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WBB • Class XI • Chemistry • Ch 5
Estimated Time: 90 minutes
Study Progress: In Progress

States of Matter: Gases and Liquids

States of matter represent the macroscopic manifestations of competing microscopic thermal energies and intermolecular attractive forces. Chapter 5 explores the physical principles governing the gaseous and liquid states of matter under the WBCHSE Class 11 curriculum. Beginning with intermolecular forces—London dispersion, dipole-dipole, and dipole-induced dipole interactions—the chapter establishes the quantitative empirical gas laws formulated by Boyle, Charles, Gay-Lussac, and Avogadro, unifying them into the Ideal Gas Equation. Dalton law of partial pressures and Graham law of effusion and diffusion offer practical methods for gas collection and isotopic separation. The microscopic perspective is formalized through the Kinetic Molecular Theory of Gases (KMTG), leading to the derivation of the kinetic gas equation and the Maxwell-Boltzmann distribution of molecular speeds (most probable, average, and root-mean-square velocities). The chapter then confronts the reality of non-ideal gas behavior, explaining why real gases deviate from ideality at high pressures and low temperatures. Johannes van der Waals equation of state introduces molecular volume and intermolecular attraction corrections, providing a theoretical foundation for the compressibility factor Z, Boyle temperature, Andrews isotherms, and critical liquefaction constants. Finally, the chapter details the liquid state, analyzing dynamic vapour pressure, normal and standard boiling points, surface tension, capillary action, and laminar viscosity with its temperature dependence.

Why This Chapter Matters

Understanding the behavior of gases and liquids is central to chemical engineering, atmospheric science, cryogenics, and physiology. In industrial synthesis, such as the Haber-Bosch process for ammonia or petrochemical cracking, gas phase pressure-volume-temperature calculations and compressibility factors dictate reactor wall thickness, compression energy costs, and safety margins. In medical physiology, Dalton law explains alveolar gas exchange of oxygen and carbon dioxide, while Graham law explains the rapid diffusion of inhalation anesthetics. In cryogenics and aerospace propulsion, liquefying industrial gases like oxygen, nitrogen, and hydrogen relies entirely on critical temperature thresholds and Joule-Thomson throttling expansion. Furthermore, liquid transport phenomena like surface tension and viscosity govern microfluidic lab-on-a-chip diagnostics, aerosol drug delivery, and automotive lubrication formulations. For WBCHSE Class 11, WBJEE, JEE Main, and NEET aspirants, this chapter provides the foundational bridge connecting classical thermodynamics with molecular kinetic theory and practical chemical engineering.

Before You Begin (Prerequisites)

  • Concept of mole, molar mass, and STP conditions (273.15 K and 1 bar or 1 atm).
  • Basic algebraic manipulation of multi-variable equations and graphical interpretation (slopes, intercepts, isotherms).
  • Fundamental concepts of force, pressure, kinetic energy, work, and vector quantities from physics.
  • Basic atomic structure and chemical bonding concepts including polarity and electronegativity.

Chapter Roadmap & Progression

1 Module 1: Intermolecular Forces, Em...
2 Module 2: Dalton's Law of Partial P...
3 Module 3: Kinetic Molecular Theory...
4 Module 4: Molecular Speeds (u_rms,...
5 Module 5: Real Gases, van der Waals...
6 Module 6: Liquid State Properties:...

Complete Concept Guide (100% Curriculum Coverage)

Module 1: Intermolecular Forces, Empirical Gas Laws & The Ideal Gas Equation

1.1 Intermolecular Forces vs. Thermal Energy

Matter exists in three primary classical states—solid, liquid, and gas—as a consequence of the dynamic competition between two fundamental microscopic factors:

  • Intermolecular Forces (Attractive Potential Energy): Forces that tend to hold molecules together, favoring condensed states (liquids and solids). These include:
    • Dispersion (London) Forces: Weak attractive forces between non-polar molecules or noble gases arising from temporary, instantaneous dipole moments that induce complementary dipoles in neighboring atoms. Potential energy varies as $V \propto - rac{1}{r^6}$.
    • Dipole-Dipole Forces: Attractive electrostatic interactions between molecules possessing permanent dipole moments (e.g., $\text{HCl}, \text{SO}_2$). For stationary polar molecules, $V \propto - rac{1}{r^3}$; for rotating polar molecules in gases or liquids, $V \propto - rac{1}{r^6}$.
    • Dipole-Induced Dipole Forces: Interactions where a permanent dipole induces a dipole in a non-polar neighbor (e.g., $\text{HCl}$ interacting with $\text{Ar}$ or $\text{Cl}_2$), also exhibiting $V \propto - rac{1}{r^6}$.
    • Hydrogen Bonding: A particularly strong, directional dipole-dipole interaction occurring between hydrogen bonded to small, highly electronegative atoms ($\text{F}, \text{O}, \text{N}$) and an electronegative lone pair on another molecule.
  • Thermal Energy (Kinetic Energy): The energy arising from molecular motion, directly proportional to absolute temperature ($T$). Thermal energy promotes randomness, separation, and chaos, favoring the gaseous state.
State Equilibrium Criterion:
$$\text{Predominance of Intermolecular Forces} \xrightarrow{\text{Cooling / Compression}} \text{Solid} \xrightleftharpoons{\text{Heating}} \text{Liquid} \xrightarrow{\text{Heating / Expansion}} \text{Gas (Predominance of Thermal Energy)}$$
1.2 Empirical Gas Laws

A gas is characterized macroscopically by four measurable state variables: Pressure ($P$), Volume ($V$), Temperature ($T$), and Amount of substance ($n$). The empirical gas laws govern these variables when two are held constant:

Gas Law Constant Variables Mathematical Relationship Graphical Behavior
Boyle's Law (1662) Temperature ($T$), Moles ($n$) $P \propto \frac{1}{V} \implies P_1 V_1 = P_2 V_2 = k_1$ $P$ vs $V$: Rectangular hyperbola (Isotherm); $P$ vs $\frac{1}{V}$: Straight line through origin; $PV$ vs $P$: Horizontal line.
Charles's Law (1787) Pressure ($P$), Moles ($n$) $V \propto T \implies \frac{V_1}{T_1} = \frac{V_2}{T_2} = k_2$ $V$ vs $t$ ($^\circ\text{C}$): Linear with intercept at $-273.15^\circ\text{C}$ ($V = 0$). $V$ vs $T$ (K): Straight line through origin (Isobar).
Gay-Lussac's Law (1802) Volume ($V$), Moles ($n$) $P \propto T \implies \frac{P_1}{T_1} = \frac{P_2}{T_2} = k_3$ $P$ vs $T$ (K): Straight line passing through origin (Isochore).
Avogadro's Law (1811) Temperature ($T$), Pressure ($P$) $V \propto n \implies \frac{V_1}{n_1} = \frac{V_2}{n_2} = k_4$ Equal volumes of all gases under identical $T$ and $P$ contain equal numbers of molecules ($N = n N_A$).
1.3 Derivation of the Ideal Gas Equation & Universal Gas Constant

Combining the four proportionalities yields the unified Ideal Gas Equation:

$$V \propto \frac{1}{P} \quad (\text{Boyle}), \quad V \propto T \quad (\text{Charles}), \quad V \propto n \quad (\text{Avogadro})$$ $$V \propto \frac{nT}{P} \implies PV = nRT$$

where $R$ is the Universal Gas Constant, representing the work done per degree Kelvin per mole of an ideal gas ($R = \frac{PV}{nT} = \frac{\text{Work}}{\text{mol} \cdot \text{K}}$).

Values of Universal Gas Constant ($R$):

  • $R = 0.082057 \approx 0.0821 \; \text{L} \cdot \text{atm} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}$ (Use when $P$ is in atm and $V$ is in liters).
  • $R = 0.08314 \; \text{L} \cdot \text{bar} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} = 0.08314 \; \text{dm}^3 \cdot \text{bar} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}$.
  • $R = 8.314 \; \text{J} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} = 8.314 \; \text{N} \cdot \text{m} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}$ (Standard SI unit).
  • $R = 8.314 \times 10^7 \; \text{erg} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}$ (CGS unit).
  • $R = 1.987 \approx 2.0 \; \text{cal} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}$ (Thermochemical unit).
1.4 Combined Gas Law, Gas Density & Molar Mass

For a fixed mass of gas ($n = \text{constant}$), the Combined Gas Law relates two arbitrary thermodynamic states:

$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$

Since the number of moles is $n = \frac{m}{M}$ (where $m$ is mass and $M$ is molar mass), substituting into $PV = nRT$ yields:

$$PV = \frac{m}{M}RT \implies P = \frac{m}{V} \frac{RT}{M} = d \frac{RT}{M}$$ $$d = \frac{PM}{RT} \iff M = \frac{dRT}{P}$$

This fundamental relation reveals that at constant temperature and pressure, the density of an ideal gas is directly proportional to its molar mass ($d \propto M$).

Module 2: Dalton's Law of Partial Pressures, Aqueous Tension & Graham's Law of Diffusion

2.1 Dalton's Law of Partial Pressures

Formulated by John Dalton in 1801, this law applies to mixtures of non-reacting gases confined in a container of fixed volume $V$ at constant temperature $T$:

Dalton's Law Statement: The total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures that each constituent gas would exert if it occupied the entire volume of the container alone at the same temperature. $$P_{\text{total}} = p_1 + p_2 + p_3 + \dots + p_k = \sum_{i=1}^k p_i$$

For an ideal mixture of $k$ gases with mole numbers $n_1, n_2, \dots, n_k$, each gas follows $p_i = \frac{n_i RT}{V}$. Therefore:

$$P_{\text{total}} = \sum_{i=1}^k \frac{n_i RT}{V} = \left(\sum_{i=1}^k n_i\right) \frac{RT}{V} = n_{\text{total}} \frac{RT}{V}$$

Dividing the partial pressure of component $i$ by the total pressure yields:

$$\frac{p_i}{P_{\text{total}}} = \frac{n_i \frac{RT}{V}}{n_{\text{total}} \frac{RT}{V}} = \frac{n_i}{n_{\text{total}}} = x_i \implies p_i = x_i \cdot P_{\text{total}}$$

where $x_i$ is the mole fraction of component $i$ ($\sum x_i = 1$). The partial pressure of any component in a gas mixture is the product of its mole fraction and the total pressure.

2.2 Aqueous Tension & Moist Gas Pressure Correction

When an insoluble or sparingly soluble gas (e.g., $\text{H}_2, \text{O}_2, \text{N}_2$) is prepared in the laboratory, it is typically collected over water by downward displacement. The collected gas inside the gas jar is saturated with water vapour:

$$P_{\text{moist gas}} = P_{\text{dry gas}} + p_{\text{H}_2\text{O}}$$

The partial pressure exerted by water vapour in dynamic equilibrium with liquid water at a specific temperature is known as the Aqueous Tension ($p_{\text{aq}}$):

$$P_{\text{dry gas}} = P_{\text{moist gas}} - p_{\text{aq}} = P_{\text{barometric}} - p_{\text{aq}}$$
Critical Exam Rule: Always subtract aqueous tension before using the pressure of the collected gas in the ideal gas equation ($PV = nRT$) or combined gas law calculations!
2.3 Graham's Law of Diffusion and Effusion

Thomas Graham (1833) experimentally investigated the rates at which gases disperse through one another (diffusion) or escape through a pinhole into a vacuum (effusion):

Graham's Law Statement: Under identical conditions of temperature and pressure, the rates of diffusion or effusion of different gases are inversely proportional to the square roots of their densities or their molar masses. $$r \propto \frac{1}{\sqrt{d}} \propto \frac{1}{\sqrt{M}}$$

For two gases 1 and 2 at identical temperature and pressure:

$$\frac{r_1}{r_2} = \sqrt{\frac{d_2}{d_1}} = \sqrt{\frac{M_2}{M_1}}$$

Operational Expressions for Rate ($r$):

  • In terms of volume ($V$) diffused per time ($t$): $r = \frac{V}{t} \implies \frac{r_1}{r_2} = \frac{V_1 / t_1}{V_2 / t_2} = \sqrt{\frac{M_2}{M_1}}$. If equal volumes diffuse ($V_1 = V_2$), then $\frac{t_2}{t_1} = \sqrt{\frac{M_2}{M_1}}$.
  • In terms of distance ($l$) travelled in a tube of uniform cross-section: $r = \frac{l}{t} \implies \frac{l_1}{l_2} = \sqrt{\frac{M_2}{M_1}}$ (for identical time $t$).
  • In terms of moles ($n$) effused per time ($t$): $\frac{n_1 / t_1}{n_2 / t_2} = \sqrt{\frac{M_2}{M_1}}$.
  • When pressures are unequal ($P_1 \neq P_2$): The rate of effusion is directly proportional to pressure: $$\frac{r_1}{r_2} = \frac{P_1}{P_2} \sqrt{\frac{M_2}{M_1}}$$

Practical Applications: Graham law is used in separation of isotopes (such as gaseous diffusion of $^{235}\text{UF}_6$ and $^{238}\text{UF}_6$), determination of unknown molar masses, and detecting explosive methane leaks in coal mines using atmolysis.

Module 3: Kinetic Molecular Theory of Gases (KMTG) & Maxwell-Boltzmann Speed Distribution

3.1 Postulates of the Kinetic Molecular Theory of Gases

Developed by Bernoulli, Clausius, Maxwell, and Boltzmann, the Kinetic Molecular Theory of Gases (KMTG) provides a microscopic mechanical explanation for macroscopic gas laws based on the following fundamental postulates:

  1. Particulate Nature & Vast Separation: A gas consists of an enormous number of extremely minute, identical spherical particles (atoms or molecules). The actual volume occupied by the individual gas molecules is negligible compared to the total volume of the container ($V_{\text{molecules}} \ll V_{\text{container}}$).
  2. Continuous Random Motion: Gas molecules are in ceaseless, chaotic, rapid straight-line motion in all possible directions. During this motion, they collide with one another and with the walls of the container.
  3. Negligible Intermolecular Forces: Under ordinary conditions, attractive and repulsive forces between gas molecules are negligible. The molecules behave completely independently of one another.
  4. Perfect Elasticity of Collisions: All collisions (molecule-molecule and molecule-wall) are perfectly elastic. Although individual molecules may exchange momentum and kinetic energy during a collision, the total translational kinetic energy of the entire system remains conserved.
  5. Origin of Gas Pressure: The pressure exerted by a gas is the cumulative macroscopic force resulting from the continuous bombardment of gas molecules per unit area of the container walls ($P = \frac{F}{A} = \frac{\Delta p / \Delta t}{A}$).
  6. Absolute Temperature and Kinetic Energy: The average kinetic energy of gas molecules is directly proportional to the absolute temperature of the gas ($E_k \propto T$). At absolute zero ($0 \; \text{K}$), molecular motion theoretically ceases.
3.2 The Kinetic Gas Equation & Kinetic Energy Relations

By analyzing the rate of change of momentum of $N$ molecules, each of mass $m$, colliding with the mutually perpendicular planar walls of a cubic container of edge length $L$ and volume $V = L^3$, Clausius derived the Kinetic Gas Equation:

$$PV = \frac{1}{3} m N u_{\text{rms}}^2 = \frac{1}{3} M_{\text{total}} u_{\text{rms}}^2$$

where $u_{\text{rms}}$ is the root-mean-square speed of the gas molecules. For 1 mole of gas ($N = N_A$, so $m N_A = M$, the molar mass):

$$PV = \frac{1}{3} M u_{\text{rms}}^2 = RT \implies u_{\text{rms}} = \sqrt{\frac{3RT}{M}}$$

Total Kinetic Energy Expressions:

  • Total Kinetic Energy for $n$ moles: $E_k = \frac{1}{2} M_{\text{total}} u_{\text{rms}}^2 = \frac{1}{2} (n M) \left(\frac{3RT}{M}\right) = \frac{3}{2} nRT$.
  • Translational Kinetic Energy per mole ($n=1$): $E_{k, \text{molar}} = \frac{3}{2} RT$. (Notice this depends strictly on temperature $T$ and is completely independent of the nature or molar mass of the gas!).
  • Translational Kinetic Energy per molecule: Dividing by Avogadro's number $N_A$: $$E_{k, \text{molecule}} = \frac{3}{2} \frac{R}{N_A} T = \frac{3}{2} k_B T$$ where $k_B = \frac{R}{N_A} = 1.3806 \times 10^{-23} \; \text{J} \cdot \text{K}^{-1}$ is the Boltzmann Constant.
3.3 Maxwell-Boltzmann Distribution of Molecular Speeds

Because gas molecules constantly undergo millions of random collisions per second, their speeds are continuously changing. In 1859, James Clerk Maxwell and Ludwig Boltzmann applied statistical mechanics to derive the probability distribution function of molecular speeds in a three-dimensional gas:

$$\frac{1}{N} \frac{dN_u}{du} = 4\pi \left(\frac{M}{2\pi RT}\right)^{3/2} u^2 e^{-\frac{Mu^2}{2RT}}$$

where $\frac{dN_u}{N}$ represents the fraction of total gas molecules possessing speeds between $u$ and $u + du$.

Salient Features of the Maxwell-Boltzmann Distribution Curve:

  • The fraction of molecules with zero speed is zero (curve starts at origin, $(0,0)$).
  • The curve rises sharply with increasing speed, reaches a distinctive maximum corresponding to the most probable speed ($u_{\text{mp}}$), and then descends asymptotically toward zero at very high speeds, forming an asymmetrical bell curve with a long tail.
  • The total area under the curve represents the total fraction of molecules and equals unity: $\int_0^\infty \frac{1}{N} \frac{dN_u}{du} du = 1$.
  • Effect of Temperature: As temperature increases ($T_2 > T_1$), the peak shifts to the right (higher speed) and simultaneously broadens and flattens (lower peak height). This proves that at higher temperatures, a significantly greater fraction of molecules possess high speeds.
  • Effect of Molar Mass: At a given temperature, lighter gases (e.g., $\text{H}_2, \text{He}$) have flatter, broader distributions shifted toward higher speeds, whereas heavier gases (e.g., $\text{O}_2, \text{Cl}_2$) have narrower, taller peaks shifted toward lower speeds.

Module 4: Molecular Speeds (u_rms, u_av, u_mp) & Quantitative Velocity Relationships

4.1 Defining the Three Characteristic Molecular Speeds

To characterize the speed distribution of billions of gas molecules moving in all directions, three mathematically distinct statistical speeds are employed:

  1. Most Probable Speed ($u_{\text{mp}}$ or $\alpha$):

    The speed possessed by the maximum fraction of gas molecules at a given temperature. It corresponds to the mathematical peak (maximum) of the Maxwell-Boltzmann distribution curve (found by setting $\frac{d}{du}\left[\frac{dN_u}{du}\right] = 0$):

    $$u_{\text{mp}} = \sqrt{\frac{2RT}{M}} = \sqrt{\frac{2k_B T}{m}} = \sqrt{\frac{2PV}{M}} = \sqrt{\frac{2P}{d}}$$
  2. Average Speed ($u_{\text{av}}$ or $\bar{u}$):

    The arithmetic mean of the speeds of all gas molecules present in the sample ($u_{\text{av}} = \frac{u_1 + u_2 + \dots + u_N}{N}$):

    $$u_{\text{av}} = \sqrt{\frac{8RT}{\pi M}} = \sqrt{\frac{8k_B T}{\pi m}} = \sqrt{\frac{8PV}{\pi M}} = \sqrt{\frac{8P}{\pi d}} \approx \sqrt{\frac{2.546 RT}{M}}$$
  3. Root-Mean-Square Speed ($u_{\text{rms}}$ or $c$):

    The square root of the arithmetic mean of the squares of the speeds of all molecules ($u_{\text{rms}} = \sqrt{\frac{u_1^2 + u_2^2 + \dots + u_N^2}{N}}$). This speed represents the direct kinetic energy equivalence of the gas:

    $$u_{\text{rms}} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3k_B T}{m}} = \sqrt{\frac{3PV}{M}} = \sqrt{\frac{3P}{d}}$$
4.2 Mutual Ratios and Speed Ordering

Comparing the mathematical expressions under identical conditions of temperature ($T$) and molar mass ($M$):

$$u_{\text{mp}} : u_{\text{av}} : u_{\text{rms}} = \sqrt{\frac{2RT}{M}} : \sqrt{\frac{8RT}{\pi M}} : \sqrt{\frac{3RT}{M}} = \sqrt{2} : \sqrt{\frac{8}{\pi}} : \sqrt{3}$$ $$u_{\text{mp}} : u_{\text{av}} : u_{\text{rms}} = 1.414 : 1.596 : 1.732 = 1.000 : 1.128 : 1.224$$
Fundamental Speed Hierarchy: $$u_{\text{mp}} < u_{\text{av}} < u_{\text{rms}}$$ $$u_{\text{rms}} = 1.085 \; u_{\text{av}} = 1.224 \; u_{\text{mp}}$$ $$u_{\text{av}} = 0.921 \; u_{\text{rms}} = 1.128 \; u_{\text{mp}}$$ $$u_{\text{mp}} = 0.816 \; u_{\text{rms}} = 0.886 \; u_{\text{av}}$$
4.3 Temperature & Molar Mass Dependences

All three speeds obey identical functional proportionalities:

$$u \propto \sqrt{\frac{T}{M}}$$
  • For the same gas at different temperatures $T_1$ and $T_2$: $$\frac{u_1}{u_2} = \sqrt{\frac{T_1}{T_2}}$$
  • For two different gases 1 and 2 at the same temperature $T$: $$\frac{u_1}{u_2} = \sqrt{\frac{M_2}{M_1}}$$
  • Equal Speed Condition: If the root-mean-square speed of gas 1 at temperature $T_1$ equals the root-mean-square speed of gas 2 at temperature $T_2$: $$\sqrt{\frac{3RT_1}{M_1}} = \sqrt{\frac{3RT_2}{M_2}} \implies \frac{T_1}{M_1} = \frac{T_2}{M_2}$$

Module 5: Real Gases, van der Waals Equation of State & Critical Liquefaction Phenomena

5.1 Deviation from Ideality & Flaws of Kinetic Molecular Theory

Under experimental conditions of high pressure and low temperature, real gases show marked deviations from ideal gas behavior. The Dutch physicist Johannes Diderik van der Waals (1873) identified two faulty assumptions in the classical Kinetic Molecular Theory:

  1. Assumption 1 (Negligible Volume of Molecules): KMT assumes that gas molecules occupy zero volume. In reality, molecules are hard, incompressible spheres possessing finite volume. At high pressure, the volume of the container becomes very small, making the volume occupied by the molecules themselves significant. Hence, the actual free space available for molecular movement is $(V - nb)$, where $b$ is the excluded volume or co-volume per mole.
  2. Assumption 2 (Zero Intermolecular Attraction): KMT assumes no forces of attraction exist between molecules. In reality, intermolecular attractive forces pull striking molecules inward, reducing the net momentum transferred per collision to the wall. The measured pressure $P_{\text{real}}$ is less than the ideal pressure $P_{\text{ideal}}$ by a pressure correction factor $p' = \frac{a n^2}{V^2}$.
5.2 The van der Waals Equation of State

Introducing the pressure correction $P_{\text{ideal}} = P + \frac{a n^2}{V^2}$ and volume correction $V_{\text{ideal}} = V - nb$ into the ideal gas equation yields the celebrated van der Waals Equation:

$$\left(P + \frac{a n^2}{V^2}\right)(V - nb) = nRT \quad (\text{for } n \text{ moles})$$ $$\left(P + \frac{a}{V_m^2}\right)(V_m - b) = RT \quad (\text{for } 1 \text{ mole, where } V_m = V/n)$$

Physical Significance and Units of van der Waals Constants:

  • van der Waals Constant $a$: Measures the magnitude of intermolecular attractive forces.
    • Units: $\text{atm} \cdot \text{L}^2 \cdot \text{mol}^{-2}$ or $\text{bar} \cdot \text{dm}^6 \cdot \text{mol}^{-2}$ or $\text{N} \cdot \text{m}^4 \cdot \text{mol}^{-2}$ (SI).
    • Higher $a$ values indicate stronger intermolecular attractions, implying easier liquefaction (e.g., $a_{\text{SO}_2} > a_{\text{NH}_3} > a_{\text{CO}_2} > a_{\text{CH}_4} > a_{\text{O}_2} > a_{\text{N}_2} > a_{\text{H}_2} > a_{\text{He}}$).
  • van der Waals Constant $b$: Measures the effective incompressible volume (co-volume) of the molecules.
    • Units: $\text{L} \cdot \text{mol}^{-1}$ or $\text{dm}^3 \cdot \text{mol}^{-1}$ or $\text{m}^3 \cdot \text{mol}^{-1}$ (SI).
    • $b = 4 N_A v = 4 N_A \left(\frac{4}{3}\pi r^3\right)$, proving that excluded volume is exactly four times the actual spherical volume of the molecules.
5.3 Compressibility Factor ($Z$) and Pressure Regimes

The deviation of a real gas from ideality is quantitatively evaluated by the Compressibility Factor ($Z$):

$$Z = \frac{PV_m}{RT} = \frac{V_{m, \text{real}}}{V_{m, \text{ideal}}}$$
  • Ideal Gas: $Z = 1$ under all conditions of temperature and pressure.
  • Case I: Low to Moderate Pressure Regime ($P < 200 \; \text{atm}$): $V_m$ is relatively large, making $b$ negligible compared to $V_m$ ($V_m - b \approx V_m$): $$\left(P + \frac{a}{V_m^2}\right)V_m = RT \implies PV_m + \frac{a}{V_m} = RT \implies Z = \frac{PV_m}{RT} = 1 - \frac{a}{V_m RT}$$ Since $\frac{a}{V_m RT} > 0$, $Z < 1$. Attractive forces dominate; the gas is more compressible than an ideal gas ($V_{\text{real}} < V_{\text{ideal}}$).
  • Case II: High Pressure Regime ($P > 500 \; \text{atm}$): $P$ is very large, so the attractive correction $\frac{a}{V_m^2}$ is negligible compared to $P$ ($P + \frac{a}{V_m^2} \approx P$): $$P(V_m - b) = RT \implies PV_m - Pb = RT \implies Z = \frac{PV_m}{RT} = 1 + \frac{Pb}{RT}$$ Since $\frac{Pb}{RT} > 0$, $Z > 1$. Molecular volume repulsions dominate; the gas is less compressible than an ideal gas ($V_{\text{real}} > V_{\text{ideal}}$).
  • Case III: Hydrogen ($\text{H}_2$) and Helium ($\text{He}$): Because of extremely small molecular masses and minimal polarizability, London dispersion forces are virtually non-existent ($a \approx 0$) at ordinary temperatures. Thus: $$Z = 1 + \frac{Pb}{RT} > 1 \quad \text{at all pressures!}$$ The curves for $\text{H}_2$ and $\text{He}$ lie entirely above $Z = 1$ without showing an initial dip.
  • Boyle Temperature ($T_b$): The specific temperature at which the attractive term $\frac{a}{V_m}$ and repulsive term $\frac{Pb}{RT}$ cancel out over an appreciable pressure range, making $Z \approx 1$: $$T_b = \frac{a}{Rb}$$
5.4 Andrews Isotherms, Liquefaction & Critical Constants

Thomas Andrews (1869) studied the $P-V$ isotherms of carbon dioxide ($\text{CO}_2$), demonstrating that a gas can only be liquefied below a specific threshold temperature known as the Critical Temperature ($T_c$).

Expanding the 1-mole van der Waals equation as a cubic equation in $V_m$:

$$\left(P + \frac{a}{V_m^2}\right)(V_m - b) = RT \implies V_m^3 - \left(b + \frac{RT}{P}\right)V_m^2 + \left(\frac{a}{P}\right)V_m - \frac{ab}{P} = 0$$

At the critical point $(P_c, T_c)$, the three roots of this cubic equation coalesce into a single real root: $(V_m - V_c)^3 = 0 \implies V_m^3 - 3V_c V_m^2 + 3V_c^2 V_m - V_c^3 = 0$. Equating coefficients yields the fundamental Critical Constants:

Critical Constant Physical Meaning Formula in Terms of $a, b$
Critical Temperature ($T_c$) The highest temperature at which a gas can be liquefied by pressure alone; above $T_c$, no pressure can liquefy the gas. $$T_c = \frac{8a}{27Rb}$$
Critical Pressure ($P_c$) The minimum pressure required to liquefy a gas at its critical temperature $T_c$. $$P_c = \frac{a}{27b^2}$$
Critical Volume ($V_c$) The volume occupied by 1 mole of gas at its critical temperature and pressure. $$V_c = 3b$$
Critical Compressibility ($Z_c$) Universal dimensionless ratio at the critical point for all van der Waals gases. $$Z_c = \frac{P_c V_c}{R T_c} = \frac{3}{8} = 0.375$$

Module 6: Liquid State Properties: Vapour Pressure, Surface Tension & Viscosity

6.1 Dynamic Vapour Pressure and Boiling Phenomena

In a closed evacuated container partially filled with liquid, molecules with kinetic energy exceeding the average escape into the vapour phase (evaporation). Simultaneously, vapour molecules collide with the liquid surface and are recaptured (condensation). Eventually, a dynamic equilibrium is reached:

$$\text{Liquid} \xrightleftharpoons[\text{condensation}]{\text{evaporation}} \text{Vapour}$$

Vapour Pressure: The equilibrium pressure exerted by the vapour phase above the liquid at a constant temperature.

  • Factors Influencing Vapour Pressure:
    • Nature of Liquid: Weaker intermolecular forces result in higher volatility and higher vapour pressure (e.g., $P_{\text{diethyl ether}} > P_{\text{ethanol}} > P_{\text{water}}$).
    • Temperature: Vapour pressure increases exponentially with temperature according to the Clausius-Clapeyron equation: $\ln\left(\frac{P_2}{P_1}\right) = \frac{\Delta H_{\text{vap}}}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$.
  • Boiling Point: The temperature at which the equilibrium vapour pressure of the liquid equals the surrounding external atmospheric pressure.
    • Normal Boiling Point: Boiling temperature when external pressure is exactly $1 \; \text{atm} = 1.01325 \; \text{bar}$ (for pure water, $100.0^\circ\text{C}$).
    • Standard Boiling Point: Boiling temperature when external pressure is exactly $1 \; \text{bar} = 0.987 \; \text{atm}$ (for pure water, $99.6^\circ\text{C}$).
6.2 Surface Tension: Molecular Origin and Consequences

Molecules within the bulk of a liquid experience isotropic (equal) cohesive forces in all directions from surrounding neighbors. However, molecules at the liquid-gas interface lack liquid neighbors above them and experience a net inward unbalanced cohesive force pulling them toward the bulk.

Definition of Surface Tension ($\gamma$ or $\sigma$): The force acting per unit length perpendicular to an imaginary line drawn on the surface of the liquid, or the work required to increase the surface area of the liquid by unit amount. $$\gamma = \frac{F}{l} = \frac{W}{\Delta A}$$
  • Units: SI unit is $\text{N} \cdot \text{m}^{-1}$ or $\text{J} \cdot \text{m}^{-2}$; CGS unit is $\text{dyn} \cdot \text{cm}^{-1}$ or $\text{erg} \cdot \text{cm}^{-2}$ ($1 \; \text{N} \cdot \text{m}^{-1} = 10^3 \; \text{dyn} \cdot \text{cm}^{-1}$).
  • Physical Consequences:
    • Spherical Shape of Liquid Drops: To minimize surface free energy, liquids naturally minimize surface area. For a given volume, a sphere possesses the minimum surface area, causing raindrops, soap bubbles, and mercury droplets to assume spherical shapes.
    • Capillary Action: The rise or fall of liquids inside narrow capillary tubes: $$h = \frac{2\gamma \cos\theta}{r \rho g}$$ where $\theta$ is the contact angle. When adhesive forces (liquid-glass) exceed cohesive forces ($\theta < 90^\circ$, e.g., water), capillary rise occurs and the meniscus is concave. When cohesive forces exceed adhesive forces ($\theta > 90^\circ$, e.g., mercury), capillary depression occurs and the meniscus is convex.
    • Effect of Temperature: As temperature increases, kinetic energy weakens cohesive forces; surface tension decreases linearly and becomes zero at the critical temperature $T_c$.
6.3 Viscosity: Laminar Flow and Newton's Law

When a real liquid flows smoothly through a pipe or across a flat surface, it moves in parallel layers or laminae with varying velocities (laminar flow). The layer in direct contact with the solid boundary is stationary ($u = 0$), while successive layers move with progressively increasing velocities up to a maximum velocity $u_{\text{max}}$ at the center.

The velocity gradient perpendicular to the direction of flow is $\frac{du}{dz}$. Isaac Newton established that the internal frictional shearing force ($F$) acting between adjacent liquid layers of contact area $A$ is proportional to the area and the velocity gradient:

Newton's Viscosity Law: $$F = -\eta A \frac{du}{dz}$$

where $\eta$ is the Coefficient of Viscosity, defined as the tangential shearing force per unit area required to maintain a unit velocity gradient between two parallel liquid layers unit distance apart.

  • Units of Viscosity:
    • SI Unit: $\text{Pa} \cdot \text{s} = \text{N} \cdot \text{s} \cdot \text{m}^{-2} = \text{kg} \cdot \text{m}^{-1} \cdot \text{s}^{-1}$.
    • CGS Unit: Poise (P) $= \text{dyn} \cdot \text{s} \cdot \text{cm}^{-2} = \text{g} \cdot \text{cm}^{-1} \cdot \text{s}^{-1}$.
    • Conversion: $1 \; \text{Pa} \cdot \text{s} = 10 \; \text{Poise} = 1000 \; \text{centipoise (cP)}$. (Water at $20^\circ\text{C}$ has $\eta \approx 1.002 \; \text{cP} = 1.002 \times 10^{-3} \; \text{Pa} \cdot \text{s}$).
  • Effect of Temperature on Viscosity:
    • Liquids: Viscosity decreases rapidly with increasing temperature because elevated thermal energy overcomes intermolecular hydrogen bonding and van der Waals attractions (empirically $\eta = A e^{E_a / RT}$).
    • Gases: In stark contrast to liquids, gas viscosity increases with temperature ($\eta \propto \sqrt{T}$) due to increased momentum transfer between rapidly colliding gas molecules.

Key Formulas, Reactions & Definitions

Ideal Gas Equation & Density-Molar Mass Relations
$$PV = nRT = \frac{m}{M}RT \implies d = \frac{PM}{RT} \iff M = \frac{dRT}{P}$$
Dalton's Law of Partial Pressures & Aqueous Tension
$$P_{\text{total}} = \sum_{i=1}^k p_i, \quad p_i = x_i P_{\text{total}}, \quad P_{\text{dry gas}} = P_{\text{moist}} - p_{\text{aq}}$$
Graham's Law of Diffusion and Effusion
$$\frac{r_1}{r_2} = \sqrt{\frac{d_2}{d_1}} = \sqrt{\frac{M_2}{M_1}} = \frac{V_1 / t_1}{V_2 / t_2} = \frac{P_1}{P_2}\sqrt{\frac{M_2}{M_1}}$$
Maxwell-Boltzmann Molecular Speeds & Kinetic Energy
$$u_{\text{rms}} = \sqrt{\frac{3RT}{M}}, \; u_{\text{av}} = \sqrt{\frac{8RT}{\pi M}}, \; u_{\text{mp}} = \sqrt{\frac{2RT}{M}}, \; E_k = \frac{3}{2}RT$$
van der Waals Equation of State & Compressibility Factor
$$\left(P + \frac{a n^2}{V^2}\right)(V - nb) = nRT, \quad Z = \frac{PV_m}{RT} = \frac{V_{\text{real}}}{V_{\text{ideal}}}, \quad T_b = \frac{a}{Rb}$$
Critical Constants & Transport Properties of Liquids
$$T_c = \frac{8a}{27Rb}, \; P_c = \frac{a}{27b^2}, \; V_c = 3b, \; \gamma = \frac{F}{l}, \; F = -\eta A \frac{du}{dz}$$

Conceptual Solved Examples & Case Studies

Example 1
At 27°C and 760 mmHg pressure, the density of a gaseous oxide of nitrogen is 1.98 g/L. Calculate the molar mass of the oxide and identify its molecular formula. [R = 0.0821 L·atm·K⁻¹·mol⁻¹]
Step-by-Step Solution:
Step 1: Identify given state variables and convert to standard units:
Temperature: $T = 27^\circ\text{C} = 27 + 273.15 = 300.15 \; \text{K} \approx 300 \; \text{K}$.
Pressure: $P = 760 \; \text{mmHg} = 1.00 \; \text{atm}$.
Density: $d = 1.98 \; \text{g/L}$.
Universal Gas Constant: $R = 0.0821 \; \text{L} \cdot \text{atm} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}$.

Step 2: Apply the density-molar mass equation:
$$M = \frac{dRT}{P}$$ $$M = \frac{1.98 \times 0.0821 \times 300}{1.00} = \frac{48.767}{1.00} \approx 48.8 \; \text{g/mol}$$
Step 3: Identify the molecular formula:
Let us evaluate nitrogen oxides: $\text{NO} = 14 + 16 = 30 \; \text{g/mol}$; $\text{NO}_2 = 14 + 32 = 46 \; \text{g/mol}$; $\text{N}_2\text{O} = 28 + 16 = 44 \; \text{g/mol}$.
Taking the slight dimerization of $\text{NO}_2$ into $\text{N}_2\text{O}_4$ into account in equilibrium, the average molar mass of $48.8 \; \text{g/mol}$ corresponds to nitrogen dioxide ($\text{NO}_2$) partially dimerized, with the principal oxide monomer being $\text{NO}_2$.
Final Answer: Molar mass = $48.8 \; \text{g/mol}$; Oxide = $\text{NO}_2$.
Example 2
A sample of 250 mL of nitrogen gas is collected over water at 25°C and 745 torr total barometric pressure. The aqueous tension of water at 25°C is 23.8 torr. Calculate the volume that this dry nitrogen gas would occupy at Standard Temperature and Pressure (STP: 0°C and 760 torr).
Step-by-Step Solution:
Step 1: Compute the pressure of dry nitrogen gas:
According to Dalton law of partial pressures: $$P_{\text{dry } \text{N}_2} = P_{\text{moist}} - p_{\text{aq}} = 745 - 23.8 = 721.2 \; \text{torr}$$
Step 2: Tabulate initial and final state variables:
Initial state (wet collection conditions): $$P_1 = 721.2 \; \text{torr}, \quad V_1 = 250 \; \text{mL}, \quad T_1 = 25 + 273.15 = 298.15 \; \text{K}$$ Final state (STP standard conditions): $$P_2 = 760.0 \; \text{torr}, \quad V_2 = ?, \quad T_2 = 273.15 \; \text{K}$$
Step 3: Apply the Combined Gas Law:
$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \implies V_2 = V_1 \times \left(\frac{P_1}{P_2}\right) \times \left(\frac{T_2}{T_1}\right)$$ $$V_2 = 250 \times \left(\frac{721.2}{760.0}\right) \times \left(\frac{273.15}{298.15}\right)$$ $$V_2 = 250 \times 0.94895 \times 0.91615 \approx 217.34 \; \text{mL}$$
Final Answer: The volume of dry nitrogen at STP is $217.34 \; \text{mL}$.
Example 3
Under identical experimental conditions of temperature and pressure, 50 mL of methane (CH₄) gas diffuses through a porous pinhole in 20 seconds. An unknown hydrocarbon gas of volume 40 mL requires 32 seconds to diffuse through the identical pinhole. Calculate the molar mass of the hydrocarbon and deduce its molecular formula if it belongs to the alkene family (C_n H_2n).
Step-by-Step Solution:
Step 1: Calculate the diffusion rates for both gases:
Rate of diffusion of methane ($\text{CH}_4$): $$r_1 = \frac{V_1}{t_1} = \frac{50 \; \text{mL}}{20 \; \text{s}} = 2.50 \; \text{mL/s}$$ Rate of diffusion of unknown hydrocarbon ($X$): $$r_2 = \frac{V_2}{t_2} = \frac{40 \; \text{mL}}{32 \; \text{s}} = 1.25 \; \text{mL/s}$$
Step 2: Apply Graham's Law of Diffusion:
$$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$$ Given molar mass of $\text{CH}_4$: $M_1 = 12 + 4(1) = 16.0 \; \text{g/mol}$.
$$\frac{2.50}{1.25} = \sqrt{\frac{M_2}{16.0}} \implies 2 = \sqrt{\frac{M_2}{16.0}}$$ Squaring both sides: $$4 = \frac{M_2}{16.0} \implies M_2 = 4 \times 16.0 = 64.0 \; \text{g/mol}$$
Step 3: Deduce the molecular formula:
For an alkene with formula $\text{C}_n\text{H}_{2n}$: $$12n + 2n = 14n = 64 \implies n = \frac{64}{14} \approx 4.57$$ If the unknown gas is a pure gaseous fluorinated/chlorinated derivative or alkene mixture, let us check cyclobutane/butene isomer $\text{C}_4\text{H}_8 = 56 \; \text{g/mol}$, sulfur dioxide $\text{SO}_2 = 64 \; \text{g/mol}$. For the calculated molar mass of $64 \; \text{g/mol}$, the gas matches sulfur dioxide (or chloroethane $\text{C}_2\text{H}_5\text{Cl} = 64.5 \; \text{g/mol}$).
Final Answer: Molar mass $M_2 = 64.0 \; \text{g/mol}$.
Example 4
For oxygen gas (O₂) at 300 K: (a) Calculate the root-mean-square speed (u_rms), average speed (u_av), and most probable speed (u_mp) in m/s. (b) Calculate the total translational kinetic energy of 16.0 g of oxygen at this temperature. [Given: R = 8.314 J·K⁻¹·mol⁻¹, Molar mass of O₂ = 32.0 g/mol = 0.032 kg/mol]
Step-by-Step Solution:

Part (a): Calculation of Molecular Speeds:
Convert molar mass to SI units: $M = 32.0 \; \text{g/mol} = 0.032 \; \text{kg/mol}$.
Universal gas constant: $R = 8.314 \; \text{J} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}$. Temperature: $T = 300 \; \text{K}$.

  1. Root-Mean-Square Speed ($u_{\text{rms}}$):

$$u_{\text{rms}} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3 \times 8.314 \times 300}{0.032}} = \sqrt{\frac{7482.6}{0.032}} = \sqrt{233831.25} \approx 483.56 \; \text{m/s}$$


  1. Average Speed ($u_{\text{av}}$):

$$u_{\text{av}} = \sqrt{\frac{8RT}{\pi M}} = \sqrt{\frac{8 \times 8.314 \times 300}{3.1416 \times 0.032}} = \sqrt{\frac{19953.6}{0.10053}} = \sqrt{198483.9} \approx 445.51 \; \text{m/s}$$


  1. Most Probable Speed ($u_{\text{mp}}$):

$$u_{\text{mp}} = \sqrt{\frac{2RT}{M}} = \sqrt{\frac{2 \times 8.314 \times 300}{0.032}} = \sqrt{\frac{4988.4}{0.032}} = \sqrt{155887.5} \approx 394.83 \; \text{m/s}$$


Verification of speed ordering: $394.83 < 445.51 < 483.56 \implies u_{\text{mp}} < u_{\text{av}} < u_{\text{rms}}$. (Check: $394.83 \times 1.2247 = 483.55 \; \text{m/s}$).

Part (b): Total Translational Kinetic Energy:
Number of moles of $\text{O}_2$:

$$n = \frac{m}{M} = \frac{16.0 \; \text{g}}{32.0 \; \text{g/mol}} = 0.50 \; \text{mol}$$

Total translational kinetic energy:

$$E_k = \frac{3}{2} nRT = \frac{3}{2} \times 0.50 \times 8.314 \times 300 = 1870.65 \; \text{J}$$


Final Answer: (a) $u_{\text{rms}} = 483.56 \; \text{m/s}$, $u_{\text{av}} = 445.51 \; \text{m/s}$, $u_{\text{mp}} = 394.83 \; \text{m/s}$; (b) $E_k = 1870.65 \; \text{J} = 1.871 \; \text{kJ}$.
Example 5
Two moles of ammonia gas (NH₃) occupy a volume of 5.0 L at 27°C. Calculate the pressure exerted by the gas using: (a) the Ideal Gas Equation, and (b) the van der Waals Equation. Given for NH₃: a = 4.17 atm·L²·mol⁻², b = 0.0371 L·mol⁻¹, R = 0.0821 L·atm·K⁻¹·mol⁻¹. Also calculate the compressibility factor Z.
Step-by-Step Solution:

Step 1: State the given variables:
Number of moles: $n = 2.0 \; \text{mol}$.
Volume: $V = 5.0 \; \text{L}$.
Temperature: $T = 27 + 273.15 = 300.15 \; \text{K} \approx 300 \; \text{K}$.
$a = 4.17 \; \text{atm} \cdot \text{L}^2 \cdot \text{mol}^{-2}$, $b = 0.0371 \; \text{L} \cdot \text{mol}^{-1}$, $R = 0.0821 \; \text{L} \cdot \text{atm} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}$.

Part (a): Pressure from Ideal Gas Equation:

$$P_{\text{ideal}} = \frac{nRT}{V} = \frac{2.0 \times 0.0821 \times 300}{5.0} = \frac{49.26}{5.0} = 9.852 \; \text{atm}$$


Part (b): Pressure from van der Waals Equation:

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

  1. Calculate effective volume term $(V - nb)$:

$$nb = 2.0 \times 0.0371 = 0.0742 \; \text{L} \implies V - nb = 5.0 - 0.0742 = 4.9258 \; \text{L}$$

  1. Calculate kinetic repulsion pressure term:

$$\frac{nRT}{V - nb} = \frac{49.26}{4.9258} = 10.0004 \; \text{atm}$$

  1. Calculate internal cohesive pressure correction $\frac{a n^2}{V^2}$:

$$\frac{a n^2}{V^2} = \frac{4.17 \times (2.0)^2}{(5.0)^2} = \frac{4.17 \times 4.0}{25.0} = \frac{16.68}{25.0} = 0.6672 \; \text{atm}$$

  1. Calculate net van der Waals pressure:

$$P_{\text{real}} = 10.0004 - 0.6672 = 9.3332 \approx 9.33 \; \text{atm}$$


Part (c): Calculate Compressibility Factor ($Z$):

$$Z = \frac{P_{\text{real}} V}{nRT} = \frac{9.3332 \times 5.0}{49.26} = \frac{46.666}{49.26} \approx 0.947$$

Because $Z = 0.947 < 1$, attractive forces dominate in ammonia at this temperature and pressure, rendering it more compressible than an ideal gas.
Final Answer: (a) $P_{\text{ideal}} = 9.85 \; \text{atm}$; (b) $P_{\text{real}} = 9.33 \; \text{atm}$; $Z = 0.947$.

Example 6
(a) The van der Waals constants for carbon dioxide (CO₂) are a = 3.59 atm·L²·mol⁻² and b = 0.0427 L·mol⁻¹. Calculate the critical temperature (T_c), critical pressure (P_c), critical volume (V_c), and Boyle temperature (T_b) for CO₂. [R = 0.0821 L·atm·K⁻¹·mol⁻¹]. (b) Explain why water rises in a glass capillary tube while mercury is depressed, stating the relationship for capillary height h.
Step-by-Step Solution:
Part (a): Critical Constants and Boyle Temperature for $\text{CO}_2$:
Given: $a = 3.59 \; \text{atm} \cdot \text{L}^2 \cdot \text{mol}^{-2}$, $b = 0.0427 \; \text{L} \cdot \text{mol}^{-1}$, $R = 0.0821 \; \text{L} \cdot \text{atm} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}$.

1. Critical Temperature ($T_c$):
$$T_c = \frac{8a}{27Rb} = \frac{8 \times 3.59}{27 \times 0.0821 \times 0.0427} = \frac{28.72}{0.09465} \approx 303.43 \; \text{K} \; (30.28^\circ\text{C})$$
2. Critical Pressure ($P_c$):
$$P_c = \frac{a}{27b^2} = \frac{3.59}{27 \times (0.0427)^2} = \frac{3.59}{27 \times 0.0018233} = \frac{3.59}{0.04923} \approx 72.92 \; \text{atm}$$
3. Critical Volume ($V_c$):
$$V_c = 3b = 3 \times 0.0427 = 0.1281 \; \text{L/mol} = 128.1 \; \text{mL/mol}$$
4. Boyle Temperature ($T_b$):
$$T_b = \frac{a}{Rb} = \frac{3.59}{0.0821 \times 0.0427} = \frac{3.59}{0.0035057} \approx 1024.05 \; \text{K} \; (750.9^\circ\text{C})$$

Part (b): Molecular Basis of Capillary Rise vs Depression:
The height $h$ of liquid column in a capillary tube of internal radius $r$ is given by Jurin's law: $$h = \frac{2\gamma \cos\theta}{r \rho g}$$ where $\gamma$ is surface tension, $\rho$ is density, $g$ is acceleration due to gravity, and $\theta$ is the contact angle between liquid and solid tube wall.
  • Water in Glass (Capillary Rise): Adhesive forces between polar water molecules and glass silanol groups ($\text{Si-OH}$) are significantly stronger than cohesive hydrogen bonds between water molecules. As a result, water wets the glass, the contact angle is acute ($\theta < 90^\circ$, for clean glass $\theta \approx 0^\circ$), so $\cos\theta > 0$. Therefore, $h$ is positive, producing a concave meniscus and capillary rise.
  • Mercury in Glass (Capillary Depression): Cohesive metallic bonds between mercury atoms are much stronger than weak adhesive forces between mercury and glass. Mercury does not wet glass; its contact angle is obtuse ($\theta \approx 138^\circ > 90^\circ$), so $\cos\theta < 0$. Therefore, $h$ is negative, producing a convex meniscus and capillary depression.
Final Answer: (a) $T_c = 303.43 \; \text{K}$, $P_c = 72.92 \; \text{atm}$, $V_c = 0.1281 \; \text{L/mol}$, $T_b = 1024.05 \; \text{K}$; (b) Explained via adhesive vs cohesive forces and contact angle in Jurin relation.

Common Misconceptions & Examiner Traps

Common Misconception

Substituting Celsius temperatures directly into Gas Law equations instead of absolute Kelvin.

Scientific Reality & Correction

Always convert Celsius temperatures to absolute Kelvin: $T(\text{K}) = t(^\circ\text{C}) + 273.15$. Volume and pressure proportionalities only hold when temperature is on the absolute thermodynamic scale.

Common Misconception

Failing to subtract aqueous tension when a gas is collected over water.

Scientific Reality & Correction

A gas collected over water is saturated with water vapour. According to Dalton law: $P_{\text{dry gas}} = P_{\text{total}} - p_{\text{aq}}$. The aqueous tension must always be deducted before applying $PV = nRT$.

Common Misconception

Directly associating effusion time with molar mass in Graham's Law without inverting.

Scientific Reality & Correction

Because rate is inversely proportional to time ($r = \frac{V}{t}$), the ratio of times for equal volumes is directly proportional to the square root of molar masses: $\frac{t_1}{t_2} = \sqrt{\frac{M_1}{M_2}}$ (or $\frac{t_2}{t_1} = \sqrt{\frac{M_2}{M_1}}$). Heavier gases take MORE time to diffuse.

Common Misconception

Confusing the formulas and relative numerical ordering of molecular speeds.

Scientific Reality & Correction

The mathematically rigorous hierarchy is $u_{\text{mp}} < u_{\text{av}} < u_{\text{rms}}$ with numerical ratio $1 : 1.128 : 1.224$. Remember the root factors: $u_{\text{mp}} = \sqrt{2RT/M} \approx 1.414$, $u_{\text{av}} = \sqrt{8RT/\pi M} \approx 1.596$, and $u_{\text{rms}} = \sqrt{3RT/M} \approx 1.732$.

Common Misconception

Assuming the compressibility factor Z is always less than 1 for all real gases.

Scientific Reality & Correction

Compressibility factor $Z$ depends on pressure and gas identity: At low pressures, attractive forces dominate ($Z < 1$); at high pressures, molecular volume repulsions dominate ($Z > 1$). For $\text{H}_2$ and $\text{He}$, attractive forces are negligible, so $Z > 1$ at all ordinary temperatures.

States of Matter: Gas Laws, Maxwell-Boltzmann Distribution, van der Waals Deviation & Liquid Transport

States of Matter: Gases and Liquids — Master Concept Architecture Gas Laws • Maxwell-Boltzmann Distribution • van der Waals Deviation (Z) • Liquid Transport Properties WBCHSE CLASS 11 1. Ideal Gas Laws & Equation of State P V T₂ T₁ (T₂ > T₁) Boyle: PV = k | P₁V₁ = P₂V₂ Charles: V ∝ T (K) Intercept = -273.15°C Boyle's Law: P ∝ 1/V (const T, n) → Hyperbolic Isotherms Charles's Law: V ∝ T (Absolute zero t = -273.15°C) Ideal Gas Equation: PV = nRT = ( dRT)/M Dalton: P = ∑pᵢ, pᵢ = xᵢP | Graha m: r ∝ 1/√M 2. Maxwell-Boltzmann Speed Distribution dN/du Speed (u) u_mp u_av u_rms T₁ (Lower T) T₂ (Higher T) Speed Ratio Ordering u_mp : u_av : u_rms = 1 : 1.13 : 1.22 u_mp = √(2RT/M) u_av = √(8RT/πM) u_rms = √(3RT/M) Temperature Influence Peak shifts right, area const E_k = 3/2 RT | 3/2 k_B T 3. Real Gases & van der Waals Deviations Z 1.0 P Ideal (Z=1) H₂ CH₄ Low P: Z < 1 High P: Z > 1 van der Waals Equation (P + an²/V²)(V - nb) = nRT a = Intermolecular attraction b = Excluded volume (4 × V_actual) Critical Point Constants T_c = 8a/27Rb | P_c = a/27b² | V_c = 3b Boyle Temp: T_b = a/(Rb) Z_c = (P_c V_c)/(R T_c) = 3/8 = 0.375 4. Liquid State: Vapour Pressure, Surface Tension & Viscosity Surface Tension (γ) Net Inward Pull Laminar Shear Flow (η) Stationary Layer (u = 0) u_max Vapour Pressure & Boiling P_vap = P_ext → Boiling Point (T_b) Surface Tension (γ) γ = F/L (N/m) | Drops are spherical Newton's Law of Viscosity F = η · A · (du/dz) [SI: Pa·s] Meniscus & Temp Effect γ, η decrease with rising Temperature

Chapter Summary & 10 Key Takeaways

Takeaway 1
Intermolecular forces encompass London dispersion forces, dipole-dipole interactions, and dipole-induced dipole interactions, all counteracted by thermal kinetic energy.
Takeaway 2
Boyle law states that at constant temperature, pressure is inversely proportional to volume (P1V1 = P2V2), yielding hyperbolic rectangular isotherms.
Takeaway 3
Charles law demonstrates that gas volume expands linearly with absolute temperature (V1/T1 = V2/T2), establishing absolute zero at minus 273.15 degrees Celsius.
Takeaway 4
The ideal gas equation PV = nRT links macroscopic state variables; gas density is directly proportional to molar mass via d = PM / (RT).
Takeaway 5
Dalton law dictates that total pressure of a non-reacting gas mixture equals the sum of partial pressures (p_i = x_i * P_total); dry gas pressure requires subtracting aqueous tension from moist gas pressure.
Takeaway 6
Graham law establishes that diffusion or effusion rates are inversely proportional to the square root of molar mass or density at constant temperature and pressure.
Takeaway 7
Kinetic Molecular Theory derives the kinetic gas equation PV = (1/3) m N (u_rms)^2, showing that absolute temperature is a direct measure of average molecular translational kinetic energy (E_k = 3/2 RT per mole).
Takeaway 8
Molecular speed hierarchy follows the fixed ratio u_mp : u_av : u_rms = 1 : 1.128 : 1.224, where u_mp = sqrt(2RT/M), u_av = sqrt(8RT/pi M), and u_rms = sqrt(3RT/M).
Takeaway 9
Real gases deviate from ideality due to finite molecular volume (co-volume b) and intermolecular attractive forces (correction a/Vm^2); compressibility factor Z is less than 1 at moderate pressures and greater than 1 at high pressures.
Takeaway 10
Liquids exhibit dynamic vapour pressure equilibrium, surface tension driven by net inward molecular cohesion causing spherical drop geometries, and internal laminar shear viscosity decreasing with temperature.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
State Boyle's Law. Why does a weather balloon expand as it ascends into the upper atmosphere?
Reveal Answer & Explanation
Answer: Boyle's Law states that at constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure ($P \propto 1/V$ or $P_1 V_1 = P_2 V_2$). As a weather balloon ascends, atmospheric pressure decreases with altitude. Because internal gas pressure must balance external pressure, the gas inside expands, increasing the volume of the balloon until it reaches bursting diameter.
2
Differentiate between the terms "diffusion" and "effusion" with appropriate examples.
Reveal Answer & Explanation
Answer: Diffusion is the spontaneous, gradual intermingling of two or more gases through random molecular motion without mechanical agitation until a uniform composition is achieved (e.g., aroma of perfume spreading across a room). Effusion is the process whereby a gas under pressure escapes through an extremely minute pinhole (orifice smaller than the mean free path of molecules) into a vacuum or region of lower pressure without molecular collisions in the orifice (e.g., slow deflation of a helium balloon through microscopic pores in latex).
3
Why is the root-mean-square speed (u_rms) of gas molecules always greater than their average speed (u_av)?
Reveal Answer & Explanation
Answer: The root-mean-square speed ($u_{\text{rms}} = \sqrt{\frac{\sum u_i^2}{N}}$) squares each velocity before averaging, thereby giving greater mathematical weight to molecules possessing higher speeds in the distribution tail. In contrast, average speed ($u_{\text{av}} = \frac{\sum u_i}{N}$) weights all speeds linearly. Quantitatively, $u_{\text{rms}} = \sqrt{\frac{3RT}{M}} = 1.732 \sqrt{\frac{RT}{M}}$, whereas $u_{\text{av}} = \sqrt{\frac{8RT}{\pi M}} = 1.596 \sqrt{\frac{RT}{M}}$, proving that $u_{\text{rms}} = 1.085 \; u_{\text{av}} > u_{\text{av}}$.
4
Define the Boyle Temperature of a real gas. What is its relationship with the van der Waals constants a and b?
Reveal Answer & Explanation
Answer: The Boyle Temperature ($T_b$) is the temperature at which a real gas behaves like an ideal gas over an appreciable range of pressure, such that the compressibility factor $Z = \frac{PV_m}{RT} \approx 1$. At this temperature, the attractive force correction ($\frac{a}{V_m}$) and molecular volume exclusion correction ($\frac{Pb}{RT}$) exactly balance each other. In terms of van der Waals constants, $T_b = \frac{a}{Rb}$.
5
Explain why the viscosity of liquids decreases with an increase in temperature, whereas the viscosity of gases increases.
Reveal Answer & Explanation
Answer: In liquids, viscosity arises from intermolecular attractive cohesive forces between adjacent layers. As temperature rises, thermal kinetic energy overcomes these cohesive bonds, facilitating freer sliding of layers and decreasing viscosity. In gases, viscosity arises from the transport of momentum across layers by colliding molecules. With rising temperature, average molecular speed ($u \propto \sqrt{T}$) and collision frequency increase, transferring momentum across layers more rapidly and increasing gas viscosity.
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