7.1 Thermal Physics, Kinetic Theory of Gases & Laws of Thermodynamics

7.1 Thermal Physics, Kinetic Theory of Gases & Laws of Thermodynamics

Thermal physics and thermodynamics form a foundational pillar of FSc Class 11 Physics and are heavily tested in the Pakistan Army Medical Cadet (AMC) entrance examination. This section explores the microscopic kinetic behavior of gas molecules and macroscopic thermodynamic transformations governing heat, work, and internal energy.

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1. Postulates of Kinetic Theory of Ideal Gases

The kinetic theory provides a microscopic explanation for the macroscopic behavior of ideal gases based on several fundamental postulates:

1. Identical Point Masses: A gas consists of a very large number of extremely small, identical particles (molecules) whose individual volumes are negligible compared to the total volume occupied by the gas.
2. Continuous Random Motion: Gas molecules are in a state of continuous, rapid, and random motion, colliding with each other and with the walls of the container.
3. Elastic Collisions: All molecular collisions (inter-molecular and wall collisions) are perfectly elastic. Total kinetic energy and momentum are conserved.
4. Negligible Intermolecular Forces: Except during collisions, there are no forces of attraction or repulsion between molecules. Consequently, potential energy is zero, and total internal energy consists entirely of kinetic energy.
5. Newtonian Mechanics: The motion of gas molecules obeys Newton's laws of motion.

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2. Microscopic Origin of Gas Pressure

When gas molecules strike the container walls, they undergo a change in momentum. The cumulative force exerted per unit area during these collisions generates macroscopic pressure.

Using Newtonian mechanics, the pressure $P$ exerted by an ideal gas of density $\rho = \frac{m N}{V}$ containing $N$ molecules of mass $m$ in a volume $V$ is given by:

$P = \frac{1}{3} \rho \langle v^2 \rangle = \frac{1}{3} \frac{N m}{V} \langle v^2 \rangle$

where $\langle v^2 \rangle$ is the mean square velocity of the gas molecules. Expressing this in terms of kinetic energy:

$P = \frac{2}{3} \left( \frac{N}{V} \right) \left( \frac{1}{2} m \langle v^2 \rangle \right) = \frac{2}{3} N_0 \langle K.E. \rangle$

where $N_0 = \frac{N}{V}$ is the number density of molecules.

Temperature Interpretation & RMS Speed

Comparing the kinetic pressure equation with the Ideal Gas Law ($PV = N k_B T$ or $PV = n R T$):

$\langle K.E. \rangle = \frac{1}{2} m \langle v^2 \rangle = \frac{3}{2} k_B T$

where $k_B = \frac{R}{N_A} \approx 1.38 \times 10^{-23} \text{ J/K}$ is Boltzmann's constant. Key Takeaway for AMC: Absolute temperature $T$ (in Kelvin) is a direct measure of the average translational kinetic energy per molecule of an ideal gas.

The root-mean-square (rms) speed $v_{rms}$ is defined as:

$v_{rms} = \sqrt{\langle v^2 \rangle} = \sqrt{\frac{3 k_B T}{m}} = \sqrt{\frac{3 R T}{M}}$

where $M = m N_A$ is the molar mass of the gas in $\text{kg/mol}$. Notice that $v_{rms}$ is directly proportional to $\sqrt{T}$ and inversely proportional to $\sqrt{M}$. At a given temperature, lighter gas molecules (e.g., $H_2$) move faster on average than heavier molecules (e.g., $O_2$).

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3. Internal Energy and the First Law of Thermodynamics

Internal Energy ($U$)

The internal energy $U$ of a system is the sum of all microscopic kinetic and potential energies of its constituent atoms or molecules. For an ideal gas (where intermolecular forces are zero), internal energy is strictly a function of absolute temperature $T$:

$U = f(T)$

For a monoatomic ideal gas with 3 translational degrees of freedom per molecule:

$U = \frac{3}{2} n R T$

First Law of Thermodynamics

The First Law of Thermodynamics is the law of conservation of energy applied to thermodynamic systems. When an amount of heat $\Delta Q$ is supplied to a system, it is partitioned between increasing the internal energy $\Delta U$ of the system and performing external work $W$:

$\Delta Q = \Delta U + W$

Sign Conventions for AMC Numerical Problems:

• $\Delta Q > 0$: Heat enters (absorbed by) the system.
• $\Delta Q < 0$: Heat leaves (rejected by) the system.
• $W > 0$: Work done by the system (gas expansion, $\Delta V > 0$).
• $W < 0$: Work done on the system (gas compression, $\Delta V < 0$).
• $\Delta U > 0$: Temperature increases.
• $\Delta U < 0$: Temperature decreases.

Work done by a gas at constant pressure $P$ during a volume change $\Delta V = V_2 - V_1$ is:

$W = P \Delta V = P (V_2 - V_1)$

On a pressure-volume ($P\text{-}V$) diagram, the work done during any process equals the area under the P-V curve.

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4. Fundamental Thermodynamic Processes

Here $\gamma = \frac{C_p}{C_v}$ is the Poisson ratio (ratio of molar specific heat capacities):

• Monoatomic gas (e.g., He, Ar): $C_v = \frac{3}{2}R$, $C_p = \frac{5}{2}R \implies \gamma = \frac{5}{3} \approx 1.67$.
• Diatomic gas (e.g., $N_2, O_2$): $C_v = \frac{5}{2}R$, $C_p = \frac{7}{2}R \implies \gamma = \frac{7}{5} = 1.40$.
• Mayer's Relation: $C_p - C_v = R$.

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5. Second Law of Thermodynamics & Carnot Engine

While the First Law asserts energy conservation, the Second Law establishes the direction of spontaneous thermal processes.

Formulations of Second Law:

1. Lord Kelvin-Planck Statement: It is impossible to construct a heat engine operating in a cycle that absorbs heat from a single reservoir and converts it completely into mechanical work without rejecting any heat to a colder reservoir.
2. Clausius Statement: It is impossible to construct a device operating in a cycle that transfers heat from a colder body to a hotter body without requiring external work input.

Carnot Heat Engine & Efficiency

An ideal reversible engine operating between a hot reservoir at temperature $T_H$ and a cold reservoir at $T_C$ performs net work $W = Q_H - Q_C$. Its thermal efficiency $\eta$ is:

$\eta = \frac{W}{Q_H} = \frac{Q_H - Q_C}{Q_H} = 1 - \frac{Q_C}{Q_H}$

For a reversible Carnot Engine, $\frac{Q_C}{Q_H} = \frac{T_C}{T_H}$ (temperatures MUST be in Kelvin):

$\eta_{Carnot} = 1 - \frac{T_C}{T_H} = \frac{T_H - T_C}{T_H}$

No real heat engine operating between two given temperatures can be more efficient than a Carnot engine.

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6. Worked Numerical Examples for AMC Candidates

Example 1: RMS Velocity Calculation

Problem: Calculate the rms velocity of nitrogen gas ($N_2$, molar mass $M = 28 \text{ g/mol} = 0.028 \text{ kg/mol}$) at a temperature of $27^\circ\text{C}$. Use $R = 8.314 \text{ J/(mol}\cdot\text{K)}$.

Solution:

1. Convert temperature to Kelvin: $T = 27 + 273.15 = 300.15 \text{ K} \approx 300 \text{ K}$.
2. Apply rms velocity formula: $v_{rms} = \sqrt{\frac{3 R T}{M}} = \sqrt{\frac{3 \times 8.314 \times 300}{0.028}} = \sqrt{\frac{7482.6}{0.028}} = \sqrt{267235.7} \approx 517 \text{ m/s}$

Example 2: Carnot Efficiency Calculation

Problem: A heat engine absorbs $2000 \text{ J}$ of heat from a hot reservoir at $327^\circ\text{C}$ and rejects heat to a cold sink at $27^\circ\text{C}$. Determine the Carnot efficiency and the net work output per cycle.

Solution:

1. Convert temperatures to Kelvin:
   • $T_H = 327 + 273 = 600 \text{ K}$
   • $T_C = 27 + 273 = 300 \text{ K}$
2. Calculate efficiency: $\eta = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{600} = 1 - 0.50 = 0.50 \text{ or } 50\%$
3. Calculate work done: $W = \eta \times Q_H = 0.50 \times 2000 \text{ J} = 1000 \text{ J}$