NotesShala Class 11 · Physics

Chapter11

Thermodynamics

Basic Concepts and the Zeroth Law · Heat, Work and the First Law · Thermodynamic Processes · Second Law and Carnot Engine

Reading key Definition Note Remember Derivation Strategy Formulas sit on graph paper · coloured words are the key terms

Part 1

Basic Concepts and the Zeroth Law

Thermodynamics is the branch of physics that studies heat, work and the changes they cause in a working substance. It is a macroscopic science. It uses a few large-scale variables and does not describe molecules. The state of a gas is given by its pressure, volume, temperature, mass and composition.

1.1System, Surroundings and Boundary

Definition

Thermodynamic System

A thermodynamic system is a collection of a very large number of particles (atoms or molecules). It has definite values of pressure, volume and temperature.

It may be a solid, a liquid, a gas, or a mix of these states.

Definition

Surroundings

The surroundings of a system are everything outside it that has a direct effect on it. All the space in the universe outside the system is its surroundings.

e.g. The environment around a system.

Definition

System Boundary

A system boundary is a closed two-dimensional surface, real or imaginary, around the space filled by the system. It keeps the system apart from its surroundings. Energy may or may not pass through it. Work can be done on or by the system across it.

A system, its boundary and its surroundings

1.2Classification of Thermodynamic Systems

Systems are classified by what can cross the boundary. A system may be open, closed or isolated.

iOpen System

An open system exchanges both energy and matter with its surroundings.

iiClosed System

A closed system exchanges only energy with its surroundings, not matter.

iiiIsolated System

An isolated system exchanges neither energy nor matter with the outside.

1.3Thermodynamic Equilibrium

Definition

A system is in thermodynamic equilibrium when its macroscopic variables do not change with time. These variables are pressure, volume, temperature, mass, composition, etc.

Example

A gas is kept in a closed, rigid container. The container is fully insulated from its surroundings. The pressure, volume, temperature, mass and composition of the gas stay the same with time. So the gas is in thermodynamic equilibrium.

Whether a system is in equilibrium depends on

iThe surroundings

What lies around the system.

iiNature of the wall

The kind of wall that separates the system from its surroundings.

Two Systems with an Adiabatic Wall Between Them

Take two gases $A$ and $B$ in two separate containers. Let their pressures and volumes be $(p_A, V_A)$ and $(p_B, V_B)$ .

Definition

An adiabatic wall is an insulating wall. It does not let energy (heat) flow from one side to the other.

The two systems are placed close together with an adiabatic wall between them. Similar adiabatic walls also insulate them from the rest of the surroundings.

A

and

B

separated by an adiabatic wall

KeyIn this case, any pair of values $(p_A, V_A)$ is in equilibrium with any pair of values $(p_B, V_B)$ .

1.4Thermal Equilibrium

Definition

Two systems are in thermal equilibrium with each other when they are at the same temperature.

Now replace the adiabatic wall between gases $A$ and $B$ with a diathermic wall.

Definition

A diathermic wall is a conducting wall. It lets energy (heat) flow from one side to the other.

Heat now flows through the wall. So the macroscopic variables of both gases change by themselves. The pressure, the volume, or both may change.

After the change, let the values be $(p'_A, V'_A)$ and $(p'_B, V'_B)$ . These new states of $A$ and $B$ are in equilibrium with each other.

Energy flow from one gas to the otherStops

KeySo the temperatures of the two gases become equal. The gases are now in thermal equilibrium.

A

and

B

in thermal equilibrium

Types of Equilibrium

iMechanical

A system is in mechanical equilibrium when no unbalanced force and no unbalanced torque acts on it.

iiChemical

A system is in chemical equilibrium when its internal structure has no spontaneous change due to chemical reaction, diffusion, etc.

iiiThermal

Two systems are in thermal equilibrium when no net heat flows between them after they are put in thermal contact.

1.5Zeroth Law of Thermodynamics

This law was stated byR. H. Fowler in 1931

Law

Zeroth Law of Thermodynamics

If two systems $A$ and $B$ are each in thermal equilibrium with a third system $C$ , then $A$ and $B$ are also in thermal equilibrium with each other.

Experimental Proof

aExperiment (a)

$A$ and $B$ are separated by a fixed adiabatic wall. Each one touches a third system $C$ through a diathermic wall. The variables of $A$ and $B$ change until each reaches thermal equilibrium with $C$ . So $A$ and $B$ are each in thermal equilibrium with $C$ .

Condition (a)

bExperiment (b)

Now the wall between $A$ and $B$ is changed to a diathermic wall. An adiabatic wall now separates $C$ from $A$ and $B$ . No further change takes place in $A$ and $B$ . They are found to be in thermal equilibrium with each other.

Condition (b)

KeyExperiments (a) and (b) together prove the zeroth law.

Concept of Temperature

Concept

What the Zeroth Law Tells Us

The zeroth law shows that temperature is a physical quantity. It has the same value for all systems that are in thermal equilibrium with each other.

So, if $A$ and $B$ are each in thermal equilibrium with $C$ , then $T_A = T_C$ and $T_B = T_C$ . $\therefore\ T_A = T_B$ . So $A$ and $B$ are in thermal equilibrium too.

Some Definitions of Temperature

Temperature is the thermodynamic variable that decides whether two systems in contact are in thermal equilibrium or not.
The temperature of a body is the quantity that tells the degree of hotness or coldness of the body.
The temperature of a body is the quantity that decides the direction of heat flow when it touches another body.

Part 2

Heat, Work and the First Law

2.1Internal Energy

Definition

The internal energy of a system is the total energy of its molecules due to their motion and their configuration (arrangement). It is written as $U$ .

iInternal Kinetic Energy

U_K

It comes from the motion of the molecules. This motion is of three kinds: translational, rotational and vibrational.

iiInternal Potential Energy

U_P

It comes from the configuration of the molecules, that is, how they are arranged.

Internal Energy

U = U_K + U_P

where, $U_K$ = internal kinetic energy, $U_P$ = internal potential energy

Note

So internal energy is the sum of the kinetic and potential energies of the molecules. These energies are measured in a frame where the system's centre of mass does not move.

Ideal Gas

The molecules of an ideal gas do not attract each other. So:

Intermolecular force = 0→No intermolecular potential energy→

U_P = 0

So the internal energy of an ideal gas is only the sum of the kinetic energies of its molecules. These come from their random translational, rotational and vibrational motions.

KeyThe internal energy of an ideal gas depends only on its temperature.

Real Gas

The molecules of a real gas pull on one another. These intermolecular forces cannot be neglected. So some work must be done to change the distance between the molecules.

\text{Internal energy of a real gas} = \text{Internal kinetic energy} + \text{Internal potential energy}

KeySo the internal energy of a real gas depends on both its temperature and its volume.

Key Points about Internal Energy

Internal energy is a macroscopic state variable. It depends on the state of the system, that is, on its pressure, volume, temperature, mass, composition, etc. Here the container is at rest (Fig. a).
It depends only on the state of the system. It does not depend on the path taken to reach that state.
The internal energy of a gas can change when its macroscopic state variables change.
The kinetic energy of the system as a whole does not count in thermodynamics. Suppose the gas container moves with some velocity (Fig. b). Then the KE of the container is not included in the internal energy.

(a) Container at rest

(b) Container moving with velocity

v

2.2Heat

Definition

Heat is the energy that flows between a system and its surroundings because of a temperature difference between them.

SI unit of heatjoule (J)

The heat $Q$ needed to raise the temperature of a body from $T_i$ to $T_f$ depends on three things:

iMass

The mass of the body.

iiNature of substance

What the body is made of.

iiiTemperature change

How much its temperature changes.

\text{i.e.}\quad \Delta Q \propto m\,\Delta T \quad \text{or} \quad \Delta Q \propto m\,(T_f - T_i)

Heat Supplied

\Delta Q = m\,s\,(T_f - T_i)

where, $m$ = mass of the body, $T_i$ = initial temperature, $T_f$ = final temperature, $s$ = specific heat of the material

2.3Work Done by a Gas

In thermodynamics, work means work done by the system or work done on the system.

Setup — Given Information

A gas is enclosed in a cylinder with a movable piston.
$p$ = pressure of the gas on the piston.
$A$ = area of the piston.

Gas in a cylinder with a movable piston

Derivation

Force of the gas on the piston:
$F = pA$
The piston moves out by a tiny distance $dx$ . Work done by the gas:
$dW = F\,dx = pA\,dx = p\,dV \quad [\because A\,dx = dV]$
For a finite change in volume from $V_i$ to $V_f$ :
$W = \displaystyle\int_{V_i}^{V_f} dW$

Work Done by a Gas

W = \int_{V_i}^{V_f} p\,dV

Here, $p$ may stay constant or may change during the process.

Indicator Diagram

Definition

An indicator diagram is a graph of the pressure ( $p$ ) of a gas against its volume ( $V$ ). A gas can change its state in many ways. Each way gives its own $p$ – $V$ graph.

On an indicator diagram, the area under the $p$ – $V$ graph gives the work done.

Area under the

p

–

V

curve = work done

\text{Work} = \displaystyle\int_{V_i}^{V_f} p\,dV = \text{area under } p\text{–}V \text{ graph} \quad\Rightarrow\quad \Delta W = \displaystyle\int_{V_i}^{V_f} p\,dV

The system expands:

V_f > V_i

W

is positive

The system contracts:

V_f < V_i

W

is negative

Remember

Work Depends on the Path

Like heat, work done depends on the path. So work is not a state function. In a cyclic process, the work done equals the area enclosed by the $p$ – $V$ loop.

Cycle goes clockwise: work done is positive.
Cycle goes anticlockwise: work done is negative.

2.4Sign Convention for Heat and Work

The internal energy of a system can be changed in four ways. Each way gives a sign to $\Delta W$ or $\Delta Q$ .

(i) Work is done on the system

\Delta W

= negative

(ii) The system does work

\Delta W

= positive

(iii) Heat is given to the system

\Delta Q

= positive

(iv) Heat is taken out of the system

\Delta Q

= negative

Important

Sign Convention Used in These Notes

$\Delta Q$ is positive when heat enters the system. It is negative when heat leaves it.
$\Delta W$ is positive when the system does work (it expands). It is negative when work is done on the system (it is compressed).
These signs are used in every topic of this chapter. With them, the first law is $\Delta Q = \Delta U + \Delta W$ .

2.5Heat and Work as Energy Transfer

A simple experiment shows how heat and work change the internal energy of a gas.

Place a cylinder containing some gas over a burner. Heat flows from the burner into the gas. So the internal energy of the gas increases.
The gas pushes the piston upward. So the gas does work. Because of this work, the internal energy of the gas decreases.

Experimental arrangement

Both of these changes can also happen in the opposite direction:

iGas Touches a Colder Body

Heat flows from the gas to the colder body. The temperature of the gas falls. So its internal energy decreases.

iiPiston Pushed Down

Now work is done on the gas. So the internal energy of the gas increases.

KeyHeat and work are two separate ways to change both the state and the internal energy of a system.

Concept

Heat Is Energy in Transit

Heat is energy on the move from one body to another. Heat is not a state variable. The state of a system is described by its internal energy.

So neither heat nor work is a state variable. Both are ways of transferring energy, and each can change the internal energy of a system.

2.6First Law of Thermodynamics

Law

First Law of Thermodynamics

The first law of thermodynamics is the law of conservation of energy applied to a system. It takes into account the energy that flows between the system and its surroundings.

Statement: Heat given to a system raises its internal energy, does external work, or both.

First Law

\Delta Q = \Delta U + \Delta W

where, $\Delta Q$ = heat given to the system, $\Delta U$ = change in its internal energy, $\Delta W$ = work done by the system

\Delta U

is decided only bythe starting and final states

Experimental Arrangement to Prove the First Law

Setup — Given Information

A gas is in a cylinder with a massless, frictionless piston. It expands by a very small volume $dV$ .
$p$ = pressure and $V$ = volume of the gas.
$A$ = area of cross-section of the piston, $dx$ = distance moved by the piston.

Arrangement to prove the first law

Derivation

Work done by the gas:
$dW = \text{force} \times dx$
Force $= pA$ . So:
$dW = pA \times dx = p\,(A\,dx) = p\,dV \quad [\because A\,dx = dV]$
$\therefore\ dW = p\,dV$
For the small displacement $dx$ , the pressure is taken as constant.
So, from the first law:
$\Delta Q = \Delta U + p\,dV$

\Delta Q

heat exchanged between the system and surroundings

\Delta U

change in internal energy of the system

2.7Heat Capacity and Specific Heat

Suppose an amount of heat $\Delta Q$ changes the temperature of a body by $\Delta T$ . Then its heat capacity is:

Heat Capacity

S = \frac{\Delta Q}{\Delta T}

Definition

The heat capacity of a body is numerically equal to the heat needed to change its temperature by one unit (1 K).

Unit of heat capacity

\text{J K}^{-1}

(J/K)

Note

The heat needed for a 1-unit rise in temperature may be different at different temperatures.

Specific Heat Capacity

Definition

The specific heat capacity of a substance is the heat needed to change the temperature of unit mass of the substance by one unit.

s = \dfrac{S}{m} = \dfrac{1}{m}\,\dfrac{\Delta Q}{\Delta T}

Specific Heat Capacity

s = \frac{1}{m}\,\frac{\Delta Q}{\Delta T}

where, $s$ = specific heat capacity of the substance, $m$ = mass

It depends on

iNature of the substance

What the substance is.

iiIts temperature

The temperature at which it is heated.

Unit of specific heat capacity

\text{J kg}^{-1}\,\text{K}^{-1}

KeyHeat capacity per kilogram is called the specific heat capacity.

2.8Molar Specific Heat Capacity

Definition

The molar specific heat capacity of a substance is its heat capacity for one mole. It is the heat that raises the temperature of one gram mole of the substance by one unit.

C = \dfrac{S}{\mu} = \dfrac{1}{\mu}\left(\dfrac{\Delta Q}{\Delta T}\right)

Molar Specific Heat Capacity

C = \frac{1}{\mu}\,\frac{\Delta Q}{\Delta T}

where, $\mu$ = number of moles

C

depends on

iNature of the substance

iiIts temperature

iiiHow heat is supplied

Unit of molar specific heat capacity

\text{J mol}^{-1}\,\text{K}^{-1}

Molar Specific Heat of Solids — Law of Equipartition

The law of equipartition of energy can predict the molar specific heat capacity of a solid.

Derivation

Take a solid made of $N$ atoms. Each atom vibrates about its mean position.

The average energy for vibration in one dimension is:
$2 \times \tfrac{1}{2}\,k_B T = k_B T \qquad \ldots\text{(i)}$
In three dimensions, the average energy is $3k_B T$ . For one mole of the solid, the total energy is:
$U = 3k_B T \times N_A = 3RT \qquad [\because\ k_B N_A = R] \quad \ldots\text{(ii)}$
At constant pressure:
$\Delta Q = \Delta U_{\text{int}} + p\,\Delta V \approx \Delta U_{\text{int}}$
For a solid, $\Delta V$ is negligible.
So, using Eq. (ii):
$C = \dfrac{\Delta Q}{\Delta T} = \dfrac{\Delta U_{\text{int}}}{\Delta T} = 3R$

Substance	Specific heat ( $\text{J kg}^{-1}\,\text{K}^{-1}$ )	Molar specific heat ( $\text{J mol}^{-1}\,\text{K}^{-1}$ )
Aluminium	900.0	24.4
Carbon	506.5	6.1
Copper	386.4	24.5
Lead	127.7	26.5
Silver	236.1	25.5
Tungsten	134.4	24.9

Values for some solids, measured at room temperature and normal atmospheric pressure

2.9Specific Heat Capacity of Water

In earlier times, the calorie was used as the unit of heat.

Good to know

Calorie — Old Definition

One calorie was the heat that makes 1 g of water warmer by 1 °C.

Later it was found that the specific heat of water changes slightly with temperature. The graph shows this change.

Definition

One Calorie — Exact Definition

One calorie is the heat that warms 1 g of water from 14.5 °C to 15.5 °C.

Specific heat of water vs temperature

Note

Heat is a form of energy. So the joule (J) is the preferred unit of heat in the SI system.

Specific Heat Capacity of Water

s_{\text{water}} = 4186\ \text{J kg}^{-1}\,\text{K}^{-1} \quad \text{or} \quad 4.186\ \text{J g}^{-1}\,\text{K}^{-1}

Mechanical Equivalent of Heat

Definition

The mechanical equivalent of heat is the work that must be done to produce 1 calorie of heat.

It is only aconversion factor

Conversion

1\ \text{calorie (heat energy)} = 4.186\ \text{J} \approx 4.2\ \text{J}

2.10Specific Heat Capacities of Gases

The specific heat capacity of a gas depends on the process. That is, it depends on the conditions under which heat flows between the gas and its surroundings.

A gas has two main molar heat capacities:

iAt Constant Volume,

C_V

The molar heat capacity at constant volume is the heat that warms one mole of a gas by 1 °C (or 1 K) at constant volume. It is written as $C_V$ .

iiAt Constant Pressure,

C_p

The molar heat capacity at constant pressure is the heat that warms one mole of a gas by 1 °C (or 1 K) at constant pressure. It is written as $C_p$ .

Remember

For a gas, $C_p > C_V$ always. One mole of a gas needs more heat for a 1 °C (1 K) rise at constant pressure than at constant volume.

2.11Mayer's Formula

For an ideal gas, $C_p$ and $C_V$ are related as:

C_p - C_V = R \qquad \ldots\text{(i)}

This relation is calledMayer's formula

Derivation

Start with the first law for 1 mole of the gas: $\Delta Q = \Delta U + p\,\Delta V$ .

Heat $\Delta Q$ is taken in at constant volume, so $\Delta V = 0$ .
$\therefore\ p\,\Delta V = 0 \quad \text{and} \quad \Delta Q = C_V\,\Delta T \ \text{(for one mole)}$
Then:
$C_V = \left(\dfrac{\Delta Q}{\Delta T}\right)_{\!V} = \left(\dfrac{\Delta U}{\Delta T}\right)_{\!V} = \left(\dfrac{\Delta U}{\Delta T}\right) \qquad \ldots\text{(ii)}$
The subscript $V$ is dropped in the last step. The $U$ of an ideal gas depends only on temperature, not on volume.
Now heat $\Delta Q$ is taken in at constant pressure, so $\Delta Q = C_p\,\Delta T$ .
$C_p = \left(\dfrac{\Delta Q}{\Delta T}\right)_{\!p} = \left(\dfrac{\Delta U}{\Delta T}\right)_{\!p} + p\left(\dfrac{\Delta V}{\Delta T}\right)_{\!p}$
The subscript $p$ is dropped from the first term. The $U$ of an ideal gas depends only on $T$ , not on pressure.
Using Eq. (ii):
$C_p = C_V + p\left(\dfrac{\Delta V}{\Delta T}\right)_{\!p} \qquad \ldots\text{(iii)}$
For one mole of an ideal gas, $pV = RT$ . At constant pressure:
$p\left(\dfrac{\Delta V}{\Delta T}\right)_{\!p} = R \qquad \ldots\text{(iv)}$
From Eqs. (iii) and (iv):
$C_p - C_V = R$

Mayer's Formula

C_p - C_V = R

where, $C_p$ = molar specific heat of an ideal gas at constant pressure, $C_V$ = its molar specific heat at constant volume, $R$ = universal gas constant

Heat Capacity Ratio

\gamma

The ratio of $C_p$ to $C_V$ is written as $\gamma$ . It is called the heat capacity ratio.

\gamma = C_p\,/\,C_V

Molar Heat at Constant Volume

C_V = \dfrac{R}{\gamma - 1}

Molar Heat at Constant Pressure

C_p = \gamma\,\dfrac{R}{\gamma - 1}

Part 3

Thermodynamic Processes

3.1Thermodynamic State Variables

Definition

The macroscopic variables whose values fully describe an equilibrium state of a system are called state variables.

e.g. Pressure, volume, temperature and mass.

Equilibrium State

A system need not be in equilibrium at every moment. Given enough time, it settles into mechanical and thermal equilibrium.

Example

Petrol vapour and air are mixed and lit by a spark. This gas mixture goes through an explosive chemical reaction. It is not in an equilibrium state. With time, the gas gets a uniform temperature and pressure. It then reaches thermal and mechanical equilibrium with its surroundings.

(a) Partition removed suddenly: free expansion

(b) A gas mixture in an explosive chemical reaction

Extensive and Intensive Variables

State variables are of two types: extensive and intensive. To tell them apart, imagine a system in equilibrium divided into two equal parts.

iExtensive Variables

Variables whose values become half in each part are called extensive variables. They show the size of the system.

e.g. Internal energy ( $U$ ), volume ( $V$ ), total mass ( $M$ ).

iiIntensive Variables

Variables whose values stay the same in each part are called intensive variables. They do not depend on the size of the system.

e.g. Pressure ( $p$ ), temperature ( $T$ ), density ( $\rho$ ).

Note

Extensive variables change when the size of the system changes. Intensive variables do not change with size.

3.2Equation of State

The state variables of a system are not always independent of each other.

Definition

The equation of state of a system is the equation that gives the relation between its state variables.

e.g. Equation of State of an Ideal Gas (Ideal Gas Equation)

pV = \mu RT

So, for a fixed amount of gas, only two variables are independent. For example, $p$ and $V$ , or $T$ and $V$ .

Definition

An isotherm is the pressure–volume curve of a gas at a fixed temperature.

3.3Quasi-static Process

Definition

Thermodynamic Process

A thermodynamic process is a change in the state of a system. In it, the state variables change with time.

Definition

Quasi-static Process

A quasi-static process is an infinitely slow process. In it, the system stays in thermal and mechanical equilibrium with its surroundings at every stage. Its variables ( $p$ , $T$ , $V$ ) change only in tiny steps.

It is an imagined (ideal) idea. In practice, processes that are slow enough are treated as quasi-static.

At every stage of a quasi-static process:

Difference between the pressure of the system and the outside pressureInfinitely small

Temperature difference between the system and its surroundingsInfinitely small

To take a gas from state $(p, T)$ to state $(p', T')$ by a quasi-static process:

Change the outside pressure by a very small amount.
Let the system make its pressure equal to that of the surroundings.
Repeat this infinitely slowly until the pressure of the system becomes $p'$ .

Small steps in pressure and temperature

To change the temperature, keep a very small temperature difference between the system and a reservoir. Use a series of reservoirs whose temperatures change step by step from $T$ to $T'$ . In the end, the system reaches the temperature $T'$ .

Note

A process is quasi-static when it has no accelerated motion of the piston, no large temperature gradient and no large pressure gradient.

3.4Types of Thermodynamic Processes

iIsothermal Process

An isothermal process is a process in which the temperature stays constant.

e.g. Water at 0 °C freezing into ice at 0 °C.

iiIsobaric Process

An isobaric process is a process in which the pressure stays constant.

e.g. Water boiling in an open vessel.

iiiIsochoric Process

An isochoric process is a process in which the volume stays constant. So the system does zero work.

e.g. Heating a gas kept in a closed, rigid container (fixed walls).

ivAdiabatic Process

An adiabatic process is a process in which no heat flows into or out of the system.

e.g. Fast compression, like filling a bicycle tube with a hand pump.

(v) Cyclic and non-cyclic processes

Cyclic Process

The initial and final states are the same.

Non-cyclic Process

The initial and final states are different.

3.5Isothermal Process

In an isothermal process, the temperature stays constant. For an ideal gas, the equation of state then gives $pV$ = constant. So the gas obeys Boyle's law.

p_i V_i = p_f V_f \qquad \text{(isothermal process)}

p

–

V

Diagram

On a $p$ – $V$ diagram, an isotherm is a curve that joins points at the same temperature.

The curve $i \to f$ shows an isothermal expansion from $V_i$ to $V_f$ . The arrow shows the direction of the process.

The shaded area under $i \to f$ is the work done by the gas.

An isothermal expansion

i \to f

Work Done

Derivation

The work done in an isothermal process is the area under the $p$ – $V$ graph.

$W = \displaystyle\int_{V_i}^{V_f} p\,dV = \mu RT \int_{V_i}^{V_f} \dfrac{dV}{V} \qquad \left[\because\ pV = \mu RT,\ \ \therefore\ p = \dfrac{\mu RT}{V}\right]$
$= \mu RT\,\big[\ln V\big]_{V_i}^{V_f} = \mu RT\,[\ln V_f - \ln V_i]$
$W_{\text{iso}} = \mu RT\,\ln\!\left(\dfrac{V_f}{V_i}\right)$

Work Done in an Isothermal Process

W_{\text{iso}} = 2.303\,\mu RT\,\log_{10}\!\left(\frac{V_f}{V_i}\right)

where, $\mu$ = number of moles, $R$ = gas constant, $T$ = temperature, $V_i$ and $V_f$ = initial and final volumes of the gas

The temperature of the ideal gas stays constant

\therefore\ \Delta U = 0

So, from the first law $\Delta Q = \Delta U + \Delta W$ :

\Rightarrow\ \Delta Q = \Delta W = \mu RT\,\ln\!\left(\dfrac{V_f}{V_i}\right)

Note

In an isothermal change, all the heat supplied is used to do work on the surroundings. If work is done on the system, the system gives out an equal amount of heat.

The sign of the work depends on the change in volume. There are two cases:

V_f > V_i

, so

W > 0

This is an isothermal expansion. The gas absorbs heat.

V_f < V_i

, so

W < 0

This is an isothermal compression. The surroundings do work on the gas, and the gas releases heat.

3.6Adiabatic Process

Definition

An adiabatic process is a process in which no heat is exchanged between the system and its surroundings.

By the first law, $\Delta Q = \Delta U + \Delta W \quad \ldots\text{(i)}$ . For an adiabatic process, $\Delta Q = 0$ .

\therefore\ \Delta U = -\Delta W

Adiabatic Process

pV^{\gamma} = \text{constant}

where, $\gamma = \dfrac{C_p}{C_V}$

So, when an ideal gas goes adiabatically from $(p_i, V_i)$ to $(p_f, V_f)$ :

p_i V_i^{\gamma} = p_f V_f^{\gamma} = \text{constant}

Example

Sound in Air

When sound travels in air, the air goes through compressions and rarefactions. These are adiabatic. The air particles oscillate very fast. So heat has no time to enter or leave the region of the oscillating particles.

In an adiabatic process, $p$ and $T$ , and also $V$ and $T$ , are related as:

Relation between p and T

p^{1-\gamma} \cdot T^{\gamma} = \text{constant}

Relation between V and T

V^{\gamma-1} \cdot T = \text{constant}

p

–

V

Diagram

On a $p$ – $V$ diagram, an adiabatic process is shown by an adiabatic curve.

Definition

An adiabat is a curve that represents an adiabatic process on a $p$ – $V$ diagram.

The gas expands from $i$ to $f$ (see the arrow). Its temperature falls from 550 K to 350 K. The adiabat is steeper than the isotherms it crosses.

An adiabatic expansion

i \to f

Slope of an Adiabat

Differentiate $pV^{\gamma}$ = constant:

dp\,V^{\gamma} + p\,\gamma V^{\gamma-1}\,dV = 0

\Rightarrow\ \dfrac{dp}{dV} = -\gamma\left(\dfrac{p}{V}\right)

Remember

Slope of an adiabat $= -\gamma\left(\dfrac{p}{V}\right)$ . This is $\gamma$ times the slope of an isotherm through the same point. Since $\gamma > 1$ , an adiabat is always steeper than an isotherm.

Specific Heat in an Adiabatic Process

By the definition of specific heat:

C = \dfrac{\Delta Q}{m\,\Delta T} = \dfrac{Q}{m\,\Delta T}, \quad \text{and } \Delta Q = 0 \text{ in an adiabatic process} \quad \Rightarrow \quad C = 0

KeySo, during adiabatic expansion or compression, a gas has zero specific heat.

Work Done

Derivation

Since $pV^{\gamma} = K$ (constant), we have $p_i V_i^{\gamma} = p_f V_f^{\gamma} = K$ .

Work done in an adiabatic process:
$W = \displaystyle\int_{V_i}^{V_f} p\,dV = \int_{V_i}^{V_f} \dfrac{K}{V^{\gamma}}\,dV$
Integrate:
$= K\left[\dfrac{V^{-\gamma+1}}{-\gamma+1}\right]_{V_i}^{V_f} = \dfrac{K}{1-\gamma}\left[\dfrac{1}{V_f^{\,\gamma-1}} - \dfrac{1}{V_i^{\,\gamma-1}}\right]$
Put $K = p_f V_f^{\gamma}$ in the first term and $K = p_i V_i^{\gamma}$ in the second:
$W = \dfrac{1}{1-\gamma}\left[\dfrac{p_f V_f^{\gamma}}{V_f^{\,\gamma-1}} - \dfrac{p_i V_i^{\gamma}}{V_i^{\,\gamma-1}}\right] = \dfrac{p_i V_i - p_f V_f}{\gamma - 1}$
Since $pV = \mu RT$ , we have $p_i V_i = \mu RT_i$ and $p_f V_f = \mu RT_f$ .

Work Done in an Adiabatic Process

W = \frac{\mu R\,(T_i - T_f)}{\gamma - 1}

By the first law, $\Delta Q = \Delta U + \Delta W$ . Since $\Delta Q = 0$ , we get $\Delta U = -\Delta W$ .

iWork Done by the System

$W > 0$ , so $T_f < T_i$ . The internal energy, and so the temperature, falls.

iiWork Done on the System

$W < 0$ , so $T_f > T_i$ . The internal energy, and so the temperature, rises.

Note

Nearly all real adiabatic processes happen very fast. So no heat can flow between the system and its surroundings.

e.g. Sudden bursting of a bicycle tube, and sound waves travelling in air.

The figure shows $p$ – $V$ curves of an ideal gas. Two adiabats join two isotherms.

p

–

V

curves of an ideal gas

3.7Isochoric Process

Definition

An isochoric process is a process in which the volume of the system stays constant.

iHeat added to the system

The pressure increases.

iiHeat taken out of the system

The pressure decreases.

Equation of State

Here $p$ and $T$ change, but $V$ stays constant. The gas obeys Gay-Lussac's law: $p \propto T \;\Rightarrow\; p_1/T_1 = p_2/T_2$ (where $V$ = constant).

\Rightarrow\ \Delta W = p\,\Delta V = 0 \qquad [\because\ \Delta V = 0]

KeySo the work done is zero: $\Delta W = 0$ .

The first law then gives $\Delta Q = \Delta U$ . So in an isochoric process, the heat given or taken changes only the internal energy and temperature.

Heating at constant volume

Change in Internal Energy (decides the change in temperature)

\Delta U = \mu\,C_V\,\Delta T

3.8Isobaric Process

Definition

An isobaric process is a thermodynamic process in which a system goes through a physical change at constant pressure.

Example

A gas is heated in a piston–cylinder. A fixed weight rests on the piston, so the pressure stays the same.

Let the pressure $p$ of a gas stay constant while its volume changes from $V_i$ to $V_f$ . The work done by the gas is:

A simple set-up for an isobaric process

W = \displaystyle\int_{V_i}^{V_f} p\,dV = p\int_{V_i}^{V_f} dV = p\,(V_f - V_i)

Work Done in an Isobaric Process

W = \mu R\,(T_f - T_i)

$[\because\ pV = \mu RT]$

Note

Here the gas temperature does not stay fixed. So its internal energy changes too. In an isobaric process, the heat absorbed goes partly into internal energy and partly into doing work.

Isobaric expansion

3.9Cyclic Process

Definition

A cyclic process is a process, or a series of processes, that brings the system back to its initial state. Heat and work are exchanged on the way.

The initial and final states of a cyclic process are the same. So:

\Delta U = U_f - U_i = 0

Clockwise cycle: work done is positive

Anticlockwise cycle: work done is negative

For a cyclic process, the $p$ – $V$ graph is a closed curve. The area enclosed by the curve equals the work done. The first law is $\Delta Q = \Delta U + \Delta W$ . For a cyclic process:

\Delta U = 0 \quad \Longrightarrow \quad \Delta Q = \Delta W

KeyThe heat supplied to the system is turned into work done, and work into heat.

Calculation of Work Done

Let $A\,(V_1, p_1)$ be the initial state of a gas and $B\,(V_2, p_2)$ its final state. The gas goes from $A$ to $B$ by the upper path $AMB$ . It comes back from $B$ to $A$ by the lower path $BNA$ .

Work done by the gas in the expansion from $A$ to $B$ :

W_1 = \text{area } AMBCDA

Work done in the compression from $B$ to $A$ is $W_2 = -\,\text{area } BNADCB$ . So the net work done by the gas in the cycle is:

\begin{aligned} W_{A \to B \to A} &= W_1 + W_2 \\ &= \text{area } AMBCDA - \text{area } BNADCB \\ &= \text{area } AMBNA \end{aligned}

Work done in a cyclic process

KeyThe work done in a cyclic process equals the area enclosed by the loop.

Cycle goes clockwiseWork done is positive

Cycle goes anticlockwiseWork done is negative

Note

Cyclic processes are very important. Engines and refrigerators work in cycles. In each cycle, the system uses incoming heat to do work and then returns to its initial state.

Definition

Non-cyclic Process

A non-cyclic process is a process whose final state differs from its starting state after a series of changes.

3.10Summary of Thermodynamic Processes

All four processes can be drawn together on a single $p$ – $V$ diagram. Each path starts from the same point on isotherm $T_2$ :

Path a: isobaric (pressure constant)
Path b: isothermal (along isotherm $T_2$ )
Path c: adiabatic (steeper than the isotherm)
Path d: isochoric (volume constant)

Four processes on one

p

–

V

diagram

Process	Constant quantity	Path	Heat / Work / Internal energy
Isobaric	$p$	a	$\Delta Q = \mu C_p\Delta T$ , $\Delta W = p\,\Delta V$ , $\Delta U = \mu C_V\Delta T$
Isochoric	$V$	d	$\Delta Q = \Delta U = \mu C_V\Delta T$ , $\Delta W = 0$
Isothermal	$T$	b	$\Delta Q = \Delta W = \mu RT\ln(V_f/V_i)$ , $\Delta U = 0$
Adiabatic	$pV^{\gamma}$	c	$\Delta Q = 0$ , $\Delta W = -\Delta U$

Thermodynamic processes at a glance ( $\mu$ = number of moles)

Part 4

Second Law and Carnot Engine

4.1Second Law of Thermodynamics

Many processes obey energy conservation but never happen in nature.

Example

A ball lies on the floor. In principle, it could take some heat from the ground and jump up by itself. The first law (energy conservation) allows this. But it never happens in nature.

So some other basic law of nature must forbid such a process, even though it conserves energy.

Definition

The second law of thermodynamics is the law of nature that limits what the first law allows.

The second law sets a basic limit on

iHeat engine

The efficiency of a heat engine.

iiRefrigerator

The coefficient of performance of a refrigerator.

The Kelvin–Planck and Clausius statements rule out a perfect heat engine and a perfect refrigerator, respectively.

Kelvin–Planck Statement

Law

No process is possible whose only result is to take heat from a reservoir and change all of it into work.

Clausius Statement

Law

No process is possible in which the only result is that heat flows from a colder body to a hotter body.

Heat engine: by Kelvin–Planck,

Q_2

can never be zero

Refrigerator: by Clausius,

W

can never be zero

PMM1 and PMM2

iPMM1

A perpetual motion machine of the first kind (PMM1) is a device that breaks the first law, because it creates energy.

iiPMM2

A perpetual motion machine of the second kind (PMM2) is a device that breaks the second law.

4.2Reversible Process

Definition

A reversible process is a process that can be run backwards so that both the system and its surroundings return exactly to their initial states. No other change is left anywhere in the universe.

iHeat

If heat is absorbed in the direct process, the same amount of heat is given out in the reverse process.

iiWork

Suppose the direct process needs work to be done on the working substance. Then, in the reverse process, the substance does the same amount of work (work is done by it).

Conditions for Reversibility

There must be no dissipative forces at all, such as friction, viscosity and electrical resistance.
The direct and reverse processes must take place infinitely slowly.
The temperature of the system must never differ much from that of its surroundings.

Examples of Reversible Processes

All isothermal and adiabatic changes, when they are done very slowly.
Ice absorbs some heat and melts. When the same heat is removed, the water formed freezes back into ice.
A spring is stretched or compressed extremely slowly, so that no oscillations start.
A perfectly elastic ball falls on a perfectly elastic horizontal floor. It rises back to its starting height.
Very slow evaporation or condensation.

Remember

The conditions for a reversible process can never be fully met in practice. So a reversible process is only an ideal idea. Real processes always lose some heat through friction, conduction, radiation, etc.

Why an Ideal Engine Is Not Possible

Concept

Rule of Nature

Irreversibility is the rule of nature. A change that has taken place in the universe can never be exactly undone.

KeyThat is why no engine can have 100% efficiency.

If all resistive forces could be removed, the efficiency of an engine would be as high as possible.

Good to know

Some trains have been proposed that will run inside vacuum tunnels. This removes air friction and so raises the efficiency of the engine.

4.3Irreversible Process

Definition

An irreversible process is any process that cannot be reversed exactly.

All Natural Processes

Conduction, radiation and radioactive decay are irreversible.

All Practical Processes

Free expansion, Joule–Thomson expansion and electrical heating of a wire are also irreversible.

Two Main Causes of Irreversibility

iNon-equilibrium States

Many processes take the system to non-equilibrium states. e.g. Diffusion of gases, decay of organic matter.

iiDissipative Forces

Many processes involve dissipative forces, such as friction, viscosity and inelasticity.

Examples of Irreversible Processes

A steel ball falls on an inelastic lead sheet. Its kinetic energy changes into heat, and this heat warms the lead sheet. The heat never changes back into kinetic energy of the ball.
A spring is stretched suddenly and fast. This sets up vibrations in it, and part of the energy is dissipated (lost).
Sudden expansion or contraction, and fast evaporation or condensation.
An electric current flowing through a resistor.
Heat flow between bodies at different temperatures.
The Joule–Thomson effect. If the gas flow is reversed, the same cooling or heating effect is not seen.

4.4Carnot Engine and Carnot Cycle

Definition

A Carnot engine is a reversible heat engine that works between two temperatures: a source at $T_1$ and a sink at $T_2$ .

Ideal heat engine suggested bySadi Carnot in 1824

Note

A Theoretical Engine

It is a theoretical engine. It has none of the defects of a real engine.

It cannot be built in practice. But it serves as a standard. The performance of any real engine is judged against it.

A Carnot engine

Main Parts of a Carnot Engine

iSource

A hot reservoir at temperature $T_1$ , with conducting walls. Its thermal capacity is infinite. So we can draw any amount of heat from it, and its temperature stays the same.

iiSink

A cold reservoir at temperature $T_2$ . Its thermal capacity is also infinite. So it can take in any amount of heat with no change in its temperature.

iiiCylinder

It has a conducting base and insulating walls. It holds the working substance. An insulating, frictionless piston is fitted in it, as shown in the figure.

ivWorking Substance

An ideal gas is used as the working substance in the cylinder.

vInsulating Stand

The base of the cylinder can be placed on this stand. It isolates the gas completely from the surroundings.

Carnot Cycle

Definition

The cycle that the working substance of a Carnot engine goes through is called the Carnot cycle. It has four strokes.

The four states of the gas are:

$A\,(p_1, V_1, T_1)$ and $B\,(p_2, V_2, T_1)$
$C\,(p_3, V_3, T_2)$ and $D\,(p_4, V_4, T_2)$

$E$ , $F$ , $G$ and $H$ are the points on the $V$ -axis just below $A$ , $D$ , $B$ and $C$ . The arrows show the direction of each stroke.

In this cycle, $Q_1$ , $Q_2$ and $W_1$ to $W_4$ are written as positive amounts. The words by/on and absorbed/released give the direction.

The four processes of the Carnot cycle

1First Stroke — Isothermal Expansion (Curve

AB

)

One mole of an ideal gas is the working substance. The cylinder is placed on the source. The gas expands slowly at the constant temperature $T_1$ . The work done by the gas equals the heat absorbed.

W_1 = Q_1 = \displaystyle\int_{V_1}^{V_2} p\,dV = RT_1\,\log_e\!\left(\dfrac{V_2}{V_1}\right)

On the indicator diagram

W_1

= area

ABGEA

2Second Stroke — Adiabatic Expansion (Curve

BC

)

The cylinder is now placed on the insulating stand. The gas expands adiabatically until its temperature falls from $T_1$ to $T_2$ .

W_2 = \displaystyle\int_{V_2}^{V_3} p\,dV = \dfrac{R}{(\gamma - 1)}\,(T_1 - T_2)

On the indicator diagram

W_2

= area

BCHGB

3Third Stroke — Isothermal Compression (Curve

CD

)

The cylinder is placed on the sink. The gas is compressed at the constant temperature $T_2$ . The work done on the gas equals the heat released by it.

W_3 = Q_2 = -\displaystyle\int_{V_3}^{V_4} p\,dV = -RT_2\,\log_e\dfrac{V_4}{V_3} = RT_2\,\log_e\dfrac{V_3}{V_4}

On the indicator diagram

W_3

= area

CDFHC

4Fourth Stroke — Adiabatic Compression (Curve

DA

)

The cylinder is placed on the insulating stand again. The compression continues until the gas returns to its initial state.

W_4 = -\displaystyle\int_{V_4}^{V_1} p\,dV = -\dfrac{R}{\gamma - 1}\,(T_2 - T_1) = \dfrac{R}{\gamma - 1}\,(T_1 - T_2)

On the indicator diagram

W_4

= area

ADFEA

4.5Efficiency of a Carnot Engine

Definition

The efficiency of an engine is the ratio of the work done by it to the heat supplied to it.

\eta = \dfrac{\text{Work done}}{\text{Heat input}} = \dfrac{W_{\text{net}}}{Q_1}

Derivation

In the four strokes, $W_1$ and $W_2$ are done by the gas. $W_3$ and $W_4$ are done on the gas.

Net work done in one complete cycle:
$W = W_1 + W_2 + (-W_3) + (-W_4) = W_1 - W_3 = \text{area } ABCDA \qquad [\text{as } W_2 = W_4]$
So:
$\therefore\ \eta = \dfrac{W}{Q_1} = \dfrac{W_1 - W_3}{W_1} = \dfrac{Q_1 - Q_2}{Q_1} = 1 - \dfrac{Q_2}{Q_1}$
Put the values of $Q_1$ and $Q_2$ from the strokes:
$\eta = 1 - \dfrac{RT_2\,\log_e(V_3/V_4)}{RT_1\,\log_e(V_2/V_1)}$
Points $B$ and $C$ lie on the same adiabat:
$\therefore\ T_1 V_2^{\,\gamma-1} = T_2 V_3^{\,\gamma-1} \quad\text{or}\quad \dfrac{T_1}{T_2} = \left(\dfrac{V_3}{V_2}\right)^{\gamma-1} \qquad \ldots\text{(i)}$
Points $D$ and $A$ also lie on the same adiabat:
$\therefore\ T_1 V_1^{\,\gamma-1} = T_2 V_4^{\,\gamma-1} \quad\text{or}\quad \dfrac{T_1}{T_2} = \left(\dfrac{V_4}{V_1}\right)^{\gamma-1} \qquad \ldots\text{(ii)}$
From Eqs. (i) and (ii):
$\dfrac{V_3}{V_2} = \dfrac{V_4}{V_1} \quad\text{or}\quad \dfrac{V_3}{V_4} = \dfrac{V_2}{V_1} \quad\Rightarrow\quad \log_e\!\left(\dfrac{V_3}{V_4}\right) = \log_e\!\left(\dfrac{V_2}{V_1}\right)$
So the two log terms in step 3 cancel, and only $T_2/T_1$ is left.

Efficiency of a Carnot Engine

\eta = 1 - \frac{T_2}{T_1}

where, $T_1$ = temperature of the source, $T_2$ = temperature of the sink

Key Points about Efficiency

The efficiency of a Carnot engine depends only on the temperatures of the source and the sink.
It does not depend on the nature of the working substance.
It would be 100% only if the sink were at 0 K. A sink at 0 K cannot be reached in practice. So 100% efficiency is not possible.

4.6Carnot's Theorem

Carnot's theorem has two parts:

No heat engine can be more efficient than a Carnot engine working between the same two temperatures. These are $T_1$ of the hot reservoir (source) and $T_2$ of the cold reservoir (sink).
The efficiency of a Carnot engine is independent of the nature of the working substance.

Proof of Part (i)

Take a reversible Carnot engine $R$ and an irreversible heat engine $I$ . Both work between the same source at $T_1$ and the same sink at $T_2$ .

The two engines are coupled. Engine $I$ runs as a heat engine. Engine $R$ runs backwards as a refrigerator.

Engine

I

and engine

R

working together

Derivation

Setup — In One Complete Cycle

Engine $I$ takes heat $Q_1$ from the source. It gives work $W'$ and releases the rest, $(Q_1 - W')$ , to the sink.
Engine $R$ returns the same heat $Q_1$ to the source when work $W$ is done on it. So it takes heat $(Q_1 - W)$ from the sink.

Assume that engine $R$ is less efficient than engine $I$ , that is, $\eta_R < \eta_I$ .
Then, for the same $Q_1$ :
$W < W' \quad \text{and} \quad (Q_1 - W) > (Q_1 - W')$
So the coupled $I$ – $R$ system, as a whole, takes the heat
$(Q_1 - W) - (Q_1 - W') = (W' - W)$
from the sink. It gives the same amount of work $(W' - W)$ in one cycle. The source has no net change.
This is against the second law of thermodynamics (Kelvin–Planck statement). So the assumption $\eta_I > \eta_R$ is wrong.

KeyNo engine can be more efficient than a Carnot engine. This proves part (i) of Carnot's theorem.

4.7Coefficient of Performance

For a Carnot refrigerator (see the refrigerator diagram in topic 4.1):

\dfrac{Q_1}{Q_2} = \dfrac{T_1}{T_2} \quad\Rightarrow\quad \dfrac{Q_1 - Q_2}{Q_2} = \dfrac{T_1 - T_2}{T_2} \quad\text{or}\quad \dfrac{Q_2}{Q_1 - Q_2} = \dfrac{T_2}{T_1 - T_2}

Coefficient of Performance

\beta = \frac{T_2}{T_1 - T_2}

where, $T_1$ = temperature of the surroundings, $T_2$ = temperature of the cold body, $Q_1 - Q_2 = W$ = work done on the refrigerator

Note

Clearly, $\beta = 0$ when $T_2 = 0$ . So the coefficient of performance is zero if the cold body is at absolute zero.

End of Chapter 11 · notesshala.in