Historically, heat was viewed as a weightless, invisible fluid called caloric that resided in the pores of matter. It was believed to flow spontaneously from hotter bodies to colder bodies until their caloric levels (temperatures) equalized. This view was eventually replaced by the modern kinetic concept of heat as a form of energy transfer associated with the random motion of microscopic constituents.
Historical Chronicle: Benjamin Thomson (Count Rumford)
While mechanical equilibrium requires the net external force and torque on a system to be zero, thermodynamic equilibrium requires that the macroscopic state variables characterizing the system (such as pressure, volume, temperature, mass, and composition) do not change with time. A system is in thermodynamic equilibrium when it simultaneously satisfies thermal, mechanical, and chemical equilibrium conditions.
Formulated by R.H. Fowler in 1931, the Zeroth Law of Thermodynamics states: "If two systems A and B are separately in thermal equilibrium with a third system C, then they must be in thermal equilibrium with each other."
This law provides the physical basis for defining temperature ($T$). It \implies that there exists a scalar state variable called temperature such that systems in thermal equilibrium have equal temperatures: if $T_A = T_C$ and $T_B = T_C$, then $T_A = T_B$.
The internal energy ($U$) of a system is the sum of the kinetic and potential energies of its molecular constituents in a frame of reference where the center of mass of the system is at rest. It is a thermodynamic state variable , meaning its value depends solely on the current state of the system and is independent of the path taken to reach that state.
Unlike internal energy, heat ($Q$) and work ($W$) are not state variables; they are modes of energy transfer in transit. A system does not "possess" heat or work. Heat is energy transfer driven by a temperature difference, whereas work is energy transfer driven by mechanical means (such as a moving piston) not involving a temperature difference.
The First Law of Thermodynamics is a statement of the general law of conservation of energy applied to thermodynamic systems. Mathematically, it is expressed as:
$$\Delta Q = \Delta U + \Delta W$$Where:
Alternatively, it can be written as:
$$\Delta Q - \Delta W = \Delta U$$This highlights that while $\Delta Q$ and $\Delta W$ depend on the path taken, their difference $(\Delta Q - \Delta W)$ is path-independent and equals the change in the state variable $\Delta U$.
For a solid consisting of $N$ atoms, each atom vibrates about its mean position in three dimensions. According to the law of equipartition of energy, each 1D harmonic oscillator has an average energy of $k_B T$. In three dimensions, the average energy per atom is $3 k_B T$. For one mole of solid ($N_A$ atoms), the total internal energy is:
$$U = 3 k_B T \times N_A = 3 R T$$At constant pressure, since the volume change of a solid is negligible, $\Delta Q \approx \Delta U$. Therefore, the molar specific heat capacity of a solid is:
$$C = \frac{\Delta U}{\Delta T} = 3R$$This theoretical value of $3R$ (\approx. 24.9 J mol -1 K -1 ) agrees well with experimental values for many solids at room temperature.
For an ideal gas, we define two molar specific heats:
Since $U$ of an ideal gas depends only on temperature, the term $\frac{\Delta U}{\Delta T}$ is identical in both cases. Substituting $C_v$ into the expression for $C_p$ gives:
$$C_p = C_v + P \left(\frac{\Delta V}{\Delta T}\right)_p$$From the ideal gas equation for 1 mole ($PV = RT$), differentiating with respect to $T$ at constant pressure yields:
$$P \left(\frac{\Delta V}{\Delta T}\right)_p = R$$Substituting this back gives Mayer's formula :
$$C_p - C_v = R$$A quasi-static process is an idealized, infinitely slow process in which the system remains in thermal and mechanical equilibrium with its surroundings at every stage. The pressure and temperature of the system differ from those of the surroundings only by infinitesimal amounts.
A process occurring at a constant temperature ($T = \text{constant}$). For an ideal gas, the equation of state is $PV = \text{constant}$.
Work Done in an Isothermal Expansion:
$$W = \int_{V_1}^{V_2} P dV = \int_{V_1}^{V_2} \frac{\mu RT}{V} dV = \mu RT \ln\left(\frac{V_2}{V_1}\right)$$Since $\Delta U = 0$ for an isothermal process of an ideal gas, the first law \implies $\Delta Q = \Delta W$.
A process in which there is no heat exchange between the system and the surroundings ($\Delta Q = 0$). The equation of state is:
$$PV^{\gamma} = \text{constant}$$Where $\gamma = C_p / C_v$ is the adiabatic index.
Work Done in an Adiabatic Process:
$$W = \int_{V_1}^{V_2} P dV = K \int_{V_1}^{V_2} V^{-\gamma} dV = \frac{P_2 V_2 - P_1 V_1}{1 - \gamma} = \frac{\mu R(T_1 - T_2)}{\gamma - 1}$$| Process Type | Constant Parameter | Equation of State / Path | Work Done Formula |
|---|---|---|---|
| Isothermal | Temperature ($T$) | $PV = \text{constant}$ | $W = \mu RT \ln\left(\frac{V_2}{V_1}\right)$ |
| Adiabatic | Heat ($Q$) | $PV^{\gamma} = \text{constant}$ | $W = \frac{\mu R(T_1 - T_2)}{\gamma - 1}$ |
| Isochoric | Volume ($V$) | $V = \text{constant} \implies dV = 0$ | $W = 0$ |
| Isobaric | Pressure ($P$) | $P = \text{constant}$ | $W = P(V_2 - V_1) = \mu R(T_2 - T_1)$ |
| Cyclic | Initial State = Final State | Closed loop on $P\text{-}V$ diagram | $W = \text{Area enclosed by the loop}$ ($\Delta U = 0$) |
The First Law of Thermodynamics asserts the conservation of energy but does not specify the direction of heat flow or whether a spontaneous process is possible. For instance, it does not forbid heat from flowing spontaneously from a cold body to a hot body, which is never observed in nature. The Second Law of Thermodynamics addresses these limitations.
A process is reversible if it can be turned back such that both the system and its surroundings return to their exact initial states with no other changes in the universe. Reversibility requires the process to be quasi-static and completely free of dissipative forces (like friction or viscosity). Real processes are always irreversible due to dissipative effects and non-equilibrium intermediate states.
The Carnot engine is an idealized, reversible thermodynamic engine operating between a hot reservoir at temperature $T_1$ and a cold reservoir at $T_2$. The cycle consists of four sequential steps:
The net work done in one cycle is $W = Q_1 - Q_2$. The efficiency $\eta$ of the Carnot engine is:
$$\eta = \frac{W}{Q_1} = 1 - \frac{Q_2}{Q_1}$$Since steps 2 $\rightarrow$ 3 and 4 $\rightarrow$ 1 are adiabatic, we have $T_1 V_2^{\gamma-1} = T_2 V_3^{\gamma-1}$ and $T_1 V_1^{\gamma-1} = T_2 V_4^{\gamma-1}$, which yields $\frac{V_2}{V_1} = \frac{V_3}{V_4}$. Thus:
$$\frac{Q_2}{Q_1} = \frac{T_2}{T_1} \implies \eta = 1 - \frac{T_2}{T_1}$$Carnot's Theorem:
Mathematical Problem Statement: Calculate the change in internal energy when 1 g of water at 100 °C is converted into steam at the same temperature and at normal atmospheric pressure ( 1.013 &× 10 5 N m -2 ). The latent heat of vaporization of water is 2256 J g -1 . The volume of 1 g of water is 1 cm³ and the volume of 1 g of steam is 1671 cm³ .
Step-by-Step Algorithmic Solution Trace:
- Identify the given parameters:
- Mass of water, $m = 1\text{ g}$
- Heat supplied, $\Delta Q = m \times L = 1\text{ g} \times 2256\text{ J g}^{-1} = 2256\text{ J}$
- Pressure, $P = 1.013 \times 10^5\text{ N m}^{-2}$
- Initial volume, $V_1 = 1\text{ cm}^3 = 1 \times 10^{-6}\text{ m}^3$
- Final volume, $V_2 = 1671\text{ cm}^3 = 1671 \times 10^{-6}\text{ m}^3$
- Calculate the work done ($\Delta W$) during expansion:$$\Delta W = P(V_2 - V_1) = 1.013 \times 10^5 \times (1671 - 1) \times 10^{-6}$$$$\Delta W = 1.013 \times 10^5 \times 1670 \times 10^{-6} = 169.2\text{ J}$$
- Apply the First Law of Thermodynamics to find the change in internal energy ($\Delta U$):$$\Delta U = \Delta Q - \Delta W = 2256\text{ J} - 169.2\text{ J} = 2086.8\text{ J}$$
Final Evaluation Output: The change in internal energy of the system is $\Delta U = 2086.8\text{ J}$.
Mathematical Problem Statement: A Carnot engine operates between a hot reservoir at 500 K and a cold reservoir at 300 K . Determine the efficiency of the engine and the work output per cycle if it absorbs 1000 J of heat from the hot reservoir.
Step-by-Step Algorithmic Solution Trace:
- Identify the given parameters:
- Source temperature, $T_1 = 500\text{ K}$
- Sink temperature, $T_2 = 300\text{ K}$
- Heat absorbed, $Q_1 = 1000\text{ J}$
- Calculate the efficiency ($\eta$):$$\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{300}{500} = 1 - 0.6 = 0.4\text{ (or 40\%)}$$
- Calculate the work output ($W$):$$W = \eta \times Q_1 = 0.4 \times 1000\text{ J} = 400\text{ J}$$
Final Evaluation Output: The efficiency of the engine is 40% and the work output per cycle is $400\text{ J}$.