Introduction
Thermodynamics is a branch of physical chemistry that deals with the study of energy changes, particularly the transformation of heat into other forms of energy and vice versa. It is a crucial topic for JEE Advanced Chemistry as it lays the foundation for understanding various chemical processes and reactions. This study note document will cover the fundamental concepts, laws, and applications of thermodynamics relevant to the JEE Advanced syllabus.
Basic Concepts of Thermodynamics
System and Surroundings
- System: The part of the universe under study.
- Surroundings: Everything outside the system.
- Types of Systems:
- Open System: Can exchange both matter and energy with surroundings.
- Closed System: Can exchange only energy with surroundings, not matter.
- Isolated System: Cannot exchange either matter or energy with surroundings.
State Functions
- State Functions: Properties that depend only on the state of the system and not on the path taken to reach that state (e.g., internal energy, enthalpy, entropy).
- Path Functions: Properties that depend on the path taken to reach a specific state (e.g., work, heat).
State functions are crucial for understanding the thermodynamic properties of a system.
Thermodynamic Equilibrium
- Thermodynamic Equilibrium: A state where macroscopic properties like pressure, temperature, and volume remain constant over time.
- Types of Equilibrium:
- Thermal Equilibrium: No temperature gradient.
- Mechanical Equilibrium: No pressure gradient.
- Chemical Equilibrium: No change in chemical composition over time.
First Law of Thermodynamics
Concept and Mathematical Formulation
The first law of thermodynamics is a statement of the conservation of energy. It states that energy cannot be created or destroyed, only transformed from one form to another.
$$ \Delta U = q + W $$
Where:
- $\Delta U$ = Change in internal energy
- $q$ = Heat added to the system
- $W$ = Work done on the system
Work Done by a Gas
For a gas undergoing expansion or compression:
$$ W = -P_{\text{ext}} \Delta V $$
Where:
- $P_{\text{ext}}$ = External pressure
- $\Delta V$ = Change in volume
Example Calculation: Calculate the work done when 2 moles of an ideal gas expand isothermally at 300 K from 10 L to 20 L against a constant external pressure of 1 atm.
Given:
- $P_{\text{ext}} = 1$ atm
- $\Delta V = 20 - 10 = 10$ L
$$ W = -P_{\text{ext}} \Delta V = -1 \text{ atm} \times 10 \text{ L} = -10 \text{ L atm} $$
Converting to Joules (1 L atm = 101.3 J):
$$ W = -10 \text{ L atm} \times 101.3 \frac{\text{J}}{\text{L atm}} = -1013 \text{ J} $$
Second Law of Thermodynamics
Concept and Statements
The second law of thermodynamics introduces the concept of entropy and states that the total entropy of an isolated system can never decrease over time.
- Clausius Statement: Heat cannot spontaneously flow from a colder body to a hotter body.
- Kelvin-Planck Statement: No process is possible whose sole result is the absorption of heat from a reservoir and the conversion of this heat into work.
Entropy
- Entropy (S): A measure of the disorder or randomness of a system.
- Change in Entropy: For a reversible process,
$$ \Delta S = \frac{q_{\text{rev}}}{T} $$
Where:
- $q_{\text{rev}}$ = Heat exchanged in a reversible process
- $T$ = Temperature
Entropy tends to increase in natural processes, leading to higher disorder.
Third Law of Thermodynamics
Concept
The third law of thermodynamics states that the entropy of a perfect crystal at absolute zero (0 K) is exactly zero.
$$ S = 0 \text{ at } T = 0 \text{ K} $$
This law implies that it is impossible to reach absolute zero in a finite number of steps.
Enthalpy (H)
Definition
Enthalpy is a state function defined as:
$$ H = U + PV $$
Where:
- $H$ = Enthalpy
- $U$ = Internal energy
- $P$ = Pressure
- $V$ = Volume
Enthalpy Changes
- Enthalpy of Reaction ($\Delta H$): Change in enthalpy during a chemical reaction.
- Standard Enthalpy of Formation ($\Delta H_f^\circ$): Enthalpy change when one mole of a compound is formed from its elements in their standard states.
Example Calculation: Calculate the standard enthalpy change of the reaction:
$$ \text{C(graphite) + O}_2(g) \rightarrow \text{CO}_2(g) $$
Given:
- $\Delta H_f^\circ (\text{CO}_2(g)) = -393.5 \text{ kJ/mol}$
Since graphite and $O_2$ are in their standard states:
$$ \Delta H_{\text{reaction}} = \Delta H_f^\circ (\text{CO}_2(g)) = -393.5 \text{ kJ/mol} $$
Gibbs Free Energy (G)
Definition
Gibbs free energy is a state function defined as:
$$ G = H - TS $$
Where:
- $G$ = Gibbs free energy
- $H$ = Enthalpy
- $T$ = Temperature
- $S$ = Entropy
Gibbs Free Energy Change
- Gibbs Free Energy Change ($\Delta G$): Determines the spontaneity of a process.
- $\Delta G
< 0$: Spontaneous process
- $\Delta G >
0$: Non-spontaneous process
- $\Delta G = 0$: System is at equilibrium
$$ \Delta G = \Delta H - T \Delta S $$
Students often confuse $\Delta G$ with $\Delta H$. Remember, $\Delta G$ includes the effect of entropy and temperature.
Hess's Law
Concept
Hess's Law states that the total enthalpy change for a reaction is the sum of all changes, regardless of the pathway taken.
$$ \Delta H_{\text{total}} = \sum \Delta H_{\text{steps}} $$
Example Calculation: Given the following reactions:
- $\text{C(graphite) + O}_2(g) \rightarrow \text{CO}_2(g)$, $\Delta H_1 = -393.5 \text{ kJ/mol}$
- $\text{CO}_2(g) \rightarrow \text{CO}(g) + \frac{1}{2} \text{O}_2(g)$, $\Delta H_2 = 283.0 \text{ kJ/mol}$
Find the enthalpy change for the reaction:
$$ \text{C(graphite) + \frac{1}{2} O}_2(g) \rightarrow \text{CO}(g) $$
Using Hess's Law:
$$ \Delta H_{\text{reaction}} = \Delta H_1 + \Delta H_2 = -393.5 \text{ kJ/mol} + 283.0 \text{ kJ/mol} = -110.5 \text{ kJ/mol} $$
Conclusion
Thermodynamics is a fundamental topic in JEE Advanced Chemistry, encompassing the study of energy transformations and the principles governing these changes. Mastery of the basic concepts, laws, and their applications is essential for solving complex problems and understanding the behavior of chemical systems. Use this study note document as a comprehensive guide to prepare for the thermodynamics section in your JEE Advanced exam.
