Basic Concepts of Thermodynamic

Q: What is the difference between a system and its surroundings in thermodynamics?

The system is the specific portion of matter under study, such as a reaction mixture in a flask. The surroundings are everything else that can potentially exchange energy or matter with it. The boundary separates them and can be real or imaginary.

Q: Why is the universe considered a perfect isolated system?

By definition the universe contains everything, so there is nothing outside it that could exchange matter or energy with it. Hence exchanges across a universe-boundary are zero, satisfying the definition of an isolated system.

Q: Are heat and work state functions or path functions?

Both heat and work are path functions - their values depend on how the process is carried out. But their sum q + w equals the change in internal energy, which is a state function.

Q: Why is reversible work maximum for an ideal gas expansion?

In a reversible expansion the external pressure is only infinitesimally less than the gas pressure at every stage, so the gas pushes against the largest possible opposing pressure throughout. Any irreversible expansion involves a lower constant external pressure, giving smaller work per unit volume.

Q: Why does an ideal gas cool during adiabatic expansion?

No heat enters in an adiabatic expansion (q = 0), but the gas still does work. By the first law the internal energy must drop. For an ideal gas, internal energy depends only on temperature, so temperature falls.

Q: What is the IUPAC sign convention for heat and work?

Any energy added to the system is positive. Heat absorbed by the system is +q, heat released is -q; work done on the system is +w, work done by the system is -w. The first law reads Delta U = q + w.

Q: What is a cyclic process and what is special about it?

A cyclic process returns the system to its initial state. For any state function, the change over the cycle is zero, so Delta U = 0 and Delta H = 0. The total heat absorbed equals the total work done by the system.

ChemistryChemical ThermodynamicsFor NEET aspirants

Thermodynamics is the branch of science that deals with different forms of energy and their interconversion. In this concept you will learn the key language of thermodynamics - system, surroundings, types of systems, state functions vs path functions, thermodynamic processes (isothermal, isochoric, isobaric, adiabatic), reversible and irreversible processes, internal energy, and the mathematical form of the first law of thermodynamics ( $Δ U = q + w$ ). These building blocks appear in almost every question in the Thermodynamics chapter of Class 11 Chemistry for JEE and NEET.

Key Formulas - Quick Reference

First law of thermodynamics: $Δ U = q + w$ (IUPAC sign convention)
Work done on the gas (P-V work): $w = - \int P_{e x t} d V$
Isothermal reversible work (ideal gas): $w = - n R T ln \frac{V _{2}}{V _{1}} = - n R T ln \frac{P _{1}}{P _{2}}$
Isothermal irreversible work against constant $P_{e x t}$ : $w = - P_{e x t} (V_{2} - V_{1})$
Adiabatic process: $q = 0$ , so $Δ U = w$
Adiabatic reversible relation: $T V^{γ - 1} =$ constant, $P V^{γ} =$ constant
Adiabatic reversible work: $w = \frac{n R ( T _{2} - T _{1} )}{γ - 1} = \frac{P _{2} V _{2} - P _{1} V _{1}}{γ - 1}$
Internal energy of an ideal gas: $U = \frac{f}{2} n R T$ (function of $T$ only)

1. What is thermodynamics

The word thermodynamics literally means the flow of heat. It deals with energy changes accompanying every kind of physical and chemical process. Thermodynamics helps you predict:

Feasibility - whether two substances will react on mixing, and in which direction the reaction will proceed.
Extent - if the reaction reaches equilibrium, what the concentrations of reactants and products will be.
Energy changes - how much heat is absorbed or released, and how much work is done.

Limitations of thermodynamics. The laws apply to matter in bulk, not to individual particles. Also, thermodynamics tells you whether a change is possible, not how fast it will occur - the time to attain equilibrium is not predicted by thermodynamics.

2. System, surroundings, and boundary

System

A system is the specific part of the universe under study for energy changes. Examples: air in a room, water in a bottle, a chemical reaction mixture in a flask.

Surroundings

Everything outside the system that can potentially exchange energy or matter with the system is called the surroundings.

Universe = System + Surroundings

Boundary

Anything that separates the system from its surroundings is called the boundary. Boundaries can be:

Real or imaginary (an imagined surface can be a boundary).
Flexible or rigid - e.g. air in a balloon has a flexible boundary; air in a sealed room has a rigid one.
Adiabatic (non-conducting to heat) or diathermic (heat-conducting).

3. Types of systems

Type of system	Exchange of matter	Exchange of energy	Example
Open	Yes	Yes	Boiling water in an open kettle; any living organism
Closed	No	Yes	Reaction in a sealed glass flask
Isolated	No	No	Hot coffee in a well-sealed thermos flask (approximately); the whole universe (ideal)

Figure: Three types of systems classified by what can cross the boundary. Matter arrows in blue, energy (heat/work) arrows in red. Hatched shading on the isolated system's boundary shows perfect insulation.

Solved Example 1

Thermodynamics is concerned with:
(A) total energy in a system (B) energy changes in a system
(C) rate of a chemical change (D) mass changes in nuclear reactions

Solution:

(B) Thermodynamics tells us about changes in energy (not absolute total values), and its laws do not deal with the rate of a chemical change or nuclear mass defects.

4. State of a system and state functions

The state of a system is fixed by specifying measurable properties like pressure, volume, temperature, amount of substance and composition. When even one such property changes, the system is said to change into a new state.

State function (point function)

A property that depends only on the current state of the system - not on the path used to reach that state - is called a state function. Examples:

Temperature (

T

), pressure (

P

), volume (

V

), internal energy (

U

), enthalpy (

H

), Gibbs free energy (

G

), entropy (

S

) - all state functions.

Path function

A quantity whose value depends on the path taken by the system between two states is called a path function. Path functions are defined only during a process; they do not have a fixed value for a state.

Heat ( $q$ ) and Work ( $w$ ) are path functions. Heat capacity is also a path function.

5. Extensive vs intensive properties

A convenient test: divide the system into two parts (real or imaginary). Properties that keep the same value are intensive; properties whose values differ are extensive.

Property type	Depends on amount?	Additive?	Examples
Extensive	Yes	Yes (add up on combining parts)	Mass, volume, $U$ , $H$ , $S$ , $G$ , total heat capacity
Intensive	No	No	Temperature, pressure, density, molar heat capacity, specific heat, refractive index, vapour pressure, concentration

6. Thermodynamic equilibrium

A system is in thermodynamic equilibrium when no observable property changes with time. This has three components which must all hold:

Mechanical equilibrium - no pressure gradient in the system (for a gas).
Thermal equilibrium - no temperature gradient with time; temperature is constant.
Chemical equilibrium - no concentration gradient for any species; composition is uniform.

7. Thermodynamic processes

A thermodynamic process is any method by which the system moves from one equilibrium state to another. The important types are:

Process	What is constant	Mathematical condition
Isothermal	Temperature ( $T$ )	$d T = 0, Δ T = 0$
Isochoric	Volume ( $V$ )	$d V = 0, Δ V = 0$
Isobaric	Pressure ( $P$ )	$d P = 0, Δ P = 0$
Adiabatic	Heat exchange	$q = 0$
Cyclic	All state functions return to start	$Δ U = 0, Δ H = 0$

Figure: P-V diagrams of four important processes from a common initial state (

P_{1}, V_{1}

). Isobaric: horizontal (constant

P

). Isochoric: vertical (constant

V

). Isothermal:

P V =

constant (hyperbola). Adiabatic:

P V^{γ} =

constant (steeper hyperbola).

8. Reversible vs irreversible processes

Reversible process

A process that can be reversed by an infinitesimally small change in the driving force is called reversible. Practically:

The system is in thermodynamic equilibrium at every intermediate stage.
Driving force $-$ opposing force $= d F \to 0$ (infinitesimally small).
The process takes infinite time to complete.
A true reversible process is an idealisation - no real process is perfectly reversible.

Sand-piston picture. Imagine gas enclosed by a massless piston loaded with fine sand grains. Remove grains one at a time. Each removal shifts the piston by an infinitesimal amount; the gas is essentially in equilibrium at every stage. The expansion is reversible.

Irreversible process

System is in equilibrium only at initial and final states, not in between.
Driving force $-$ opposing force is a finite quantity, so the process happens at a finite rate.
Completed in finite time. All real processes are irreversible.
At intermediate stages, state functions like $P$ and $T$ are not defined for the system as a whole.

Quick tests for irreversibility. A process is irreversible if (a) it is carried out against a constant external pressure, or (b) it involves any sudden change in

P

V

T

9. Modes of energy exchange: heat and work

A system can exchange energy with its surroundings only in two ways:

Heat ( $q$ ) - energy transfer that happens because of a temperature difference between system and surroundings.
Work ( $w$ ) - any energy transfer that is not because of a temperature difference. Work can be mechanical (P-V work), electrical, magnetic, gravitational, etc.

IUPAC sign convention (used throughout chemistry)

Any energy added to the system is taken as positive.
\bullet\ Heat given to system:

q > 0

(positive)
\bullet\ Heat given out by system:

q < 0

(negative)
\bullet\ Work done on system:

w > 0

(positive)
\bullet\ Work done by system:

w < 0

(negative)

Chemistry vs Physics. Physics textbooks often use

Δ U = q - w

(with

w

= work done by the system). Chemistry (and JEE/NEET) uses

Δ U = q + w

with

w

= work done on the system. The physical answer is the same - only the sign of

w

differs.

10. Internal energy (U)

Every system contains a definite amount of energy stored in its molecules - kinetic (translation, rotation, vibration), potential (bonds, intermolecular forces), electronic, etc. The sum of all these microscopic energies is called internal energy, denoted $U$ (or $E$ ).

U = U_{translational} + U_{rotational} + U_{vibrational} + U_{bonding} + \dots

Key properties of $U$ :

Internal energy is a state function and an extensive property.
The absolute value of $U$ cannot be measured; only the change $Δ U$ is meaningful.
For an ideal gas, $U$ depends only on temperature: $(\frac{\partial U}{\partial V})_{T} = 0$ and $(\frac{\partial U}{\partial P})_{T} = 0$ .
By the equipartition theorem, for an ideal gas $U = \frac{f}{2} n R T$ , where $f$ is the degree of freedom ( $f = 3$ monoatomic, $5$ diatomic, $6$ non-linear polyatomic).

11. First law of thermodynamics

Statement. Energy can neither be created nor destroyed; it can only be transformed from one form to another. Equivalently, the total energy of an isolated system is constant. This is the law of conservation of energy.

Mathematical form

Let a system in state $A$ have internal energy $U_{A}$ . It absorbs $q$ from the surroundings and $w$ amount of work is done on it. In the new state $B$ the energy becomes $U_{B}$ . Then:

Δ U = U_{B} - U_{A} = q + w

This is the mathematical form of the first law. $Δ U$ is a state function; individually, $q$ and $w$ are not.

Solved Example 2

1

mole of an ideal monoatomic gas at

2 7^{\circ} C

expands adiabatically against a constant external pressure of

1.5 atm

from a volume of

4 dm^{3}

16 dm^{3}

. Calculate (i)

q

(ii)

w

and (iii)

Δ U

Solution:

(i) The process is adiabatic, so no heat exchange: $q = 0$ .

(ii) The gas expands against constant external pressure $P_{e x t} = 1.5 atm$ :

$w = - P_{e x t} (V_{2} - V_{1}) = - 1.5 \times (16 - 4) = - 18 atm dm^{3}$

(iii) By the first law: $Δ U = q + w = 0 + (- 18) = - 18 atm dm^{3}$ .

The negative sign tells us the internal energy of the gas decreased - consistent with the gas doing work and, being adiabatic, having no way to replace that energy from heat.

12. P-V work: expression for work done

Consider gas enclosed in a cylinder with a piston of area $A$ . If external pressure $P_{e x t}$ acts on the piston and the piston moves out by $d x$ , the work done on the gas is:

d w = - F_{e x t} d x = - P_{e x t} A d x = - P_{e x t} d V

Integrating between initial and final volumes:

w = - \int_{V_{1}}^{V_{2}} P_{e x t} d V

Here $P_{e x t}$ is the pressure the surroundings apply on the gas. For a reversible process $P_{e x t} = P_{g a s}$ at every instant; for an irreversible process $P_{e x t}$ is a specified constant that differs from $P_{g a s}$ .

Useful unit conversion

1 L atm = 101.325 J = 24.206 cal

13. Isothermal reversible expansion of an ideal gas

For an isothermal process on an ideal gas, $Δ T = 0$ so $Δ U = 0$ (since $U$ depends only on $T$ ). By the first law $q = - w$ : heat absorbed equals work done by the gas.

For reversible expansion, $P_{e x t} = P_{g a s} = n R T / V$ throughout:

w = - \int_{V_{1}}^{V_{2}} \frac{n R T}{V} d V = - n R T ln \frac{V _{2}}{V _{1}} = - n R T ln \frac{P _{1}}{P _{2}}

For an expansion $V_{2} > V_{1}$ , so $w < 0$ (work done by the gas). Correspondingly $q > 0$ (heat absorbed).

14. Isothermal irreversible expansion

If the gas expands isothermally against a constant external pressure $P_{e x t}$ (single step):

w = - P_{e x t} (V_{2} - V_{1})

If the same expansion is done in multiple steps (external pressure reduced gradually in $n$ finite jumps), the total work is the sum of stepwise contributions - and this total is larger in magnitude than the single-step work.

Key ranking (magnitude of work done by the gas in expansion):

∣ w_{single step} ∣ < ∣ w_{two step} ∣ < \dots < ∣ w_{reversible} ∣

Reversible expansion gives the maximum work. This is why heat engines are designed to approach reversibility.

Figure: Work done by the gas is the area under the

P

V

path. Between the same initial and final states, a reversible isothermal expansion (right, green) always gives a larger area - and hence more work - than any irreversible expansion (left, pink) against a constant external pressure.

Solved Example 3

10 g

of helium at

12 7^{\circ} C

is expanded isothermally from

100 atm

1 atm

. Calculate the work done when the expansion is carried out (i) in a single step, (ii) in three steps through intermediate pressures

60

and

30 atm

, and (iii) reversibly.

Solution:

Moles of He: $n = 10/4 = 2.5 mol$ . Temperature $T = 400 K$ . Use $R = 8.314 J K^{- 1} mol^{- 1}$ .

Initial volume: $V_{1} = \frac{n R T}{P _{1}} = \frac{2.5 \times 8.314 \times 400}{100 \times 1 0 ^{5}} m^{3} \approx 8.314 \times 1 0^{- 4} m^{3}$ .

Final volume: $V_{2} = \frac{n R T}{P _{2}} = 100 \times V_{1} = 8.314 \times 1 0^{- 2} m^{3}$ .

(i) Single step against $P_{e x t} = 1 atm$ :

$w = - P_{e x t} (V_{2} - V_{1}) = - (1 \times 1 0^{5}) (8.314 \times 1 0^{- 2} - 8.314 \times 1 0^{- 4}) \approx - 8231 J$

(ii) Three step: through $P_{e x t, 1} = 60$ , $P_{e x t, 2} = 30$ , $P_{e x t, 3} = 1 atm$ :

Compute volumes at each stage using $P V = n R T$ :

$V_{1} \approx 8.31 \times 1 0^{- 4} m^{3}$ , $V^{'} \approx 1.39 \times 1 0^{- 3} m^{3}$ , $V^{''} \approx 2.77 \times 1 0^{- 3} m^{3}$ , $V_{2} \approx 8.31 \times 1 0^{- 2} m^{3}$ .

$w_{1} = - (60 \times 1 0^{5}) (V^{'} - V_{1}) \approx - 3326 J$

$w_{2} = - (30 \times 1 0^{5}) (V^{''} - V^{'}) \approx - 4157 J$

$w_{3} = - (1 \times 1 0^{5}) (V_{2} - V^{''}) \approx - 8037 J$

$w_{total} = w_{1} + w_{2} + w_{3} \approx - 15519 J$

(iii) Reversibly:

$w = - n R T ln \frac{P _{1}}{P _{2}} = - 2.303 \times 2.5 \times 8.314 \times 400 \times lo g \frac{100}{1} \approx - 38294 J$

Note how $∣ w_{single} ∣ < ∣ w_{three step} ∣ < ∣ w_{reversible} ∣$ - the reversible case gives maximum work.

15. Adiabatic process

An adiabatic process has no heat exchange with the surroundings: $q = 0$ . By the first law $Δ U = w$ .

For an ideal gas $Δ U = n C_{V} Δ T$ , so:

w = Δ U = n C_{V} (T_{2} - T_{1})

In adiabatic expansion, gas does work at the cost of its internal energy - so $T$ falls.
In adiabatic compression, work is done on the gas, raising $T$ .

Reversible adiabatic relations (for an ideal gas)

Using $d U = - P d V$ and $P = n R T / V$ with $C_{p} - C_{v} = R$ (Mayer's relation), integrating gives the following relations for a reversible adiabatic process:

T V^{γ - 1} =

constant

P V^{γ} =

constant

T P^{(1 - γ) / γ} =

constant

where $γ = C_{p} / C_{v}$ is the heat capacity ratio (adiabatic index).

Work done in reversible adiabatic:

w = \frac{n R ( T _{2} - T _{1} )}{γ - 1} = \frac{P _{2} V _{2} - P _{1} V _{1}}{γ - 1}

Irreversible adiabatic against constant $P_{e x t}$

For an irreversible adiabatic expansion/compression against a constant external pressure, $P V^{γ} =$ constant does not apply. Use:

w = - P_{e x t} (V_{2} - V_{1}) = n C_{V} (T_{2} - T_{1})

These two expressions can be solved simultaneously (with the ideal gas law) for $T_{2}, V_{2}, P_{2}$ .

Two states can be linked by only one adiabatic path. If two states are connected by a reversible adiabatic path, they cannot also be connected by an irreversible adiabatic - because

U

is a state function (

Δ U_{r e v} = Δ U_{i r r}

q = 0

for both, so

w

would be identical - contradicting the fact that

w

is path-dependent.

16. Free expansion

An ideal gas expanding into vacuum has $P_{e x t} = 0$ , so:

$w = - \int P_{e x t} d V = 0$ .
If also adiabatic ( $q = 0$ ), then $Δ U = 0$ , so $Δ T = 0$ (temperature unchanged).
Free expansion is always irreversible.

Solved Example 4

A system absorbs

5 kJ

of heat and does

1 kJ

of work on the surroundings. Calculate the change in internal energy.

Solution:

Heat given to the system: $q = + 5 kJ$ .

Work done by the system: this means work done on the system is $w = - 1 kJ$ (IUPAC convention).

By first law: $Δ U = q + w = 5 + (- 1) = + 4 kJ$ .

Solved Example 5

Calculate the maximum work done when the pressure on

10 g

of hydrogen is reduced from

20 atm

1 atm

at a constant temperature of

273 K

. The gas behaves ideally. Also calculate

q

and

Δ U

Solution:

Maximum work $\Rightarrow$ reversible isothermal process. Moles: $n = 10/2 = 5 mol$ .

$w = - n R T ln \frac{P _{1}}{P _{2}} = - 2.303 \times 5 \times 2 \times 273 \times lo g \frac{20}{1}$ cal $\approx - 8180 cal$

Isothermal, ideal gas $\Rightarrow Δ U = 0$ .

By first law: $q = Δ U - w = 0 - (- 8180) = + 8180 cal$ (absorbed by the gas).

17. Common Mistakes to Avoid

Watch out

Mixing the IUPAC (chemistry) and physics sign conventions. In IUPAC, $Δ U = q + w$ where $w$ is work done on the system. If a problem says "work done by the gas is $X$ ", then $w = - X$ in your formula.
Using $w = - n R T ln (V_{2} / V_{1})$ for irreversible isothermal expansion. This log formula applies only to reversible isothermal. For irreversible constant- $P_{e x t}$ expansion use $w = - P_{e x t} (V_{2} - V_{1})$ .
Applying $P V^{γ} =$ constant to irreversible adiabatic processes. This relation holds only for reversible adiabatic changes on an ideal gas.
Treating $q$ or $w$ as state functions. Both are path functions. Only $Δ U$ , $Δ H$ , $Δ S$ etc. depend on states alone.
Forgetting that $Δ U = 0$ for an ideal gas isothermal process only. For real gases or liquids/solids, isothermal $Δ U \neq = 0$ .
Confusing "isothermal" with "adiabatic." Isothermal $\Rightarrow T$ constant, heat can flow. Adiabatic $\Rightarrow$ no heat flow, $T$ generally changes.

Frequently Asked Questions

Q1. What is the difference between a system and its surroundings in thermodynamics?

The system is the specific portion of matter under study - for example a reaction mixture in a flask. The surroundings are everything else that can potentially exchange energy or matter with it. The boundary separates them and can be real (walls of a container) or imaginary (a chosen surface in space).

Q2. Why is the universe considered a perfect isolated system?

By definition the universe contains everything - there is nothing "outside" it that could exchange matter or energy with it. Hence exchanges of matter and energy across a universe-boundary are zero, satisfying the definition of an isolated system.

Q3. Are heat and work state functions or path functions?

Both heat ( $q$ ) and work ( $w$ ) are path functions. Their individual values depend on how the process is carried out (reversible vs irreversible, single step vs multi-step). But the sum $q + w = Δ U$ is a state function - a remarkable result of the first law.

Q4. Why is reversible work maximum for an ideal gas expansion?

In a reversible expansion the external pressure is kept just infinitesimally less than the gas pressure at every stage, so the gas pushes against the largest possible opposing pressure throughout. In an irreversible expansion against a lower constant external pressure, the gas does less work per unit volume change - so the total work by the gas is smaller in magnitude.

Q5. Why does an ideal gas cool during adiabatic expansion?

In an adiabatic expansion no heat enters the gas ( $q = 0$ ). The gas still does work against the surroundings, and by the first law $Δ U = w$ . Since $w < 0$ (work done by the gas), $Δ U < 0$ . For an ideal gas $U$ depends only on $T$ , so the temperature must fall.

Q6. What is the IUPAC sign convention for heat and work?

Any energy added to the system is taken as positive. So heat absorbed by the system is $+ q$ , heat released is $- q$ ; work done on the system is $+ w$ , work done by the system is $- w$ . The first law then reads $Δ U = q + w$ .

Q7. What is a cyclic process and what is special about it?

A cyclic process is one in which the system returns to its initial state after a series of steps. For any state function $X$ , $Δ X_{cy c l e} = 0$ - so $Δ U = 0$ and $Δ H = 0$ over a full cycle. Consequently the total heat absorbed equals the total work done by the system: $q_{cy c l e} = - w_{cy c l e}$ .

Q8. Can the internal energy of a system decrease even if heat is added to it?

Yes. By the first law $Δ U = q + w$ . If the system does a large amount of work on the surroundings (large negative $w$ ) that exceeds the heat added, then $Δ U$ can become negative. A common example is isothermal expansion of a gas where heat is absorbed but internal energy stays constant, and if a process is more expansion-heavy, $Δ U$ can even fall.

Previous year questions on Basic Concepts of Thermodynamic

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1. What is thermodynamics

2. System, surroundings, and boundary

System

Surroundings

Boundary

3. Types of systems

4. State of a system and state functions

State function (point function)

Path function

5. Extensive vs intensive properties

6. Thermodynamic equilibrium

7. Thermodynamic processes

8. Reversible vs irreversible processes

Reversible process

Irreversible process

9. Modes of energy exchange: heat and work

IUPAC sign convention (used throughout chemistry)

10. Internal energy (U)

11. First law of thermodynamics

Mathematical form

12. P-V work: expression for work done

Useful unit conversion

13. Isothermal reversible expansion of an ideal gas

14. Isothermal irreversible expansion

15. Adiabatic process

Reversible adiabatic relations (for an ideal gas)

Irreversible adiabatic against constant Pext​

16. Free expansion

17. Common Mistakes to Avoid

Frequently Asked Questions

Q1. What is the difference between a system and its surroundings in thermodynamics?

Q2. Why is the universe considered a perfect isolated system?

Q3. Are heat and work state functions or path functions?

Q4. Why is reversible work maximum for an ideal gas expansion?

Q5. Why does an ideal gas cool during adiabatic expansion?

Q6. What is the IUPAC sign convention for heat and work?

Q7. What is a cyclic process and what is special about it?

Q8. Can the internal energy of a system decrease even if heat is added to it?

Previous year questions on Basic Concepts of Thermodynamic

Ready to master Chemical Thermodynamics?

Irreversible adiabatic against constant $P_{e x t}$