being used to produce work.

“ A cycle in contact with only one source cannot produce any work. ” — Kelvin-Planck principle.

It can be seen that at least two reservoirs, a relatively hotter one, and a relatively colder reservoir are need to drive a process to produce any work.

Entropy (S)

• Isolated systems and physical processes tend towards disorder, and entropy is a measure of this disorder.
• E.g. If all molecules of a gas in a room move together, this is a very ordered, unlikely state. If the molecules move randomly in all directions, changing speed after collisions, this is a very disordered, likely state.
• In a reversible process between 2 equilibrium states, change in entropy is given by:

$dS={\frac {dQ_{r}}{T}}$

Where $dS={\text{change in entropy}}$, $dQ_{r}={\text{heat absorbed or expelled by the system in a reversible process}}$, and $T={\text{absolute }}T$.

• Heat absorbed by the system = +ve dQ_r and S↑. Heat lost = –ve dQ_r and S↓.
• The most useful statement of the 2nd law of thermodynamics:

  “The entropy of the universe increases in all natural spontaneous processes.”

• For reversible adiabatic process or reversible reactions, $\Delta S>0$.

• For irreversible process or irreversible reactions, $\Delta S>0$.

Where $\Delta S$ = change in entropy of the system + surroundings (the universe).

$\Delta S=\int dS=\int {\frac {dQ_{r}}{T}}$

Function of state for entropy

$S=\sum _{i}S_{i}({\frac {J}{K}})$

See state functions for more information. Essentially, entropy, like internal energy and enthalpy state quantities that through state functions descrbie equilibrium states of the system that depend only on the cur4rent thermodynamic state of the system and not the path taken by the system to reach that state. You can compare entropy, enthalpy and enthalpy to processes such as heat and work, which are path functions.

Entropy exchange types

Entropy changes due to either of the two processes:

1. External processes
2. Internal processes

$\delta S=\delta S_{e}+\delta S_{gen}$

Types of external entropy exchange

• Heat exchange
• Mass exchange

Types of internal entropy exchange

• Internal dissipation
  • Friction
  • Dissipation
  • Chemical reaction

Entropy change due to internal processes, are always, even in ideal system equal to or greater than 0. Entropy change due to internal processes can never be negative.

$\delta S_{gen}\geq 0{\begin{cases}=0,&{\text{if process is reversible}}\\>0,&{\text{if processs is irreversible}}\end{cases}}$

For reversible adiabatic process, no heat is transferred between system and surroundings, so $\Delta S>0$.

For Carnot engine, $\Delta S={\frac {Q_{h}}{T_{h}}}-{\frac {Q_{c}}{T_{c}}}$. Since ${\frac {Q_{c}}{Q_{h}}}={\frac {T_{c}}{T_{h}}}$, then $\Delta S>0$.

Entropy steady-state power formulation

$0={\frac {\delta S}{\delta t}}={\frac {\sum _{i}{\dot {Q}}_{i}}{T_{i}}}+\sum _{k}{{\dot {m}}_{k}s_{k}}+{\dot {S}}_{gen}$

Entropy closed system formulation

$\Delta S=\sum _{i}{Q_{i}}+\sum _{k}{m_{k}s_{k}}+S_{gen}$

Adiabatic system

${\frac {\delta S}{\delta t}}=\sum _{k}{{\dot {m}}_{k}s_{k}}+{\dot {S}}_{gen}$

The Carnot Engine

   Theorem:
   Carnot's Theorem

   Carnot’s theorem: No real (irreversible) engine operating between 2 heat reservoirs can be more efficient than a Carnot (reversible) engine operating between the same 2 reservoirs.

• Carnot made the most efficient heat engine. Net work taken from the Carnot cycle is the largest possible for a given amount of heat supplied.
• Carnot’s theorem: No real (irreversible) engine operating between 2 heat reservoirs can be more efficient than a Carnot (reversible) engine operating between the same 2 reservoirs.
• Carnot used an ideal gas contained in a cylinder with a movable piston at one end. The cylinder walls and the piston are thermally non–conducting.
• The Carnot cycle consists of 2 adiabatic and 2 isothermal processes, all reversible:

$A\longrightarrow B$

Isothermal expansion of a gas placed in thermal contact with a heat reservoir at temperature $T_{h}$.

During the process, the gas absorbs heat $Q_{h}$ from the base of the cylinder and does a work $WAB$ in raising the piston.

$B\longrightarrow C$

Adiabatic expansion by replacing the base of the cylinder by a thermally non–conducting wall.

During the process, temperature, $T$, falls from $T_{h}$ to $T_{c}$ and the gas does work $WBC$ in raising the piston.

$C\longrightarrow D$

Isothermal compression by placing the gas in thermal contact with a heat reservoir at temp. Tc.

During the process, the gas expels heat $Q_{c}$ to the reservoir and the work $WCD$ is done on the gas by external agent.

$D\longrightarrow A$

Adiabatic compression by replacing the base of the cylinder by a non–conducting wall.

During the process, temperature, $T$, increases from $T_{c}$ to $T_{h}$ and the work $WDA$ is done on the gas by external agent.

• Net work done in this reversible cyclic process = area enclosed by the path $ABCDA$ = net heat transferred into the system, since $\Delta U=0$.
• Since the internal $E$. of an ideal gas depends only on absolute $T$, then in $AB$ and $CD$, $T$ and hence $U$ remain constant. From the 1st law of thermodynamics:

$Q_{h}=W_{AB}=n\ R\ T_{h}ln{\frac {V_{2}}{V_{1}}}$

$Q_{c}=-W_{CD}=n\ R\ T_{c}ln{\frac {V_{3}}{V_{4}}}$

• By dividing the equations,

${\frac {Q_{h}}{Q_{c}}}={\frac {T_{h}ln({\frac {V_{2}}{V_{1}}})}{T_{c}ln({\frac {V_{3}}{V_{4}}})}}$

• For $BC$ and $DA$,

$T_{h}\ V_{2}^{(\gamma -1)}=T_{c}\ V_{3}^{(\gamma -1)}$

$T_{h}V_{1}^{(\gamma -1)}=T_{c}\ V_{4}^{(\gamma -1)}$

• By dividing the equations and taking the $(\gamma -1)$th root,

${\frac {V_{3}}{V_{4}}}={\frac {V_{2}}{V_{1}}}$

• Thus,

${\frac {Q_{h}}{Q_{c}}}={\frac {T_{h}}{T_{c}}}$

$\eta =1-{\frac {T_{c}}{T_{h}}}$

(And this of course applies also to heat pumps and refrigerators. i.e., in their formulas, you can switch the $Q_{h}$ and $Q_{c}$ by $T_{h}$ and $T_{c}$ respectively).

Heat engines

Device that converts thermal energy, ${\textstyle E}$ (i.e. ${\textstyle Q_{h}}$) into mechanical or electrical ${\textstyle E}$.

During the cycle heat (${\textstyle Q_{h}}$) is absorbed from a source at a high $T$ and work, ${\textstyle W}$, is done by the engine.

Heat (${\textstyle Q_{h}}$) is expelled to a source at a lower $T$.

Since it is a cyclic process, $\Delta U=0$. Thus, $W=Q_{total}\longrightarrow W=Q_{h}-Q_{c}$

${\textstyle \eta ={\text{efficiency}}={\text{ratio of work done to heat absorbed}}}$

${\textstyle \eta =W=Q_{h}={\frac {(Q_{h}-Q_{c})}{Q_{h}}}=1-{\frac {Q_{c}}{Q_{h}}}}$

${\textstyle \eta =100\%}$ only if ${\textstyle Q_{c}=0}$, i.e. a perfect heat engine would convert all of the absorbed heat ${\textstyle Q_{h}}$ into mechanical work. From the 2nd law of thermodynamics, this is impossible: only a fraction of heat is converted to mechanical E.

E.g. η of automobile engines = 20%.

“ It is impossible by a cycle to take heat from a hot reservoir and convert it into work without, at the same time, transferring an amount of heat from hot to cold reservoir. ” — Kelvin–Planck form of the 2nd law of thermodynamics.

Heat pumps

Heat pumps are devices that convert work into useful thermal energy transfer (${\textstyle Q_{h}}$), which are used to heat/cool homes. During the cycle, heat, (${\textstyle {Q_{c}}}$), is absorbed from a source at low ${\textstyle \scriptstyle {T}}$ (e.g. outside air or food) by a circulating fluid, usually a refrigerant.

Work is done on the engine by a compressor. Heat (${\textstyle Q_{h}}$) is expelled to a source at higher ${\textstyle \scriptstyle {T}}$ (e.g. room).

Since it is a cyclic process: ${\textstyle \scriptstyle {\Delta U=0}}$. Thus,

$W=Q_{\text{total}}\longrightarrow W=Q_{h}-Q_{c}$

${\textstyle COP={\text{coefficient of performance}}={\text{ratio of heat transferred to work required to transfer the heat}}}$

$COP={\frac {Q_{h}}{W}}={\frac {Q_{h}}{Q_{h}-Q_{c}}}$

$COP$ can be much larger than 100% (denominator < 1).

Refrigerators

“ It is impossible to use a cyclic process to transfer heat from a colder to a hotter body without doing work on the system. ” — Clausius form of the 2nd law of thermodynamics

In other words heat will not flow spontaneously from a cold to a hot object.

Refrigerators are devices that convert work into thermal energy, ${\textstyle E}$, (${\textstyle {Q_{c}}}$).

During the cycle,

Heat (${\textstyle {Q_{c}}}$) is absorbed from a source at low $T$ (e.g. outside air or food) by a circulating fluid.

Work is done on the engine by a compressor.

Heat (${\textstyle Q_{h}}$) is expelled to a source at higher $T$ (e.g. room ambient temperature).

Since it is a cyclic process, $\Delta U=0$. Thus, $W=Q_{total}\Longrightarrow W=Q_{h}-Q_{c}$

${\textstyle COP={\text{coefficient of performance}}={\text{ratio of heat transferred to work required to transfer the heat}}}$

$COP={\frac {Qc}{W}}={\frac {Qc}{(Qh-Qc)}}$

$COP$ can be much larger than 100% (denominator < 1).

A perfect refrigerator would transfer heat from a colder body to a hotter body without doing any work. From 2nd law of thermodynamics, this is impossible. For example the $COP$ of a refrigerator is around 5 or 6.

The thermodynamic temperature scale (absolute T)

Since the efficiency of a Carnot engine is:

$\eta =1-{\frac {Q_{c}}{Q_{h}}}=1-{\frac {T_{c}}{T_{h}}}$

The zero point of the thermodynamic scale must be fixed as the $T$ of the cold reservoir $T_{c}$ at which η = 1 and hence $Q_{c}=0$ and $W=Q_{h}$. T must be the absolute scale because otherwise it may have –ve values, and hence the engine will perform work more than the heat given by the source. i.e. we would create E. from nothing, which is in contradiction to 1st law of thermodynamics. Thus, T = 0 is the lowest T in all scales i.e. the absolute zero.

Quasi–static reversible process for an ideal gas

For an ideal gas undergoing a quasi–static reversible process from $T_{i}\land V_{i}$ to $T_{f}\land V_{f}$,

$dQ_{r}=dU+dW$; $dQ=PdV$. Since it’s an ideal gas,

$dU=nC_{v}dT$; $P={\frac {nRT}{V}}$. Thus,

$dQ_{r}=nC_{v}dT+{\frac {nRTdV}{V}}$. Dividing by $T$,

${\frac {dQ_{r}}{T}}=nC_{v}{\frac {dT}{T}}+nRT{\frac {dV}{V}}$. Assuming $C_{v}$ is constant,

by integration, $\Delta S=\int {\frac {dQ_{r}}{T}}=nC_{v}\ln({\frac {T_{f}}{T_{i}}})+nR\ln({\frac {V_{f}}{V_{i}}})$. Thus, $\Delta S$ is independent of the reversible path and depends only on initial and final states. For a cyclic process, $T_{i}=T_{f}$ and $V_{i}=V_{f}$, so $\Delta S=0$.

Heat conduction

• When heat transfers from a hot $T_{h}$ to cold $T_{c}$ reservoirs, the entropy of the cold reservoir increases by ${\frac {Q}{T_{c}}}$ and the entropy of the hot reservoir decreases by ${\frac {Q}{T_{h}}}$.

• Since $T_{c}<T_{h}$, total change in entropy of the system (universe) > 0

• $\Delta S_{u}={\frac {Q}{T_{c}}}-{\frac {Q}{T_{h}}}>0$
• Thus, if $\Delta S_{u}<0$, the process cannot occur. For example heat transfer from cold to hot body will never occur without artificial assistance.

Free expansion

• An ideal gas in an insulated container occupies a volume $V_{i}$. A membrane separating it from a vacuum is suddenly broken and the gas expands irreversibly to $V_{f}$.
• Work done against the vacuum = 0. Since walls are perfectly insulated, $Q=0$. Thus, $\Delta U=0$ and $U_{i}=U_{f}$. Since the gas is ideal, $U$ depends only on $T$. Thus, $T_{i}=T_{f}$.
• Since $\Delta S=\int dS=\int {\frac {dQ_{r}}{T}}$ only applies to reversible processes, we cannot use it directly. Thus, we imagine a reversible process with the same initial and final states: an isothermal reversible expansion. Since T is constant, $\Delta S=\int {\frac {dQ_{r}}{T}}={\frac {1}{T}}\int dQ_{r}$.
• Since it is isothermal, $\int dQ_{r}=W=nRT\ln({\frac {V_{f}}{V_{i}}})$. Thus,

$\Delta S=nR\ln {({\frac {V_{f}}{V_{i}}})}$

• Since $V_{f}>V_{i},\Delta S_{u}>0$. This can also be obtained by equation *** by setting $T_{f}=T_{i}$.