Unit 1: Introduction to Thermodynamics

Zeroth Law of Thermodynamics & Concept of Temperature

Conceptual Explanation

The Zeroth Law of Thermodynamics establishes the basis for the measurement of temperature. It provides a formal definition of thermal equilibrium and justifies the use of thermometers.

Zeroth Law of Thermodynamics: If two thermodynamic systems, A and B, are each in thermal equilibrium with a third system, C, then they are also in thermal equilibrium with each other.

Thermal equilibrium means there is no net transfer of heat energy between systems when they are placed in thermal contact. This law implies that there exists a physical property of state, called temperature, which determines whether or not systems are in thermal equilibrium.

The Concept of Temperature

When system A is in thermal equilibrium with system C, their temperatures are equal: T_A = T_C. Similarly, if B is in thermal equilibrium with C, then T_B = T_C. Consequently, T_A = T_B, showing that system A and system B are also in thermal equilibrium. Temperature is thus a scalar state variable that indicates the direction of heat flow when systems are not in equilibrium. Heat flows from a body of higher temperature to one of lower temperature.

Practical Example & Real-World Application

This law is the working principle behind thermometers. Let system C be a mercury-in-glass thermometer. When we place the thermometer (C) in contact with a body (A) until thermal equilibrium is reached, the mercury column stabilizes at a certain height. If we then place the thermometer (C) in contact with another body (B) and the mercury stabilizes at the exact same height, we can conclude that body A and body B are at the same temperature, without ever placing them in direct contact.

First Law of Thermodynamics & Differential Form

The Principle of Conservation of Energy

The First Law of Thermodynamics is a statement of the law of conservation of energy applied to thermodynamic systems. It establishes that energy can change form but cannot be created or destroyed.

First Law of Thermodynamics: The net heat energy supplied to a system is equal to the sum of the increase in its internal energy and the external work done by the system on its surroundings.

Mathematical Form

If a system absorbs a quantity of heat dQ, this heat is utilized in two ways: changing the internal energy of the system by dU, and performing external work dW.

dQ = dU + dW

In differential form, considering work done during expansion (dW = P * dV, where P is pressure and dV is change in volume):

dQ = dU + P * dV

Here, U is a state function (depends only on the state of the system, not the path taken), while Q and W are path functions (depend on the path taken to reach the state).

Sign Conventions

Common Mistakes

• Confusing Path and State Functions: Assuming that because dU is a state function, dQ and dW are also independent of path. Remember: internal energy change depends only on initial and final states, but the heat absorbed and work done depend on the specific thermodynamic process.
• Sign Errors: Forgetting to use a negative sign for work done during compression. If gas is compressed, dW is negative, making dQ = dU - P * dV.

Applications of First Law: General Relation between Cp and Cv

Definitions of Specific Heat Capacities

Specific heat capacity is the amount of heat required to raise the temperature of a unit mass of substance by one degree.

• Cp (Molar Heat Capacity at Constant Pressure): The amount of heat required to raise the temperature of one mole of gas by one Kelvin at constant pressure.
• Cv (Molar Heat Capacity at Constant Volume): The amount of heat required to raise the temperature of one mole of gas by one Kelvin at constant volume.

Derivation of Meyer's Relation (Cp - Cv = R)

Let us consider one mole of an ideal gas. According to the First Law of Thermodynamics:

dQ = dU + P * dV

Case 1: At Constant Volume (dV = 0)

Since volume is constant, work done dW = P * dV = 0. Therefore:

dQ_v = dU

By definition, dQ_v = Cv * dT. Thus, we get:

dU = Cv * dT

Since the internal energy of an ideal gas depends only on its temperature, this expression for dU holds true for any process, whether volume is constant or not.

Case 2: At Constant Pressure

By definition, the heat absorbed at constant pressure is:

dQ_p = Cp * dT

Using the First Law: dQ_p = dU + P * dV. Substituting the value of dU from Case 1:

Cp * dT = Cv * dT + P * dV

From the ideal gas equation for one mole of gas: PV = R * T. Differentiating with respect to temperature at constant pressure:

P * dV = R * dT

Substitute this back into the heat equation:

Cp * dT = Cv * dT + R * dT

Dividing throughout by dT:

Cp - Cv = R

This is known as Meyer's Relation. Since R (the universal gas constant) is always positive, Cp is always greater than Cv. This is because, at constant pressure, some supplied heat is spent in doing work against external pressure, whereas at constant volume, all supplied heat is used solely to increase internal energy.

Applications of First Law: Isothermal and Adiabatic Processes

Isothermal Process

An isothermal process is a thermodynamic process in which the temperature of the system remains constant (T = constant, dT = 0).

Since the internal energy of an ideal gas is a function of temperature only, dU = 0 in an isothermal process. Therefore, from the First Law of Thermodynamics:

dQ = dW

This means all the heat supplied to the system is converted completely into work. The equation of state is given by Boyle's Law: PV = constant (or P1 * V1 = P2 * V2).

Work Done during Isothermal Expansion

Let n moles of an ideal gas expand isothermally from an initial volume V1 to a final volume V2 at temperature T. The work done W is given by:

W = integral of (P * dV) from V1 to V2

From the ideal gas equation, P = n * R * T / V. Substituting this in the integral:

W = integral of (n * R * T / V) * dV from V1 to V2

Since T is constant, we can pull n * R * T out of the integral:

W = n * R * T * integral of (1/V) * dV from V1 to V2

W = n * R * T * [ln(V)] from V1 to V2

W = n * R * T * ln(V2 / V1)

Converting to base 10 logarithm:

W = 2.303 * n * R * T * log10(V2 / V1)

Since P1 * V1 = P2 * V2, we can also write: V2 / V1 = P1 / P2. Therefore:

W = n * R * T * ln(P1 / P2)

Adiabatic Process

An adiabatic process is a thermodynamic process in which there is no transfer of heat between the system and its surroundings (dQ = 0).

According to the First Law of Thermodynamics: dQ = dU + dW. Since dQ = 0:

dU = -dW

This implies that if a gas expands adiabatically (dW is positive), its internal energy decreases, causing the temperature to drop. Conversely, if the gas is compressed adiabatically, its temperature rises. The equation of state is: P * V^gamma = constant, where gamma = Cp / Cv.

Work Done during Adiabatic Expansion

Let n moles of an ideal gas expand adiabatically from (P1, V1, T1) to (P2, V2, T2). The work done W is:

W = integral of (P * dV) from V1 to V2

Since P * V^gamma = K (constant), we can write P = K / V^gamma = K * V^(-gamma). Substituting this in:

W = integral of (K * V^(-gamma) * dV) from V1 to V2

W = K * [V^(1 - gamma) / (1 - gamma)] from V1 to V2

W = [1 / (1 - gamma)] * [K * V2^(1 - gamma) - K * V1^(1 - gamma)]

Since K = P1 * V1^gamma = P2 * V2^gamma, we substitute K dynamically:

W = [1 / (1 - gamma)] * [P2 * V2^gamma * V2^(1 - gamma) - P1 * V1^gamma * V1^(1 - gamma)]

W = [1 / (1 - gamma)] * [P2 * V2 - P1 * V1]

Rearranging the denominator to gamma - 1:

W = (P1 * V1 - P2 * V2) / (gamma - 1)

Using the ideal gas equation (P * V = n * R * T):

W = n * R * (T1 - T2) / (gamma - 1)

Comparison: Isothermal vs Adiabatic Processes

Heat Engines & Efficiency

Definition and Components

A heat engine is a device operating in a cycle that absorbs heat from a high-temperature reservoir (source), converts a portion of it into useful work, and rejects the remaining heat to a low-temperature reservoir (sink).

Every heat engine consists of three essential parts:

1. Source: A hot reservoir at a constant high temperature T1. It has infinite heat capacity, meaning drawing heat from it does not change its temperature.
2. Working Substance: The medium (e.g., steam in steam engines, air-gas mixture in internal combustion engines) that undergoes cyclic expansion and compression.
3. Sink: A cold reservoir at a constant lower temperature T2. It has infinite heat capacity, meaning rejecting heat to it does not raise its temperature.

Thermal Efficiency

The efficiency (eta) of a heat engine is defined as the ratio of the net work done by the engine in one cycle to the total heat absorbed from the source in that cycle.

Let Q1 be the heat absorbed from the source, and Q2 be the heat rejected to the sink. Since the working substance returns to its initial state, the net change in internal energy over a complete cycle is zero (dU = 0). From the First Law, the net work done W is:

W = Q1 - Q2

The efficiency is:

eta = W / Q1 = (Q1 - Q2) / Q1

Or in simplified form:

eta = 1 - Q2 / Q1

Since Q2 is always positive, the efficiency eta is always less than 1 (or 100%).

Refrigerators & Coefficient of Performance

Definition and Principles

A refrigerator (or heat pump) is essentially a heat engine operating in reverse. It extracts heat from a low-temperature body (cooling chamber) and rejects it to a high-temperature body (surrounding environment) with the aid of external work performed on the system.

Let Q2 be the heat extracted from the cold reservoir (sink at T2), W be the external work done on the working substance, and Q1 be the heat rejected to the hot reservoir (source at T1).

By the principle of conservation of energy:

Q1 = Q2 + W

Or: W = Q1 - Q2

Coefficient of Performance (COP)

The performance of a refrigerator is measured by its Coefficient of Performance (COP), denoted by beta.

beta = Heat extracted from cold body / Work spent on the system = Q2 / W

Substituting W = Q1 - Q2:

beta = Q2 / (Q1 - Q2)

Unlike efficiency, COP can be greater than 1 (or 100%).

Relation between Efficiency (eta) and COP (beta)

We can establish a direct mathematical link between the efficiency of a heat engine working between T1 and T2 and the COP of a refrigerator working between the same limits:

eta = 1 - Q2/Q1, which implies Q2/Q1 = 1 - eta

Dividing the numerator and denominator of the COP equation by Q1:

beta = (Q2 / Q1) / (1 - Q2 / Q1)

Substitute Q2/Q1 = 1 - eta:

beta = (1 - eta) / eta

beta = (1 / eta) - 1

Second Law of Thermodynamics: Kelvin-Planck and Clausius Statements

Introduction

While the First Law asserts energy conservation, it does not specify the direction of heat flow or whether a spontaneous process is possible. The Second Law of Thermodynamics defines these limitations.

Kelvin-Planck Statement

Kelvin-Planck Statement: It is impossible to construct an engine operating in a cycle that produces no other effect than the extraction of heat from a single reservoir and the performance of an equivalent amount of work.

This means that a heat engine cannot have a thermal efficiency of 100% (eta = 1). Some heat must always be rejected to a colder reservoir (sink).

Clausius Statement

Clausius Statement: It is impossible to construct a self-acting device, operating in a cycle, whose sole effect is to transfer heat from a colder body to a hotter body without the aid of an external agent.

This implies that heat cannot flow spontaneously from a lower temperature body to a higher temperature body. External work must be performed to achieve this transfer.

Equivalence of Kelvin-Planck and Clausius Statements

Although these statements look different, they are completely equivalent. If one statement is violated, the other is automatically violated.

Proof Part 1: Violation of Clausius leads to violation of Kelvin-Planck

Suppose we have a refrigerator R that violates the Clausius statement. It transfers a heat Q2 from a cold reservoir at T2 to a hot reservoir at T1 without any external work (W = 0). Thus, it extracts Q2 from the sink and delivers Q2 to the source.

Now, couple a heat engine E to the same reservoirs. Let E absorb heat Q1 from the source at T1, perform work W = Q1 - Q2, and reject heat Q2 to the sink at T2.

Consider the combined system of the engine and refrigerator:

• The net heat extracted from the sink is Q2 - Q2 = 0.
• The net heat absorbed from the source is Q1 - Q2.
• The net work done is W = Q1 - Q2.

The combined device acts as an engine that extracts heat (Q1 - Q2) from a single reservoir (the source) and converts it entirely into work without rejecting any heat to the sink. This directly violates the Kelvin-Planck statement.

Proof Part 2: Violation of Kelvin-Planck leads to violation of Clausius

Suppose we have a heat engine E that violates the Kelvin-Planck statement. It extracts heat Q1 from a hot reservoir at T1 and converts it completely into work W = Q1, rejecting no heat to the sink (Q2 = 0).

Now, use this work W to drive a standard refrigerator R. The refrigerator extracts heat Q2 from the cold reservoir at T2, has work W done on it, and rejects heat Q1' = Q2 + W to the hot reservoir at T1.

Consider the combined system of the engine and refrigerator:

• The work produced by the engine is used entirely to run the refrigerator, so no net external work is required.
• The net heat extracted from the cold reservoir is Q2.
• The net heat delivered to the hot reservoir is Q1' - Q1 = (Q2 + W) - Q1 = Q2.

The combined device transfers heat Q2 from the cold reservoir to the hot reservoir without any external work input. This directly violates the Clausius statement.

Thus, the two statements are equivalent expressions of the same fundamental law of nature.

Carnot's Theorem

Introduction

Sadi Carnot proposed an idealized, reversible cycle (the Carnot cycle) that achieves the maximum possible efficiency. Based on this, Carnot formulated a key theorem regarding the limits of engine efficiencies.

Carnot's Theorem:

1. No engine operating between two given temperatures can be more efficient than a reversible (Carnot) engine operating between the same two temperatures.
2. All reversible engines operating between the same two temperatures have the same efficiency, regardless of the working substance used.

Conceptual Explanation and Proof Outline

Let us consider two engines operating between a source at T1 and a sink at T2. One is an irreversible engine (I) and the other is a reversible engine (R).

Suppose, for contradiction, that the irreversible engine is more efficient than the reversible engine: eta_I > eta_R.

Let both engines extract the same amount of heat Q1 from the source. Since eta_I > eta_R:

W_I / Q1 > W_R / Q1, which implies W_I > W_R

Let engine I reject heat Q2_I to the sink, and engine R reject heat Q2_R to the sink. Since W = Q1 - Q2:

Q1 - Q2_I > Q1 - Q2_R, which implies Q2_I < Q2_R

Now, run the reversible engine R in reverse as a refrigerator, driven by the work output of engine I. Since W_I > W_R, the irreversible engine can easily drive the refrigerator and still have a surplus work of (W_I - W_R) left over.

The refrigerator R extracts heat Q2_R from the sink, receives work W_R, and rejects heat Q1 back to the source.

Analyzing the combined system:

• Net heat drawn from the source: Q1 (by engine I) - Q1 (rejected by refrigerator R) = 0.
• Net heat drawn from the sink: Q2_R - Q2_I. Since Q2_I < Q2_R, this is a positive quantity.
• Net work produced: W_I - W_R. Since W = Q1 - Q2, this net work is equal to Q2_R - Q2_I.

The combined device extracts a net heat of (Q2_R - Q2_I) from a single reservoir (the sink) and converts it entirely into work without any other effect. This violates the Kelvin-Planck statement of the Second Law of Thermodynamics.

Hence, our assumption that eta_I > eta_R must be false. Thus, no engine can be more efficient than a reversible engine working between the same temperatures.

Thermodynamic Scale of Temperature & Perfect Gas Scale Equivalence

The Thermodynamic Scale of Temperature (Kelvin Scale)

Traditional temperature scales depend on the thermometric properties of specific materials (such as the thermal expansion of mercury, or pressure variations of a gas). Lord Kelvin proposed a temperature scale based on the Carnot engine, which is completely independent of the properties of any working substance.

The efficiency of a Carnot engine depends only on the temperatures of the source and the sink: eta = 1 - Q2 / Q1 = f(T1, T2). This implies that the ratio of heat absorbed to heat rejected is a function of the temperatures:

Q1 / Q2 = phi(T1, T2)

By analysis of cascaded Carnot engines, it can be mathematically shown that the function phi can be represented as a ratio of single-variable temperature functions:

Q1 / Q2 = theta1 / theta2

Here, theta is the temperature on the thermodynamic scale. This scale is defined such that the ratio of two thermodynamic temperatures is equal to the ratio of heat absorbed and heat rejected by a Carnot engine operating between those two temperatures.

Equivalence to the Perfect (Ideal) Gas Scale

To prove that the thermodynamic scale (theta) is identical to the perfect gas scale (T), we look at the efficiency of a Carnot cycle using a perfect gas as the working substance.

For a perfect gas undergoing a Carnot cycle, the heat absorbed from the source at ideal gas temperature T1 is Q1, and the heat rejected to the sink at temperature T2 is Q2. The ratio of these heat values is calculated using isothermal expansion equations:

Q1 / Q2 = T1 / T2

However, by definition of the thermodynamic scale:

Q1 / Q2 = theta1 / theta2

Equating the two expressions:

theta1 / theta2 = T1 / T2

If we define the triple point of water to have the same value on both scales (273.16 K), then for any temperature:

theta = T

Thus, the thermodynamic temperature scale is mathematically equivalent to the perfect gas scale. The advantage of the thermodynamic scale is its theoretical independence from material properties, which defines an absolute zero (where Q2 = 0, meaning efficiency becomes 100%, and theta = 0, the absolute lowest limit of temperature).