Introduction to thermodynamics/The second law of thermodynamics: Kelvin-Planck and Clausius statements

**Introduction to thermodynamics**
Lesson : The second law of thermodynamics: Kelvin-Planck and Clausius statements
Chapter 8
Previous chapter:	Refrigerator and heat pump
Next chapter :	Carnot cycles

Introduction

The first law of thermodynamics tells us two facts verified by observations about our physical reality. The first is that an increase in one form of energy, E, must be accompanied by a decrease in another form of energy, E. The second 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. As with other physical laws, the two possible statements of the second law are based on experimental observations.

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?

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.

“

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.

Heat engines

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

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

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

Since it is a cyclic process, $Δ U = 0$ . Thus, $W = Q_{t o t a l} ⟶ W = Q_{h} - Q_{c}$

$η = efficiency = ratio of work done to heat absorbed$

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

$η = 100 %$ only if $Q_{c} = 0$ , i.e. a perfect heat engine would convert all of the absorbed heat $Q_{h}$ 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.

Either statement independently supports the second law

Both the Kelvin-Planck and the Clausius statement are sufficiently independent to support the second law. This means that any cycle that violates the Kelvin-Planck statement, also violates the Clausius statement and vice-versa.

The thermodynamic temperature scale (absolute T)

Since the efficiency of a Carnot engine is:

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

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 $T_{f} \land V_{f}$ ,

$d Q_{r} = d U + d W$ ; $d Q = P d V$ . Since it’s an ideal gas,

$d U = n C_{v} d T$ ; $P = \frac{n R T}{V}$ . Thus,

$d Q_{r} = n C_{v} d T + \frac{n R T d V}{V}$ . Dividing by $T$ ,

$\frac{d Q_{r}}{T} = n C_{v} \frac{d T}{T} + n R T \frac{d V}{V}$ . Assuming $C_{v}$ is constant,

by integration, $Δ S = \int \frac{d Q_{r}}{T} = n C_{v} \ln (\frac{T_{f}}{T_{i}}) + n R \ln (\frac{V_{f}}{V_{i}})$ . Thus, $Δ S$ 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 $Δ S = 0$ .

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

$Δ S_{u} = \frac{Q}{T_{c}} - \frac{Q}{T_{h}} > 0$
Thus, if $Δ S_{u} < 0$ , 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, $Δ U = 0$ and $U_{i} = U_{f}$ . Since the gas is ideal, $U$ depends only on $T$ . Thus, $T_{i} = T_{f}$ .
Since $Δ S = \int d S = \int \frac{d Q_{r}}{T}$ 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, $Δ S = \int \frac{d Q_{r}}{T} = \frac{1}{T} \int d Q_{r}$ .
Since it is isothermal, $\int d Q_{r} = W = n R T \ln (\frac{V_{f}}{V_{i}})$ . Thus,

$Δ S = n R \ln (\frac{V_{f}}{V_{i}})$

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