Thermodynamics

Thermodynamics is a basic branch of physics that governs the behavior of heat, energy, and work in systems. Have you ever asked yourself how engines operate, why ice melts, or the way refrigerators make food chilly? All these happenings are interpreted using thermodynamics. This article will discuss the main concepts, laws, and uses of thermodynamics in everyday life.

What is Thermodynamics?

Thermodynamics is a branch of physics that deals with the study of heat, energy, and work, and the relationships between them. It focuses on understanding how energy is transferred and transformed in physical systems and how this affects matter.

At its core, thermodynamics explains how systems, from engines to refrigerators, use energy to perform work and how energy flows between different forms (like heat and mechanical work). The laws of thermodynamics help describe natural processes, like the flow of heat from hot objects to cold ones, and are fundamental in fields such as physics, chemistry, engineering, and environmental science.

Some key concepts of thermodynamics include:

1. Energy transfer: The movement of energy between systems, often as heat or work.
2. Heat: A form of energy transfer due to temperature differences.
3. Work: Energy transfer resulting from a force causing displacement.
4. Entropy: A measure of disorder or randomness in a system, which tends to increase over time.

Thermodynamics is essential in understanding how engines work, how refrigerators cool food, and how energy systems operate, among many other applications.

Thermodynamics Definition

Thermodynamics is the analysis of energy transfer and conversion between two or more forms, especially between mechanical work and heat. It is a model of understanding how energy travels in physical systems and affects matter. The topic is fundamental to physics, engineering, chemistry, and environmental science.

Different Branches of Thermodynamics

Thermodynamics is classified into the following four branches:

• Classical Thermodynamics
• Statistical Thermodynamics
• Chemical Thermodynamics
• Equilibrium Thermodynamics

Thermodynamic Process

A thermodynamic process refers to a change that a thermodynamic system undergoes from one state to another. These processes can involve changes in temperature, pressure, volume, and energy, and are central to understanding how systems behave in different conditions. The specific properties of the system—like the pressure, volume, and temperature—change during a thermodynamic process.

There are several types of thermodynamic processes based on the conditions under which these changes occur:

1. Isothermal Process

• Definition: A process that occurs at constant temperature.
• Key Property: The internal energy of an ideal gas remains unchanged because temperature is constant.
• Example: The expansion or compression of an ideal gas in a piston at constant temperature.

2. Adiabatic Process

• Definition: A process in which there is no heat exchange between the system and its surroundings (Q = 0).
• Key Property: All the energy in the system is used to do work or is stored as internal energy.
• Example: The compression or expansion of a gas in an insulated container.

3. Isochoric (Iso-volumetric) Process

• Definition: A process that occurs at constant volume.
• Key Property: The volume of the system remains constant, and thus no work is done (W = 0).
• Example: Heating a gas in a sealed, rigid container, where the volume does not change.

4. Isobaric Process

• Definition: A process that occurs at constant pressure.
• Key Property: The pressure remains constant throughout the process, and the system may expand or contract.
• Example: Heating or cooling water in an open container where the pressure remains constant at atmospheric pressure.

5. Polytropic Process

• Definition: A process where the pressure and volume are related by the equation PVn=constantP V^n = \text{constant}, where nn is the polytropic index.
• Key Property: The value of nn determines the nature of the process (e.g., isothermal, adiabatic).
• Example: Real-life compression processes in engines that may follow a polytropic relationship.

6. Cyclic Process

• Definition: A process where the system returns to its initial state after completing a cycle.
• Key Property: The net change in the internal energy of the system is zero because it returns to the original state.
• Example: The working cycle of an engine, such as the Carnot cycle, which operates between two heat reservoirs.

7. Reversible Process

• Definition: A process that can be reversed by infinitesimal changes in the system’s conditions.
• Key Property: It occurs slowly and without dissipative effects like friction or turbulence, allowing the system to return to its initial state.
• Example: A slow, isothermal expansion of a gas.

8. Irreversible Process

• Definition: A process that cannot return the system to its initial state due to the presence of dissipative forces like friction or rapid changes.
• Key Property: Entropy increases in the system, and it cannot return to its starting point without additional work.
• Example: The rapid compression or expansion of a gas, like in a real engine.

Each of these processes can be analyzed using the laws of thermodynamics, and understanding how energy behaves in these processes is crucial in fields such as engineering, chemistry, and physics.

Thermodynamic Properties

Thermodynamic properties are distinctive characteristics that describe the state of a system. These properties can be classified as either intensive or extensive.

• Intensive properties are independent of the amount of matter present in the system. Examples include pressure and temperature.
• Extensive properties, on the other hand, depend on the system’s mass. Common examples are volume, energy, and enthalpy.

Laws of Thermodynamics

Below are the laws of Thermodynamics-

1. Zeroth Law of Thermodynamics- Defines thermal equilibrium. Two systems are in equilibrium with one another if they are each in equilibrium with a third system.

2. First Law of Thermodynamics- Otherwise referred to as the law of conservation of energy. It maintains that energy cannot be created or annihilated, but rather transferred or transformed.

Mathematical expression: ΔU = Q - W, where:

• ΔU is the internal energy change,
• Q is the heat supplied to the system,
• W is the work that the system performs.

3. Second Law of Thermodynamics- Entropy is defined as energy spontaneously dispersing and systems tending to become disordered. Example: Spontaneous heat flows from a hot body to a cold one but not vice versa in the absence of external work.

4. Third Law of Thermodynamics- It gives the fact that entropy goes to a minimum as temperature goes to absolute zero, rendering perfect order theoretically unattainable.

Applications of Thermodynamics

• Engines & Power Plants: Thermodynamics accounts for how combustion of fuel is responsible for producing energy to drive automobiles and turbines.
• Refrigerators & Air Conditioners: Operate by moving heat from a cooler to a hotter area, contrary to the natural flow.
• Weather & Climate: Accounts for atmospheric change, oceanic heat transfer, and the greenhouse effect.
• Human Metabolism: Converts food into energy for body functions.

Solved Examples on Thermodynamics

Question: How much heat is required to raise the temperature of 2 kg of water from 25°C to 75°C? The specific heat capacity of water is 4184 J/kg·K.

Solution: We can use the formula for heat transfer:

Q = m · c · ΔT

Where:

Q is the heat energy (in joules),

m is the mass of the substance (in kg),

c is the specific heat capacity of the substance (in J/kg·K),

ΔT is the change in temperature (in K or °C).

Given values:

m = 2 kg

c = 4184 J/kg·K

ΔT = 75°C - 25°C = 50°C

Substitute the values into the formula:

Q = (2) · (4184) · (50) = 418400 J

The amount of heat required is 418,400 J or 418.4 kJ.

Question: A system absorbs 500 J of heat and does 200 J of work on its surroundings. What is the change in the internal energy of the system?

Solution: We will use the First Law of Thermodynamics, which states:

ΔU = Q - W

Where:

ΔU is the change in internal energy,

Q is the heat absorbed by the system,

W is the work done by the system.

Given values:

Q = 500 J (heat absorbed)

W = 200 J (work done on the surroundings)

Substitute the values into the equation:

ΔU = 500 - 200 = 300 J

The change in internal energy of the system is 300 J.

Question: A gas expands isothermally (at constant temperature) from an initial volume of 1 L to a final volume of 3 L. The external pressure is constant at 2 atm. How much work is done by the gas during the expansion?

Solution: For an isothermal process, the work done by the gas is given by:

W = P · ΔV

Where:

W is the work done,

P is the external pressure,

ΔV is the change in volume.

Since we are working with constant pressure, the work is given by:

W = P · (V_f - V_i)

Given values:

P = 2 atm

V_i = 1 L (initial volume)

V_f = 3 L (final volume)

First, convert the pressure to pascals and the volume to cubic meters:

1 atm = 101325 Pa, 1 L = 1 × 10^-3 m³

Thus, the pressure is:

P = 2 atm × 101325 Pa/atm = 202650 Pa

And the volume change is:

ΔV = (3 L - 1 L) × 1 × 10^-3m³/L = 2 × 10^-3 m³

Now, calculate the work done:

W = 202650 Pa × 2 × 10^-3 m³ = 405.3 J

The work done by the gas during the isothermal expansion is 405.3 J.

Question: How much entropy change occurs when 1 kg of ice at -10°C is heated to 0°C and then melted into water at 0°C? The specific heat of ice is 2.1 kJ/kg·K, and the latent heat of fusion of ice is 334 kJ/kg.

Solution: First, calculate the heat required to raise the temperature of the ice from -10°C to 0°C:

Q₁ = m · c · ΔT

Where:

m = 1 kg

c = 2.1 kJ/kg·K

ΔT = 0°C - (-10°C) = 10°C

Substitute the values into the formula:

Q₁ = 1 · 2.1 · 10 = 21 kJ

Next, calculate the heat required to melt the ice at 0°C:

Q₂ = m · L_f

Where:

L_f = 334 kJ/kg (latent heat of fusion)

Substitute the values:

Q₂ = 1 · 334 = 334 kJ

Now, calculate the change in entropy for both processes.

For the temperature change (ice heating from -10°C to 0°C):

ΔS₁ = Q₁ / T

Where:

Q₁ = 21 kJ

T = 273.15 K (temperature at 0°C)

Substitute the values:

ΔS₁ = 21 / 273.15 = 0.077 kJ/K

For the phase change (melting of the ice):

ΔS₂ = Q₂ / T

Where:

Q₂ = 334 kJ

T = 273.15 K (temperature at 0°C)

Substitute the values:

ΔS₂ = 334 / 273.15 = 1.223 kJ/K

Finally, the total change in entropy is:

ΔS_total = ΔS₁ + ΔS₂ = 0.077 + 1.223 = 1.3 kJ/K

The total change in entropy is 1.3 kJ/K.