States of Matter and Thermodynamics — NEET Chemistry

These two chapters are the calculation heart of physical chemistry, worth 3–4 NEET questions a year. The gaseous state gives you clean numerical problems from a handful of laws; thermodynamics gives you the single most powerful idea in chemistry — why reactions happen at all, answered by Gibbs free energy. This chapter derives the gas laws, builds kinetic theory, corrects it for real gases, then develops the first and second laws to the point where you can predict, from $\Delta H$ and $\Delta S$, whether any reaction is spontaneous. Every formula is worked, not just quoted.

Part A — States of Matter (Gaseous State)

1. The gas laws

A gas is described by four variables: pressure $P$, volume $V$, temperature $T$ (always in kelvin) and amount $n$. Four experimental laws each hold two variables fixed:

• Boyle's law (fixed $T$, $n$): $P\propto \frac{1}{V}$, so $P_1{V}_1=P_2{V}_2$. (Isothermal.)
• Charles's law (fixed $P$, $n$): $V\propto T$, so $\frac{V_1}{T_1}=\frac{V_2}{T_2}$. (Volume → 0 at 0 K — the basis of absolute zero.)
• Gay-Lussac's law (fixed $V$, $n$): $P\propto T$, so $\frac{P_1}{T_1}=\frac{P_2}{T_2}$.
• Avogadro's law (fixed $P$, $T$): $V\propto n$ — equal volumes of gases have equal numbers of molecules.

Combined, these give the ideal gas equation:

$PV=nRT,R=8.314{\text{J K}}^{-1}{\text{mol}}^{-1}=0.0821{\text{L atm K}}^{-1}{\text{mol}}^{-1}$

Dalton's law of partial pressures: the total pressure of a non-reacting gas mixture is the sum of partial pressures, $P_{\text{total}}=\sum p_i$, where $p_i=x_i{P}_{\text{total}}$ ($x_i$ = mole fraction).

Worked example 1.1. A gas occupies 2 L at 300 K and 1 atm. What volume at 600 K and 2 atm? Combined law $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$: $\frac{1\times 2}{300}=\frac{2\times V_2}{600}\Rightarrow V_2=\frac{1\times 2\times 600}{300\times 2}=2$ L. (Doubling $T$ doubles $V$; doubling $P$ halves it — they cancel.)

Worked example 1.2. Mass of 5.6 L of O₂ at STP? At STP 22.4 L = 1 mol, so $n=5.6/22.4=0.25$ mol; mass $=0.25\times 32=8$ g.

2. Kinetic theory and molecular speeds

Kinetic theory pictures a gas as tiny, fast, randomly moving molecules with negligible volume and no forces between collisions. Its central result links temperature to molecular motion:

$PV=\frac{1}{3}mN\overline{u^2},\text{average KE per molecule}=\frac{3}{2}{k}_{B}T$

So temperature is a measure of average molecular kinetic energy — it depends only on $T$, not on the gas's identity.

Three characteristic speeds describe the molecular distribution:

$u_{mp}=\sqrt{\frac{2RT}{M}}<{\overline{u}}_{avg}=\sqrt{\frac{8RT}{\pi M}}<u_{rms}=\sqrt{\frac{3RT}{M}}$

Their ratio is fixed: $u_{mp}:{\overline{u}}_{avg}:u_{rms}=1:1.128:1.224=\sqrt{2}:\sqrt{8/\pi }:\sqrt{3}$. All rise with $\sqrt{T}$ and fall with $\sqrt{M}$ — lighter gases move faster (why H₂ effuses quickest).

The Maxwell–Boltzmann distribution shows the spread of speeds: raising temperature broadens and flattens the curve and shifts its peak to higher speed.

Worked example 2.1. By what factor does the rms speed change when temperature rises from 300 K to 1200 K? $u_{rms}\propto \sqrt{T}$, so the factor is $\sqrt{1200/300}=\sqrt{4}=2$. Quadrupling $T$ doubles the rms speed.

Worked example 2.2. Which is faster at the same temperature, H₂ or O₂, and by what factor? $u_{rms}\propto 1/\sqrt{M}$. Ratio $=\sqrt{M_{O_2}/M_{H_2}}=\sqrt{32/2}=\sqrt{16}=4$. Hydrogen is four times faster.

Graham's law of diffusion/effusion follows directly: rate $\propto 1/\sqrt{M}$ (at equal $T,P$), so $\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}$.

3. Real gases and the van der Waals equation

Real gases deviate from ideality because molecules do have volume and do attract one another. Deviation is largest at high pressure and low temperature (molecules crowded and slow).

The compressibility factor $Z=\frac{PV}{nRT}$ measures deviation: $Z=1$ for an ideal gas. $Z<1$ (attraction dominates, gas more compressible) at moderate pressure; $Z>1$ (repulsion/finite volume dominates) at high pressure.

van der Waals equation corrects both defects:

$\left(P+\frac{a{n}^2}{V^2}\right)(V-nb)=nRT$

• $a$ corrects for intermolecular attraction (larger $a$ → more easily liquefied, e.g. NH₃, CO₂).
• $b$ corrects for molecular volume (the space molecules themselves occupy).

Real gases behave most ideally at high temperature and low pressure. Every gas has a critical temperature $T_c$ above which it cannot be liquefied by pressure alone; a higher $a$ means a higher $T_c$.

Worked example 3.1. Why does hydrogen show $Z>1$ at all pressures at 0 °C, while CO₂ shows $Z<1$ at moderate pressure? H₂ has very weak intermolecular attraction (tiny $a$), so its finite molecular volume ($b$) dominates and $Z>1$. CO₂ has strong attractions (large $a$) that pull molecules together, making it more compressible than ideal ($Z<1$) until high pressure, where volume effects take over.

4. The liquid state (brief)

Liquids have definite volume but no definite shape; molecules are close-packed but mobile. Three tested properties:

• Vapour pressure — the pressure of vapour in equilibrium with its liquid; rises with temperature. Boiling occurs when vapour pressure equals external pressure (so water boils below 100 °C on a mountain).
• Surface tension — energy per unit area of surface; makes drops spherical and causes capillary rise. Decreases with temperature.
• Viscosity — resistance to flow; decreases with temperature (why oil flows better when warm).

Part B — Chemical Thermodynamics

5. Basic terms and the first law

Thermodynamics studies energy changes in chemical and physical processes.

• System — the part under study; surroundings — everything else. Open (matter + energy exchange), closed (energy only), isolated (neither).
• State functions depend only on the current state, not the path: internal energy $U$, enthalpy $H$, entropy $S$, Gibbs energy $G$, plus $P,V,T$. Path functions (work $w$, heat $q$) depend on how the change happens.

First law of thermodynamics (energy conservation):

$\Delta U=q+w$

with the convention: heat added to the system is $+q$; work done on the system is $+w$. For pressure–volume work at constant external pressure, $w=-P_{\text{ext}}\Delta V$ (gas expanding does work, $\Delta V>0$, so $w<0$).

Worked example 5.1. A gas absorbs 500 J of heat and does 200 J of work on the surroundings. Change in internal energy? $\Delta U=q+w=+500+(-200)=+300$ J. (Work done by the gas is negative $w$.)

6. Enthalpy and heats of reaction

Most reactions occur at constant pressure, where the heat exchanged is the enthalpy change:

$H=U+PV,\Delta H=\Delta U+\Delta (PV)=\Delta U+\Delta n_{g}RT(\text{gases})$

• $\Delta H<0$: exothermic (heat released); $\Delta H>0$: endothermic (heat absorbed).
• At constant volume, $q_V=\Delta U$; at constant pressure, $q_P=\Delta H$.

Heat capacities: $C_V={\left(\frac{\partial U}{\partial T}\right)}_V$, $C_P={\left(\frac{\partial H}{\partial T}\right)}_P$, and for an ideal gas $C_P-C_V=R$.

Standard enthalpies (at 298 K, 1 bar): of formation ${\Delta }_f{H}^{\ominus }$ (from elements; zero for an element in its standard state), of combustion, of neutralisation (≈ −57.1 kJ/mol for strong acid + strong base).

Hess's law — enthalpy is a state function, so the total $\Delta H$ of a reaction is the same whatever the route:

${\Delta }_{r}H=\sum {\Delta }_{f}H(\text{products})-\sum {\Delta }_{f}H(\text{reactants})$

Worked example 6.1. For a reaction, $\Delta U=-100$ kJ and $\Delta n_g=-1$ at 300 K. Find $\Delta H$. $\Delta H=\Delta U+\Delta n_{g}RT=-100+(-1)(8.314\times 10^{-3})(300)=-100-2.49=-102.5$ kJ.

Worked example 6.2 (Hess's law). Given ${\Delta }_{f}H$: CO₂ = −393.5, H₂O = −285.8, CH₄ = −74.8 kJ/mol, find $\Delta H$ for CH₄ + 2O₂ → CO₂ + 2H₂O. ${\Delta }_{r}H=[(-393.5)+2(-285.8)]-[(-74.8)+0]=(-965.1)-(-74.8)=-890.3$ kJ/mol (O₂ is an element, ${\Delta }_{f}H=0$).

Bond enthalpy method: ${\Delta }_{r}H=\sum (\text{bonds broken})-\sum (\text{bonds formed})$ — energy is absorbed breaking bonds and released forming them.

7. Entropy and the second law

The first law says energy is conserved but not which way a process goes. That is the job of entropy $S$ — a measure of disorder or the number of ways energy can be arranged.

$\Delta S=\frac{q_{\text{rev}}}{T}$

Second law: for any spontaneous process, the total entropy of the universe increases: $\Delta S_{\text{universe}}=\Delta S_{\text{system}}+\Delta S_{\text{surroundings}}>0$.

Entropy rises with: solid → liquid → gas (gases have the highest entropy), dissolving, mixing, and an increase in the number of gas molecules. Third law: the entropy of a perfect crystal at 0 K is zero.

Worked example 7.1. Predict the sign of $\Delta S$ for: (i) melting of ice, (ii) 2NO₂(g) → N₂O₄(g). (i) Solid → liquid increases disorder → $\Delta S>0$. (ii) 2 mol gas → 1 mol gas, fewer molecules → $\Delta S<0$.

8. Gibbs free energy — the master criterion of spontaneity

Judging spontaneity from the universe's entropy is awkward. Gibbs free energy folds the surroundings into a single system property:

$\Delta G=\Delta H-T\Delta S$

The decisive rule (at constant $T,P$):

• $\Delta G<0$ → spontaneous (feasible).
• $\Delta G=0$ → equilibrium.
• $\Delta G>0$ → non-spontaneous (reverse is spontaneous).

Because $\Delta G$ has two competing terms, temperature can flip spontaneity:

Link to equilibrium: $\Delta G^{\ominus }=-RT\ln K$ — a negative standard free energy means $K>1$ (products favoured).

Worked example 8.1. For a reaction $\Delta H=+40$ kJ/mol and $\Delta S=+100$ J K⁻¹ mol⁻¹. Above what temperature is it spontaneous? Spontaneous when $\Delta G=\Delta H-T\Delta S<0\Rightarrow T>\frac{\Delta H}{\Delta S}=\frac{40000}{100}=400$ K. Below 400 K it is non-spontaneous; above, entropy wins.

Worked example 8.2. At the melting point of ice (273 K), what is $\Delta G$ for melting? At the melting point, solid and liquid are in equilibrium, so $\Delta G=0$ (and hence $\Delta S=\Delta H/T$ for the phase change).

9. Common traps NEET sets here

• Always use kelvin in gas laws and thermodynamics — never Celsius.
• $u_{rms}\propto \sqrt{T/M}$ — quadruple $T$ doubles speed; lighter gas is faster.
• Speed order: $u_{mp}<u_{avg}<u_{rms}$ (never reverse).
• $Z<1$ attraction, $Z>1$ finite volume — and $a$ = attraction, $b$ = volume in van der Waals.
• Sign of work: $\Delta U=q+w$ with $w=-P\Delta V$; expansion does negative work on the system.
• $\Delta H=\Delta U+\Delta n_{g}RT$ — count only the change in gaseous moles.
• $C_P-C_V=R$, and $C_P>C_V$ always.
• Spontaneity is about $\Delta G$, not $\Delta H$ — endothermic reactions can be spontaneous if $T\Delta S$ is large enough.
• $\Delta S_{universe}>0$, not $\Delta S_{system}$, is the true spontaneity criterion behind Gibbs energy.

10. Memory aids

• "Kelvin always" — every gas-law and thermodynamics temperature.
• "mp < avg < rms, ratio √2 : √(8/π) : √3" — the three speeds.
• "$a$ attracts, $b$ is bulk" — van der Waals constants.
• "$q$ in plus, $w$ on plus" — first-law sign convention.
• "Break costs, form pays" — bond-enthalpy sign of $\Delta H$.
• "$\Delta G$ negative = go" — the spontaneity test.
• "Low-$T$ enthalpy, high-$T$ entropy" — which term wins the $\Delta G=\Delta H-T\Delta S$ tug of war.

11. Exam protocol

1. Convert temperatures to kelvin and choose the right gas law (which variable is fixed?); use $PV=nRT$ for full problems.
2. For speeds, apply $u\propto \sqrt{T/M}$ and Graham's law; remember the $mp<avg<rms$ order.
3. Deviations: use $Z$ and van der Waals ($a$ = attraction, $b$ = volume); ideal at high $T$, low $P$.
4. First law: $\Delta U=q+w$, $w=-P\Delta V$; convert between $q_P=\Delta H$ and $q_V=\Delta U$ via $\Delta n_{g}RT$.
5. Compute ${\Delta }_{r}H$ from formation enthalpies (Hess's law) or bond enthalpies (broken − formed).
6. Predict $\Delta S$ sign from disorder/gas-mole change; apply the second law to the universe.
7. Decide spontaneity from $\Delta G=\Delta H-T\Delta S$; find the switch temperature $T=\Delta H/\Delta S$; link to $K$ via $\Delta G^{\ominus }=-RT\ln K$.