§ 5.1NCERT Class 11 · Chemistry · Chapter 5
Thermodynamic Terms & First Law of Thermodynamics
Chemical Thermodynamics is the branch of physical chemistry that deals with the energy transformations, heat transfers, and work exchanges accompanying physical and chemical processes. It governs whether a chemical reaction can occur and what energy equilibrium it establishes.
1. Basic Concepts: System, Surroundings & Boundaries
Thermodynamics divides the physical universe into two distinct parts separated by a boundary:
- System: The specific macroscopic portion of the universe under thermodynamic observation (e.g., reacting chemical mixture in a flask).
- Surroundings: Everything in the universe outside the system that can interact with it (in practice, the immediate neighborhood).
- Boundary: The real or imaginary surface separating the system from its surroundings. A boundary may be rigid or movable, and diathermic (conducting heat) or adiabatic (thermally insulating).
| System Type | Exchange of Matter | Exchange of Energy (Heat/Work) | Real-world Chemical Example |
|---|---|---|---|
| Open System | Yes | Yes | Reactants in an open beaker; boiling water in an open pan |
| Closed System | No | Yes | Water in a sealed metal cylinder; gas in a piston flask |
| Isolated System | No | No | Liquid in a perfectly insulated, sealed thermos flask |
2. State Functions vs Path Functions & Properties
A. State of a System & State Functions
The state of a thermodynamic system is described by its macroscopic variables such as pressure (P), volume (V), temperature (T), and amount of substance (n). A State Function (or State Variable) is a property whose value depends solely on the present equilibrium state of the system, completely independent of how that state was reached.
Key State Functions: Pressure (P), Volume (V), Temperature (T), Internal Energy (U), Enthalpy (H), Entropy (S), Gibbs Free Energy (G), Helmholtz Free Energy (A).
For any cyclic process involving state function Z: ∮ dZ = 0, so ΔZ = Zfinal − Zinitial.
For any cyclic process involving state function Z: ∮ dZ = 0, so ΔZ = Zfinal − Zinitial.
B. Path Functions
A Path Function is a property whose value depends explicitly on the path or mechanistic route taken between initial and final states. The two principal path functions in thermodynamics are Heat (q) and Work (w). We write ΔU or ΔH, but never Δq or Δw.
C. Extensive vs Intensive Properties
| Property Class | Definition | Examples |
|---|---|---|
| Extensive Properties | Depend directly on the quantity or mass of matter present in the system. Additive in nature. | Mass (m), Volume (V), Internal Energy (U), Enthalpy (H), Entropy (S), Gibbs Energy (G), Heat capacity (C). |
| Intensive Properties | Independent of the total mass or size of the system. Non-additive. | Temperature (T), Pressure (P), Density (ρ), Refractive index, Molar volume (Vm), Molar heat capacity (Cm), Specific heat (c), Viscosity, Electromotive force (E°). |
NEET/JEE Trap: The ratio of any two extensive properties is always an intensive property! For example, Mass (ext.) / Volume (ext.) = Density (int.); Enthalpy (ext.) / Moles (ext.) = Molar Enthalpy (int.).
3. Thermodynamic Processes & Reversible vs Irreversible
- Isothermal Process: Temperature remains constant throughout (ΔT = 0; for an ideal gas, ΔU = 0).
- Adiabatic Process: No heat enters or leaves the system (q = 0; ΔU = wad).
- Isobaric Process: Pressure remains constant (ΔP = 0; work w = −PΔV).
- Isochoric Process: Volume remains constant (ΔV = 0; work w = 0; ΔU = qv).
- Cyclic Process: System returns to its initial state (ΔU = 0, ΔH = 0).
Thermodynamic Work: Reversible vs Irreversible P-V Diagrams
Comparison of work done during gas expansion against external pressure
Reversible Isothermal Expansion Irreversible Expansion (Single Step)
Pressure (P) Pressure (P)
^ ^
P1|* P1|*
| \ |
| \ P_ext = P_int - dP |
| \ (Infinite Steps) |
| \ |
P2|-----\* P2|-------* (P_ext = P2 constant)
| | |///////|
+------+----> Volume (V) +-------+----> Volume (V)
0 V1 V2 0 V1 V2
Area under curve = Maximum Work Area of rectangle = -P_ext * (V2 - V1)
|w_rev| > |w_irrev| Less work produced by system
Key Physical Insight: Reversible expansion is carried out infinitesimally slowly so that Pext ≈ Pint at every stage. It produces the maximum work done by the system. Irreversible single-step expansion against constant external pressure produces significantly less useful work, dissipating energy into the surroundings.
4. Work, Heat & The First Law of Thermodynamics
A. Expansion Work Formulation
When a gas expands from volume V1 to V2 against an opposing external pressure Pext:
General Formula: w = − ∫ Pext dV
- Constant External Pressure (Irreversible): w = −Pext (V2 − V1) = −Pext ΔV
- Free Expansion in Vacuum (Pext = 0): w = 0 (no work is done in vacuum expansion!)
- Reversible Isothermal Expansion of Ideal Gas (T = const):
Since P = nRT / V:
wrev = − ∫ (nRT / V) dV = − 2.303 nRT log10(V2 / V1) = − 2.303 nRT log10(P1 / P2)
B. Mathematical Statement of First Law
The First Law of Thermodynamics is the law of conservation of energy: Energy can neither be created nor destroyed, although it may be converted from one form to another.
First Law Formulation: ΔU = q + w
Where:
- ΔU = Change in internal energy of the system (in Joules or kJ)
- q = Heat transferred into the system (+ve if absorbed, −ve if evolved)
- w = Work done on the system (+ve if compression, −ve if expansion)
5. Special Cases of First Law for Ideal Gases
| Process Condition | Constraint | Consequence on First Law (ΔU = q + w) |
|---|---|---|
| Isochoric Process | ΔV = 0 ⇒ w = 0 | ΔU = qv (Heat exchanged at constant volume equals change in internal energy) |
| Adiabatic Process | q = 0 | ΔU = wad (Work done equals internal energy change; expansion cools the gas) |
| Isothermal Process | ΔT = 0 ⇒ ΔU = 0 | q = −w (All absorbed heat is converted to expansion work) |
| Free Expansion (Isothermal) | Pext = 0, ΔT = 0 | w = 0, q = 0, ΔU = 0 |
| Cyclic Process | State returns to start ⇒ ΔU = 0 | qnet = −wnet |
6. Solved High-Yield Numerical Examples & Exam MCQs
Example 1 (JEE Main): A gas expands from a volume of 2.0 L to 6.0 L against a constant external pressure of 3.0 atm. The heat absorbed by the gas during this expansion is 800 J. Calculate the change in internal energy (ΔU) of the gas. (Take 1 L·atm = 101.3 J)
Step-by-step Solution:
1. Calculate expansion work: w = −Pext ΔV
ΔV = V2 − V1 = 6.0 L − 2.0 L = 4.0 L
w = −(3.0 atm) × (4.0 L) = −12.0 L·atm
2. Convert work to Joules: w = −12.0 × 101.3 J = −1215.6 J
3. Heat absorbed: q = +800 J
4. Apply First Law: ΔU = q + w = +800 J + (−1215.6 J) = −415.6 J.
The negative sign indicates that the internal energy of the system decreases by 415.6 J.
1. Calculate expansion work: w = −Pext ΔV
ΔV = V2 − V1 = 6.0 L − 2.0 L = 4.0 L
w = −(3.0 atm) × (4.0 L) = −12.0 L·atm
2. Convert work to Joules: w = −12.0 × 101.3 J = −1215.6 J
3. Heat absorbed: q = +800 J
4. Apply First Law: ΔU = q + w = +800 J + (−1215.6 J) = −415.6 J.
The negative sign indicates that the internal energy of the system decreases by 415.6 J.
Practice MCQ 1 (NEET): Two moles of an ideal gas undergo isothermal reversible expansion from 2 L to 20 L at 300 K. What is the work done on the gas, w (IUPAC sign convention)? (R = 8.314 J K−1 mol−1)
(A) −11.49 kJ
(B) +11.49 kJ
(C) −5.74 kJ
(D) zero
Show answer
Correct Answer: (A) −11.49 kJ
Explanation:
For isothermal reversible expansion of an ideal gas:
wrev = −2.303 nRT log10(V2 / V1)
Here n = 2 mol, T = 300 K, V2/V1 = 20/2 = 10, log10(10) = 1.
w = −2.303 × 2 × 8.314 × 300 × 1 = −11488 J = −11.49 kJ.
The gas does 11.49 kJ of work on the surroundings, so w (work done on the gas) = −11.49 kJ.
For isothermal reversible expansion of an ideal gas:
wrev = −2.303 nRT log10(V2 / V1)
Here n = 2 mol, T = 300 K, V2/V1 = 20/2 = 10, log10(10) = 1.
w = −2.303 × 2 × 8.314 × 300 × 1 = −11488 J = −11.49 kJ.
The gas does 11.49 kJ of work on the surroundings, so w (work done on the gas) = −11.49 kJ.
Practice MCQ 2 (JEE Main): Which one among the following sets contains ONLY intensive properties?
(A) Volume, Temperature, Density
(B) Internal energy, Enthalpy, Heat capacity
(C) Pressure, Molar heat capacity, Refractive index
(D) Mass, Gibbs energy, Entropy
Show answer
Correct Answer: (C) Pressure, Molar heat capacity, Refractive index
Explanation:
Intensive properties are independent of system size/mass. Volume, internal energy, enthalpy, heat capacity, mass, Gibbs energy, and entropy are extensive. In option (C), Pressure, Molar heat capacity (Cm = C/n), and Refractive index are all strictly intensive.
Intensive properties are independent of system size/mass. Volume, internal energy, enthalpy, heat capacity, mass, Gibbs energy, and entropy are extensive. In option (C), Pressure, Molar heat capacity (Cm = C/n), and Refractive index are all strictly intensive.
7. Frequently Asked Questions (FAQs)
1. Does temperature change during the adiabatic expansion of an ideal gas?
Yes. In an adiabatic expansion, q = 0. Therefore, ΔU = w. Since the gas expands, work is done by the gas (w < 0), which implies ΔU < 0. For an ideal gas, internal energy is proportional to temperature (ΔU = nCvΔT), so the temperature drops (ΔT < 0), causing a cooling effect.
2. What is the difference between molar heat capacity and specific heat capacity?
Specific heat capacity (c) is the amount of heat required to raise the temperature of 1 gram of substance by 1 K (or 1 °C), with units J g−1 K−1. Molar heat capacity (Cm) is the heat required to raise the temperature of 1 mole of substance by 1 K, with units J mol−1 K−1. Both are intensive properties.
3. Can a system do work without absorbing heat?
Yes, in an adiabatic process (q = 0). The work is performed entirely at the expense of the system's own internal energy (−w = −ΔU).
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