Thermodynamics and Kinetic Theory — NEET Physics

Thermodynamics is the physics of heat, work and internal energy; kinetic theory explains it all from the motion of molecules. The two together are a steady 3–4 NEET marks and they reappear inside Physical Chemistry. The marks hinge on a firm sign convention for the first law, the shortcuts for the four standard processes, the link temperature ↔ molecular kinetic energy, and the efficiency of heat engines. This chapter derives each of these and works them through, so you can apply them to an unfamiliar setup rather than only recognise a formula.

1. Temperature, heat and the zeroth law

• Temperature measures how hot a body is — physically, the average kinetic energy of its molecules.
• Heat is energy in transit because of a temperature difference; it is not "stored" — what a body stores is internal energy.
• Zeroth law: if A is in thermal equilibrium with C, and B is with C, then A and B are in equilibrium with each other. This is what lets a thermometer (C) define temperature.

Always use absolute temperature (kelvin) in thermodynamic relations: $T(\text{K})=T(°\text{C})+273$.

2. Specific heat, latent heat and calorimetry

To change a substance's temperature, or its phase:

$Q=mc\Delta T(\text{temperature change}),Q=mL(\text{phase change})$

• $c$ is specific heat (J kg⁻¹ K⁻¹); water's is a high 4200, which is why it moderates climate and cools engines.
• $L$ is latent heat; during melting or boiling, heat is absorbed at constant temperature (it breaks bonds, not raises $T$).

Worked example 2.1. Heat to raise 2 kg of water by 10 °C: $Q=mc\Delta T=2\times 4200\times 10=84,000$ J. Worked example 2.2. Heat to melt 0.1 kg of ice at 0 °C ($L=3.36\times 10^5$ J/kg): $Q=mL=0.1\times 336000=33,600$ J — all absorbed with no temperature rise.

3. The first law of thermodynamics

Energy conservation for a gas: heat added goes into raising internal energy and doing work.

$\Delta Q=\Delta U+\Delta W$

Sign convention: heat added to the system is positive; work done by the gas is positive (so $\Delta U=\Delta Q-\Delta W$).

• Add 100 J of heat while the gas does 40 J of work: $\Delta U=100-40=60$ J.
• The work done by a gas is the area under the P–V curve; at constant pressure, $\Delta W=P\Delta V$.

For an ideal gas, internal energy depends only on temperature: $U=\frac{f}{2}nRT$, where $f$ is the degrees of freedom.

4. The four processes

In an adiabatic expansion the gas does work with no heat input, so it cools — the mechanism behind cloud formation and the cooling of gas leaving a nozzle. In an isothermal expansion, temperature is held fixed, so $\Delta U=0$ and every joule of heat becomes work.

5. Molar specific heats and degrees of freedom

For an ideal gas the two molar specific heats differ by exactly the gas constant (Mayer's relation):

$C_p-C_v=R,\gamma =\frac{C_p}{C_v}$

The values follow from the degrees of freedom $f$ (equipartition gives each mode $\frac{1}{2}kT$ per molecule):

6. The gas laws

All are special cases of $PV=nRT$ (temperature in kelvin):

• Boyle's law (constant $T$): $PV=$ const — double the pressure, halve the volume.
• Charles's law (constant $P$): $V/T=$ const — double the absolute temperature, double the volume.
• Gay-Lussac's law (constant $V$): $P/T=$ const.

7. Kinetic theory of gases

Kinetic theory derives pressure from molecular collisions:

$P=\frac{1}{3}\rho v_{\text{rms}}^2$

and connects temperature to molecular energy:

${\overline{KE}}_{\text{molecule}}=\frac{3}{2}kT,v_{\text{rms}}=\sqrt{\frac{3RT}{M}}=\sqrt{\frac{3kT}{m}}$

• Average translational KE depends only on temperature, not on the gas — at a given $T$ every ideal gas has the same molecular KE.
• $v_{\text{rms}}\propto \sqrt{T}$ (quadruple $T$ → double $v_{\text{rms}}$) and $\propto 1/\sqrt{M}$ (lighter molecules move faster). At the same temperature hydrogen ($M=2$) is $\sqrt{32/2}=4$ times faster than oxygen ($M=32$).

8. The second law, heat engines and refrigerators

The first law allows any energy-conserving process; the second law says heat flows spontaneously only hot → cold, and no engine can convert heat entirely into work.

A heat engine takes $Q_1$ from a hot source, does work $W$, rejects $Q_2$:

$\eta =\frac{W}{Q_1}=1-\frac{Q_2}{Q_1},{\eta }_{\text{Carnot}}=1-\frac{T_2}{T_1}$

• Between 400 K and 300 K, the maximum (Carnot) efficiency is $1-\frac{300}{400}=25\%$. No real engine beats this.
• A refrigerator runs the cycle backwards; its coefficient of performance is $\text{COP}=\frac{Q_2}{W}=\frac{T_2}{T_1-T_2}$.

Efficiency can never be 100% for a finite cold-reservoir temperature — some heat must always be dumped. This is the deep asymmetry the second law encodes.

9. Common traps NEET sets here

• Wrong first-law signs — heat in +, work by gas +; a slip flips the answer.
• Assuming $\Delta U\ne 0$ isothermally — for an ideal gas $\Delta U=0$ whenever $T$ is constant.
• Using Celsius in gas or kinetic-theory relations — always kelvin.
• $v_{\text{rms}}\propto T$ — it goes as $\sqrt{T}$, not $T$.
• Thinking temperature rises during melting/boiling — latent heat is absorbed at constant temperature.
• Believing an engine can be 100% efficient — forbidden by the second law.

10. Memory aids

• "ΔQ = ΔU + ΔW, in-plus, by-plus" — the first law with its sign convention.
• "Iso-T: no ΔU; Adia: no Q; Iso-V: no W" — the process shortcuts.
• "Cp minus Cv is R" — Mayer's relation.
• "3 for mono, 5 for di" — degrees of freedom (→ γ = 5/3, 7/5).
• "rms goes as root-T, inverse-root-M" — lighter and hotter means faster.

11. Exam protocol

1. Convert every temperature to kelvin before substituting.
2. Apply the first law with a firm sign convention (heat in +, work by gas +).
3. Use the process shortcuts: isothermal $\Delta U=0$, adiabatic $\Delta Q=0$ and $P{V}^{\gamma }$ const, isochoric $\Delta W=0$.
4. For phase change use $Q=mL$ at constant temperature; for temperature change $Q=mc\Delta T$.
5. Use $C_p-C_v=R$ and the $f$ / $\gamma$ values for gas type.
6. Molecular KE $=\frac{3}{2}kT$; $v_{\text{rms}}\propto \sqrt{T/M}$.
7. Engine efficiency $\eta =1-T_2/T_1$ (Carnot, kelvin); refrigerator COP $=T_2/(T_1-T_2)$.