Definition: The Gaseous State is the simplest state of matter characterized by the complete absence of definite shape and volume. Constituent molecules possess high kinetic energy and are separated by vast intermolecular distances.
In competitive examinations like JEE and NEET, mastery of this chapter requires a firm grasp of empirical gas laws, the molecular kinetic model, and deviations from ideal behavior under extreme thermodynamic conditions.
Empirical Gas Laws and Temperature Scales
The behavior of gases is fundamentally described through relationships relating pressure, volume, temperature, and moles. Temperature scales serve as the baseline for these calculations, anchored by absolute zero where molecular motion theoretically ceases. The absolute thermodynamic scale is the Kelvin scale, linked to the Celsius scale by the conversion equation: \(T(K) = ^\circ C + 273.15\).
When dealing with unknown or custom scales, the standard linear conversion formula \(\frac{R – R_0}{R_{100} – R_0} = \frac{C – 0}{100 – 0}\) allows aspirants to easily cross-calibrate readings. Empirical gas laws form the cornerstone of numerical problem-solving.
Boyle’s Law establishes that at constant temperature and mass, the volume of a given mass of gas is inversely proportional to its pressure: \(P_1V_1 = P_2V_2\). This principle is heavily applied in barometric calculations and underwater bubble expansion problems.
Charles’s Law dictates that at constant pressure, volume is directly proportional to absolute temperature: \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\). Similarly, Gay-Lussac’s Law states that pressure varies directly with absolute temperature at constant volume: \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\).
Combining these relationships yields the unified Ideal Gas Equation, expressed as \(PV = nRT\). Here, \(R\) is the universal gas constant (approximately \(8.314 \text{ J K}^{-1} \text{mol}^{-1}\) or \(0.0821 \text{ L atm K}^{-1} \text{mol}^{-1}\)). Aspirants frequently manipulate this equation to solve for gas density (\(\rho\)) and molar mass (\(M\)) using the vital derivative: \(PM = dRT\).
Recognizing how density changes inversely with temperature at constant pressure is a classic trap in competitive physical chemistry problems.
Dalton’s Law and Graham’s Law of Diffusion
When non-reacting gas mixtures occupy a vessel, calculating individual contributions becomes essential. Dalton’s Law of Partial Pressures states that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of individual gases.
The partial pressure of any component gas is directly related to its mole fraction: \(P_i = x_i \cdot P_{total}\), where \(x_i = \frac{n_i}{n_{total}}\). A common application tested in exams is finding the pressure of a gas collected over water, which requires subtracting the aqueous tension (vapor pressure of water at that specific temperature) from the total observed barometric pressure.
Gaseous mixing and movement are further governed by Graham’s Law of Diffusion and Effusion. Diffusion is the spontaneous intermingling of gases, while effusion is the escape of gas molecules through a tiny pinhole into an evacuated chamber.
Graham’s law establishes that the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molar mass or density under identical conditions of temperature and pressure:
\(\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} = \sqrt{\frac{d_2}{d_1}} = \sqrt{\frac{Vapour \ Density_2}{Vapour \ Density_1}}\)
Exam questions frequently apply this law to determine the molecular weight of an unknown gas by comparing its effusion rate against a standard reference gas like hydrogen or helium. Alternatively, it is used to calculate the exact point where two diffusing vapors meet inside a glass tube to form a solid white ring (e.g., \(\text{NH}_3\) and \(\text{HCl}\)).
Kinetic Theory of Gases and Molecular Speeds
The Kinetic Theory of Gases (KTG) bridges microscopic particle behavior with macroscopic thermodynamic properties. It assumes that gas molecules are point masses with negligible actual volume compared to the total container volume, undergo completely elastic collisions, and exert no intermolecular forces of attraction or repulsion.
The fundamental kinetic gas equation derived from these postulates is \(PV = \frac{1}{mN\overline{u^2}}\), which successfully explains macroscopic pressure as the cumulative momentum transfer from molecular wall collisions.
From KTG, the average kinetic energy for one mole of an ideal gas is derived as \(KE = \frac{3}{2}RT\). This demonstrates that absolute temperature is directly proportional to the average translational kinetic energy, completely independent of the gas’s chemical nature.
To analyze the distribution of molecular velocities within a gas sample, physical chemistry relies on three distinct statistical speed parameters:
- Root Mean Square Speed (\(U_{rms}\)): \(\sqrt{\frac{3RT}{M}}\) — Represents the square root of the mean of squared speeds.
- Average Speed (\(U_{av}\)): \(\sqrt{\frac{8RT}{\pi M}}\) — Represents the arithmetic mean of the speeds of all molecules.
- Most Probable Speed (\(U_{mps}\)): \(\sqrt{\frac{2RT}{M}}\) — Represents the speed possessed by the maximum fraction of molecules at a given temperature.
Students must memorize the strict mathematical ratio for these speeds: \(U_{mps} : U_{av} : U_{rms} = \sqrt{2} : \sqrt{\frac{8}{\pi}} : \sqrt{3} \approx 1 : 1.128 : 1.224\). Molar mass must always be substituted in kilograms per mole (\(\text{kg/mol}\)) when computing absolute speed values in SI units.
Real Gases and van der Waals’ Equation
Real gases deviate significantly from ideal behavior under conditions of high pressure and low temperature because two core assumptions of the kinetic theory fail: real molecules do possess a measurable volume, and intermolecular attractive forces do exist.
To account for these deviations, Johannes van der Waals introduced correction terms to the ideal gas equation, formulating the famous van der Waals’ Equation for \(n\) moles of a real gas:
\(\left(P + \frac{an^2}{V^2}\right)(V – nb) = nRT\)
In this equation, the pressure correction term \(\frac{an^2}{V^2}\) compensates for intermolecular attractions that pull molecules inward, reducing the effective collision force against container walls. The constant \(a\) reflects the magnitude of intermolecular forces, where a higher \(a\) means easier liquefaction.
The volume correction term \((V – nb)\) subtracts the excluded volume, accounting for the actual physical space occupied by the gas molecules themselves. Here, \(b\) represents the co-volume, which is four times the actual molar volume of the spherical molecules.
At very low pressures, volume is large, making the volume correction term \(nb\) negligible, and the gas behaves nearly ideally. At intermediate pressures, attractive forces dominate, making the compressibility factor \(Z = \frac{PV}{nRT} < 1\). At very high pressures, molecular repulsion dominates due to the incompressible nature of the particles, causing \(Z > 1\).
The temperature at which a real gas exhibits ideal behavior over a wide range of pressure is designated as the Boyle Temperature (\(T_B\)), mathematically expressed as \(T_B = \frac{a}{Rb}\).
Important Facts / Formulas
| Constant / Parameter | Symbol / Expression | Key Significance |
|---|---|---|
| Compressibility Factor | \(Z = \frac{PV}{nRT}\) | Measures deviation from ideality; \(Z = 1\) for ideal gases. |
| Critical Temperature | \(T_c = \frac{8a}{27Rb}\) | Highest temperature at which a gas can be liquefied. |
| Critical Pressure | \(P_c = \frac{a}{27b^2}\) | Minimum pressure required to liquefy a gas at \(T_c\). |
| Critical Volume | \(V_c = 3b\) | Molar volume of a gas at the critical point. |
| Boyle Temperature | \(T_B = \frac{a}{Rb}\) | Temperature where real gas obeys ideal gas law over low-to-moderate pressure ranges. |
Previous Year Question Hints
- Question Type 1 (Graham’s Law): You may be asked to find the time taken for equal volumes of two different gases to effuse through an orifice. Remember that \(\frac{t_1}{t_2} = \frac{r_2}{r_1} = \sqrt{\frac{M_1}{M_2}}\). Always double-check whether vapor density or molecular weight is provided in the problem statement.
- Question Type 2 (Compressibility & van der Waals ‘a’ and ‘b’): JEE examiners frequently test the physical units of van der Waals constants. Dimensional analysis reveals that the unit of \(a\) is \(\text{atm L}^2 \text{ mol}^{-2}\) (reflecting pressure correction) and the unit of \(b\) is \(\text{L mol}^{-1}\) (reflecting volume exclusion).
- Question Type 3 (Partial Pressure Calculations): When a gas is collected over water, aspirants must remember to subtract aqueous tension from the total measured pressure to find the dry gas pressure before applying \(PV = nRT\). Neglecting this subtraction is the most common student error.
Quick Revision Summary
- Temperature conversions rely on absolute Kelvin scales where \(T(K) = ^\circ C + 273.15\).
- Boyle’s law (\(P_1V_1 = P_2V_2\)) applies at constant temperature; Charles’s law governs isobaric conditions.
- The ideal gas equation \(PV = nRT\) integrates all empirical laws; density and molar mass relate via \(PM = dRT\).
- Dalton’s law dictates that total pressure is the sum of partial pressures, where partial pressure equals mole fraction times total pressure.
- Graham’s law states that rates of effusion and diffusion are inversely proportional to the square root of molar masses or densities.
- Root mean square speed is given by \(U_{rms} = \sqrt{\frac{3RT}{M}}\), requiring molar mass in SI units (\(\text{kg/mol}\)).
- Real gases deviate from ideality due to intermolecular attractions (\(a\)) and molecular volume (\(b\)).
- The van der Waals equation incorporates corrections: \(\left(P + \frac{an^2}{V^2}\right)(V – nb) = nRT\).
- Compressibility factor \(Z = \frac{PV}{nRT}\) drops below 1 when attractive forces dominate and rises above 1 when repulsive forces dominate.
- Critical constants define the distinct phase transition boundaries where liquid and vapor states merge into a supercritical fluid.
