"Thermodynamics"

About

Thermodynamics deals with heat, work, temperature, and their relationship to energy. Unlike kinetic theory which looks at microscopic molecular behaviour, thermodynamics takes a macroscopic approach — it studies bulk systems using a few measurable variables. This chapter covers the four laws of thermodynamics, thermodynamic processes, heat engines, and the all-important Carnot engine.

Key Concepts

11.1 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 in thermal equilibrium with each other.

The zeroth law provides the basis for the concept of temperature. A thermometer works on this principle.

11.2 Heat and Internal Energy

Heat: Energy transferred between a system and its surroundings due to temperature difference. SI unit: joule (J). Also measured in calories (1 cal = 4.186 J).

Internal energy ($U$): The sum total of kinetic energy and potential energy of all molecules of a system.

• Internal energy depends only on temperature for an ideal gas
• It is a state function — depends only on the current state, not the path

11.3 Work in Thermodynamics

Work done BY a gas during expansion: $W={\int }_{V_i}^{V_f}PdV$

In a PV diagram (indicator diagram):

• Work done = Area under the P-V curve
• Expansion: positive work (done BY system)
• Compression: negative work (done ON system)

Sign convention:

• $+W$: Work done BY the system
• $-W$: Work done ON the system

11.4 First Law of Thermodynamics

The heat supplied to a system equals the sum of the increase in its internal energy and the external work done by it.

$\Delta Q=\Delta U+\Delta W$

This is essentially the law of conservation of energy for thermodynamic systems.

Special cases:

11.5 Thermodynamic Processes

Isothermal process ($T$ = constant):

• $PV=\text{constant}$ (Boyle's law)
• $\Delta U=0$
• Work: $W=nRT\ln \frac{V_f}{V_i}$

Adiabatic process ($\Delta Q=0$):

• $P{V}^{\gamma }=\text{constant}$
• $T{V}^{\gamma -1}=\text{constant}$
• Work: $W=\frac{P_i{V}_i-P_f{V}_f}{\gamma -1}$

Isobaric process ($P$ = constant):

• $V/T=\text{constant}$ (Charles' law)

Isochoric process ($V$ = constant):

• $P/T=\text{constant}$ (Gay-Lussac's law)

11.6 Reversible and Irreversible Processes

11.7 Second Law of Thermodynamics

Kelvin-Planck statement: It is impossible to construct a heat engine that converts ALL heat from a source into work, rejecting no heat to the sink.

Clausius statement: Heat cannot flow from a colder body to a hotter body without the aid of an external agency.

11.8 Heat Engine and Carnot Engine

A heat engine converts heat into mechanical work:

• Takes heat $Q_1$ from a hot source at $T_1$
• Converts part to work $W$
• Rejects remaining heat $Q_2$ to a cold sink at $T_2$

Efficiency: $\eta =\frac{W}{Q_1}=1-\frac{Q_2}{Q_1}$

Carnot engine: The most efficient theoretical heat engine, operating between $T_1$ and $T_2$ on a reversible cycle.

${\eta }_{\text{Carnot}}=1-\frac{T_2}{T_1}$

• Efficiency depends ONLY on source and sink temperatures
• $\eta <1$ always (cannot be 100%)
• $\eta =1$ only if $T_2=0$ K (impossible)

Triple point of water: The unique temperature and pressure (273.16 K, 611 Pa) where solid, liquid, and vapour phases of water coexist in equilibrium.

INTEXT QUESTIONS 11.1

Q1. Fill in the blanks:

(i) Zeroth law of thermodynamics provides the basis for the concept of _________.

Ans: Temperature.

(ii) If a system A is in thermal equilibrium with a system B and B is in thermal equilibrium with another system C, then system A will also be in thermal equilibrium with system __________.

Ans: C.

(iii) The unit of heat is _____________.

Ans: Joule or Calorie.

Q2. An indicator diagram shows a thermodynamic process. Calculate the work done by the system in the process: (a) along the path ABC from A to C, (b) If the system is returned from C to A along the same path, how much work is done by the system.

Ans: (a) Work done by the system = Area under the curve from A to C = $P_2\times (V_2-V_1)$

(b) Returned from C to A along the same path (opposite direction): Work done by the system = $-P_2\times (V_2-V_1)$ (negative because work is done ON the system)

Q3. Fill in the blanks:

(i) A reversible process is that which can be _________ in the opposite direction from its final state to its initial state.

Ans: Retrace.

(ii) An _________ process is that which cannot be retraced along the same equilibrium states from final state to the initial state.

Ans: Irreversible.

Q4. State the basic difference between isothermal and adiabatic processes.

Ans: An isothermal process occurs at constant temperature, whereas an adiabatic process occurs without any heat exchange between the system and surroundings.

Q5. State one characteristic of the triple point.

Ans: At the triple point, all three states of matter — solid, liquid, and vapour — can coexist in equilibrium.

INTEXT QUESTIONS 11.2

Q1. Fill in the blanks:

(i) The total of kinetic energy and potential energy of molecules of a system is called its _________.

Ans: Internal energy.

(ii) Work done = –W indicates that work is done _____________ the system.

Ans: On.

Q2. The first law of thermodynamics states that _____________.

Ans: The amount of heat given to a system is equal to the sum of the change in internal energy of the system and the external work done: $\Delta Q=\Delta U+\Delta W$.

INTEXT QUESTIONS 11.3

Q1. State whether the following statements are true or false.

(i) In a Carnot engine, when heat is taken by a perfect gas from a hot source, the temperature of the source decreases.

Ans: False.

(ii) In Carnot engine, if temperature of the sink is decreased the efficiency of engine also decreases.

Ans: True. (Lower $T_2$ → higher $\eta =1-T_2/T_1$)

Q2. (i) A Carnot engine has the same efficiency between 1000 K and 500 K and between T K and 1000 K. Calculate T.

Ans: ${\eta }_1=1-\frac{500}{1000}=0.5$ ${\eta }_2=1-\frac{1000}{T}$

Since ${\eta }_1={\eta }_2$: $0.5=1-\frac{1000}{T}⟹\frac{1000}{T}=0.5⟹T=2000\text{ K}$

(ii) A Carnot engine working between an unknown temperature T and ice point gives an efficiency of 0.68. Deduce the value of T.

Ans:

• Ice point = 273 K
• $\eta =1-\frac{273}{T}=0.68$
• $\frac{273}{T}=1-0.68=0.32$
• $T=\frac{273}{0.32}=853.1$ K

Terminal Exercise

1. State the zeroth law of thermodynamics and explain how it leads to the concept of temperature.

2. State and explain the first law of thermodynamics. Write its mathematical form and explain the sign convention.

3. Derive the expression for work done in: (a) an isothermal process, (b) an adiabatic process.

4. Distinguish between isothermal and adiabatic processes. Draw PV diagrams for each.

5. Define molar specific heats $C_P$ and $C_V$. Derive Mayer's relation $C_P-C_V=R$.

6. State the second law of thermodynamics (both Kelvin-Planck and Clausius statements).

7. Describe the construction and working of a Carnot engine. Derive the expression for its efficiency.

8. A Carnot engine working between 300 K and 600 K has a work output of 800 J per cycle. Find the heat supplied.

9. Explain why the efficiency of a Carnot engine is always less than 1. Can it ever be 100%?

10. 5 moles of an ideal gas expand isothermally at 300 K from 2 L to 10 L. Calculate the work done. ($R=8.31$ J/mol⋅K)

11. A gas is compressed adiabatically such that its volume decreases to half. If initial temperature is 27°C and $\gamma =1.4$, find the final temperature.

12. Distinguish between reversible and irreversible processes with examples. Why are all natural processes irreversible?

Worked Examples

Example 1: First Law

Problem: A system absorbs 500 J of heat and does 200 J of work. Find the change in internal energy.

Solution: $\Delta U=\Delta Q-\Delta W=500-200=300\text{ J}$

Example 2: Isothermal Work

Problem: 2 moles of an ideal gas at 300 K expand isothermally from 1 L to 5 L. Find the work done. ($R=8.31$ J/mol⋅K)

Solution: $W=nRT\ln \frac{V_f}{V_i}=2\times 8.31\times 300\times \ln 5$ $=4986\times 1.609=8022\text{ J}$

Example 3: Carnot Efficiency

Problem: A Carnot engine operates between 500 K and 300 K. Find its efficiency. If it receives 2000 J of heat, how much work does it do?

Solution: $\eta =1-\frac{T_2}{T_1}=1-\frac{300}{500}=0.4=40\%$

$W=\eta \times Q_1=0.4\times 2000=800\text{ J}$

Common Mistakes

1. Confusing isothermal and adiabatic: Isothermal = constant T (heat exchanged); adiabatic = no heat exchange ($\Delta Q=0$).
2. Using Celsius instead of Kelvin in Carnot efficiency: $T$ must be in kelvin.
3. Forgetting sign convention: $W$ positive when done BY system, negative when done ON system.
4. Thinking heat engines can be 100% efficient: Second law forbids this — some heat must always be rejected.
5. Applying gas law equations to adiabatic processes incorrectly: Use $P{V}^{\gamma }=\text{const}$, not $PV=\text{const}$.

Quick Revision