1. Basic Concepts & Systems

Thermodynamics is the study of energy transformations. For NEET, understanding the boundary between the system and the universe is step one.

System: The specific part of the universe under observation.
Surroundings: Everything else (Universe - System).

1.1 Types of Systems

Open System: Exchanges both matter and energy (e.g., Boiling water in an open pan).
Closed System: Exchanges energy but NOT matter (e.g., Hot water in a sealed metallic flask).
Isolated System: Exchanges neither energy nor matter (e.g., A perfectly insulated thermos).

Unknown Fact:

The measurement of temperature (Thermometry) is based on the Zeroth Law of Thermodynamics, which was actually formulated after the First Law but is so fundamental it was given the number zero.

1.2 Thermodynamic Properties

1. Extensive Properties: Depend on quantity of matter.

Examples: Mass, Volume, Internal Energy (U), Enthalpy (H), Heat Capacity (C).

2. Intensive Properties: Independent of quantity.

Examples: Temperature, Pressure, Density, Molar Heat Capacity.

NEET Quick Byte:

Ratio of two Extensive properties is always Intensive.
Example: Mass / Volume = Density (Intensive).

2. First Law & Internal Energy

Statement: Energy can neither be created nor destroyed. The energy of an isolated system is constant.

ΔU = q + w

(Mathematical Form of First Law)

IUPAC Sign Convention

Process	Sign
Heat absorbed by system	+ve (q > 0)
Heat released by system	-ve (q < 0)
Work done ON system (Compression)	+ve (w > 0)
Work done BY system (Expansion)	-ve (w < 0)

Work Done (w) Formulas

Irreversible Process: w = -P_ext ΔV
Reversible Isothermal: w = -2.303 nRT log(V₂/V₁)
Free Expansion (Vacuum): P_ext = 0, so w = 0.

3. Enthalpy (H)

Enthalpy is the total heat content of a system at constant pressure.

ΔH = ΔU + PΔV
For Gaseous reactions:
ΔH = ΔU + Δn_gRT

Where Δn_g = (Moles of gaseous products) - (Moles of gaseous reactants).

Heat Capacity (C)

Relationship between constant pressure (C_p) and constant volume (C_v) heat capacities for 1 mole of ideal gas:

C_p - C_v = R

4. Entropy & Second Law

Entropy (S): Measure of randomness or disorder.
Order: Gas (Max) > Liquid > Solid (Min).

Second Law Statement:

For any spontaneous process, the total entropy of the universe (System + Surroundings) must increase.
ΔS_total > 0

5. Gibbs Free Energy (G)

The single parameter to decide spontaneity at constant T and P.

ΔG = ΔH - TΔS

Criteria for Spontaneity

ΔG Value	Result
Negative (ΔG < 0)	Process is Spontaneous
Positive (ΔG > 0)	Non-Spontaneous
Zero (ΔG = 0)	System at Equilibrium

Relation with Equilibrium Constant:

ΔG° = -2.303 RT log K

Thermodynamics: Numericals & HOTS

Q1. Isothermal Expansion Work

Two moles of an ideal gas are expanded isothermally and reversibly from a volume of 10 L to 100 L at 300 K. Calculate the work done. (R = 8.314 J K^-1 mol^-1)

Solution:
Formula: w = -2.303 nRT log(V₂/V₁)

Given:
n = 2 mol, T = 300 K, V₁ = 10 L, V₂ = 100 L

w = -2.303 × 2 × 8.314 × 300 × log(100/10)
w = -2.303 × 4988.4 × log(10)
w = -11488 J
w = -11.488 kJ

Q2. First Law Application

In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy (ΔU)?

Solution:
Sign Convention:
Heat absorbed (q) = +701 J
Work done by system (w) = -394 J (Expansion)

Formula: ΔU = q + w
ΔU = 701 + (-394)
ΔU = +307 J

Q3. ΔH vs ΔU Relation

For the reaction N₂(g) + 3H₂(g) → 2NH₃(g) at 298 K, calculate the difference between ΔH and ΔU. (R = 8.314 J K^-1 mol^-1)

Solution:
Formula: ΔH - ΔU = Δn_gRT

Step 1: Calculate Δn_g (Gaseous Products - Reactants)
Δn_g = 2 - (1 + 3) = 2 - 4 = -2 mol

Step 2: Calculate Value
Value = -2 × 8.314 × 298
Value = -4955 J
ΔH - ΔU = -4.955 kJ

Q4. Bomb Calorimeter (HOTS)

1g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atm pressure. The temperature rises from 298 K to 299 K. If the heat capacity of the calorimeter is 20.7 kJ/K, what is the enthalpy change for the combustion of 1 mole of graphite?

Solution:
Step 1: Calculate Heat (q) absorbed by calorimeter.
q = C × ΔT = 20.7 kJ/K × (299 - 298) K = 20.7 kJ
Heat released by system (q_v) = -20.7 kJ (Exothermic)

Step 2: Convert to Molar basis.
Moles of C = 1g / 12g mol^-1 = 1/12 mol
ΔU (per mole) = -20.7 × 12 = -248.4 kJ/mol

Step 3: Calculate ΔH
Reaction: C(s) + O₂(g) → CO₂(g)
Δn_g = 1 - 1 = 0
Since Δn_g = 0, ΔH = ΔU = -248.4 kJ/mol

Q5. Hess's Law

Calculate the standard enthalpy of formation of CH₃OH(l) from the following data:
1. CH₃OH(l) + 1.5O₂(g) → CO₂(g) + 2H₂O(l); ΔH = -726 kJ
2. C(graphite) + O₂(g) → CO₂(g); ΔH = -393 kJ
3. H₂(g) + 0.5O₂(g) → H₂O(l); ΔH = -286 kJ

Solution:
Target Eq: C + 2H₂ + 0.5O₂ → CH₃OH

Operation:
Eq(2) + 2 × Eq(3) - Eq(1)
= (-393) + 2(-286) - (-726)
= -393 - 572 + 726
= -965 + 726
ΔH_f = -239 kJ/mol

Q6. Entropy of Vaporization

The enthalpy of vaporization of a liquid is 30 kJ/mol and the entropy of vaporization is 75 J K^-1 mol^-1. Calculate the boiling point of the liquid at 1 atm.

Solution:
At boiling point, the system is in equilibrium, so ΔG = 0.
Therefore, ΔH = TΔS

T = ΔH / ΔS
(Ensure units match! 30 kJ = 30,000 J)
T = 30000 / 75
T = 400 K

Q7. Spontaneity (Conceptual)

For the reaction 2A(g) + B(g) → 2D(g), ΔU = -10.5 kJ and ΔS = -44.1 J/K. Determine if the reaction is spontaneous at 25°C.

Solution:
Step 1: Calculate ΔH
Δn_g = 2 - (2+1) = -1
ΔH = ΔU + Δn_gRT
ΔH = -10500 + (-1)(8.314)(298) = -12977 J

Step 2: Calculate ΔG
ΔG = ΔH - TΔS
ΔG = -12977 - (298)(-44.1)
ΔG = -12977 + 13141.8
ΔG = +164.8 J

Since ΔG is positive, the reaction is Non-Spontaneous.

Q8. Bond Energy Calculation

Calculate the C-C bond energy from the following data:
(i) 2C(s) + 3H₂(g) → C₂H₆(g); ΔH = -84.6 kJ
(ii) Sublimation energy of C(s) = 716 kJ/mol
(iii) B.E. (H-H) = 436 kJ/mol
(iv) B.E. (C-H) = 413 kJ/mol

Solution:
For reaction (i):
ΔH = [Energy required to form atoms] - [Energy released forming bonds]

Reactant Side (Breaking/Sublimation):
2 C(s) → 2 C(g) : 2 × 716 = 1432
3 H-H → 6 H(g) : 3 × 436 = 1308
Total Input = 2740 kJ

Product Side (Forming C₂H₆):
1 C-C bond + 6 C-H bonds
Energy Released = 1(BE_CC) + 6(413)

Equation:
-84.6 = 2740 - [BE_CC + 2478]
-84.6 = 2740 - 2478 - BE_CC
-84.6 = 262 - BE_CC
BE_CC = 262 + 84.6
C-C Bond Energy = 346.6 kJ/mol

Q9. Adiabatic Expansion (HOTS)

1 mole of an ideal monoatomic gas (γ = 1.66) at 300 K expands adiabatically against a constant external pressure of 1 atm from 10 L to 20 L. Calculate the final temperature.

Solution:
Note: This is an Irreversible Adiabatic expansion.
q = 0, so ΔU = w
nC_v(T₂ - T₁) = -P_ext(V₂ - V₁)

For monoatomic gas, C_v = 3R/2.
1 × 1.5R × (T₂ - 300) = -1 atm × (20 - 10) L

Convert R to L-atm units: R = 0.0821 L atm K^-1
1.5(0.0821)(T₂ - 300) = -10
0.123 (T₂ - 300) = -10
T₂ - 300 = -81.3
Final Temp (T₂) = 218.7 K

Q10. ΔG and K

Calculate the equilibrium constant for a reaction at 298 K if ΔG° = -13.6 kJ/mol. (R = 8.314 J K^-1 mol^-1)

Solution:
Formula: ΔG° = -2.303 RT log K

Convert kJ to J:
-13600 = -2.303 × 8.314 × 298 × log K
-13600 = -5705.8 × log K
log K = 13600 / 5705.8
log K = 2.38
K = antilog(2.38)
K = 2.4 × 10² = 240

Thermodynamics Formulas

1. First Law of Thermodynamics

ΔU = q + w

2. Work Done (w)

Irreversible (Constant Ext. Pressure):

w = -P_ext ΔV

Reversible Isothermal (Ideal Gas):

w = -2.303 nRT log(V₂/V₁)

3. Enthalpy & Internal Energy

ΔH = ΔU + Δn_gRT

(Δn_g = gaseous products moles - gaseous reactants moles)

4. Mayer's Relation

C_p - C_v = R

5. Entropy Change (ΔS)

ΔS = q_rev / T

Total Entropy (Spontaneity):

ΔS_total > 0

6. Gibbs Free Energy (ΔG)

ΔG = ΔH - TΔS

Relation with Equilibrium Constant:

ΔG° = -2.303 RT log K

20 Golden Facts for NEET

1. Free Expansion: Expansion of a gas into a vacuum (P_ext = 0) involves Zero Work (w = 0). If the gas is ideal, ΔT = 0 and ΔU = 0 as well.
2. Intensive Ratio: The ratio of two extensive properties is always an Intensive Property. (e.g., Mass/Volume = Density, Enthalpy/Mole = Molar Enthalpy).
3. Adiabatic Cooling: When a gas expands adiabatically, it does work at the cost of its internal energy. This causes the temperature to FALL.
4. Standard Enthalpy: By convention, the standard enthalpy of formation (Δ_fH°) of an element in its standard reference state (e.g., O₂, Cl₂, C-graphite, S-rhombic) is taken as ZERO.
5. Bomb Calorimeter: This device operates at constant volume. Therefore, the heat measured (q_v) corresponds directly to the change in Internal Energy (ΔU), not Enthalpy.
6. Cyclic Process: For a cyclic process, the system returns to its initial state. Hence, change in all state functions is zero: ΔU = 0, ΔH = 0.
7. Third Law: The entropy of a perfectly crystalline solid at Absolute Zero (0 K) is ZERO. This is the reference point for calculating absolute entropies.
8. Mayer's Relation: For 1 mole of an ideal gas, C_p - C_v = R. This means C_p is always greater than C_v because work is done during expansion at constant pressure.
9. Bond Energy Trick: Unlike formation enthalpies, when calculating reaction enthalpy using Bond Energies: ΔH = ΣB.E.(Reactants) - ΣB.E.(Products).
10. Strong Acid-Base: The enthalpy of neutralization of a strong acid and strong base is constant: -57.1 kJ/mol (or -13.7 kcal/mol).
11. Path Functions: Heat (q) and Work (w) are path functions. Their value depends on the route taken. All other thermodynamic properties (P, V, T, U, H, S, G) are State Functions.
12. Equilibrium Condition: At thermodynamic equilibrium, Gibbs Free Energy change is minimum and constant, so ΔG = 0.
13. Solids & Liquids: For reactions involving only solids and liquids, volume change is negligible. Thus, ΔH ≈ ΔU.
14. Spontaneity Check: If ΔH is negative (Exothermic) and ΔS is positive (Disorder increases), the reaction is Spontaneous at ALL temperatures.
15. Zeroth Law: This law defines Temperature. If A is in thermal equilibrium with B, and B with C, then A is in equilibrium with C.
16. Reversible Work: The maximum amount of work can only be obtained from a Reversible process. Irreversible processes yield less work.
17. Combustion: Enthalpy of combustion (Δ_cH) is always Negative (Exothermic), as burning releases heat.
18. Boiling Point: At the boiling point, the liquid and vapor are in equilibrium. Therefore, ΔG = 0, which implies ΔH = TΔS.
19. Atomization: Enthalpy of atomization is always Positive (Endothermic) because bonds must be broken to separate atoms.
20. Poisson's Ratio: The ratio γ = C_p/C_v determines the curve steepness in P-V graphs. Adiabatic curves are steeper than Isothermal curves by a factor of γ.

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