States of Matter: CBSE Class 11 Chemistry

The Unseen Glue: Intermolecular Forces

Intermolecular forces (IMFs) are the attractive or repulsive forces that exist between neighboring molecules. They are much weaker than the intramolecular forces (like covalent bonds) that hold atoms together within a molecule, but they are responsible for the physical properties of matter, especially boiling points and melting points. There are three main types:

1. London Dispersion Forces: These are the weakest IMFs and exist in all molecules, whether polar or nonpolar. They arise from temporary, instantaneous fluctuations in electron distribution around a molecule, creating a temporary dipole. This temporary dipole can then induce a dipole in a neighboring molecule, leading to a weak attraction. The strength of these forces increases with the size and surface area of the molecule.

1. Dipole-Dipole Interactions: These forces occur between polar molecules that have permanent dipoles. The positive end of one molecule is attracted to the negative end of another. They are stronger than London dispersion forces but weaker than hydrogen bonds.

1. Hydrogen Bonding: This is a special, stronger type of dipole-dipole interaction. It occurs when a hydrogen atom is bonded to a highly electronegative atom (Nitrogen, Oxygen, or Fluorine). This creates a very strong partial positive charge on the H atom, which is then strongly attracted to a lone pair of electrons on an N, O, or F atom of a neighboring molecule. This is why water (H₂O) has an unusually high boiling point.

Decoding Gas Behavior: The Fundamental Gas Laws

Boyle's Law

At constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure. Mathematically: P₁V₁ = P₂V₂ (where T and n are constant).

Charles's Law

At constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature (in Kelvin). Mathematically: V₁/T₁ = V₂/T₂ (where P and n are constant).

Gay-Lussac's Law

At constant volume, the pressure of a fixed mass of gas is directly proportional to its absolute temperature (in Kelvin). Mathematically: P₁/T₁ = P₂/T₂ (where V and n are constant).

Avogadro's Law

At constant temperature and pressure, equal volumes of all gases contain an equal number of moles (or molecules). Mathematically: V₁/n₁ = V₂/n₂ (where P and T are constant).

Putting It All Together: The Ideal Gas Equation in Action

• Example 1: Calculating Volume Calculate the volume occupied by 8.8 g of CO₂ at 31.1°C and 1 bar pressure. (R = 0.083 L bar K⁻¹ mol⁻¹) Step 1: Convert all units to match the gas constant 'R'. Moles (n): Molar mass of CO₂ = 12 + 2(16) = 44 g/mol. n = given mass / molar mass = 8.8 g / 44 g/mol = 0.2 mol. Temperature (T): The temperature must be in Kelvin. T = 31.1°C + 273.15 = 304.25 K. Pressure (P): The pressure is already in bar, which matches the units of R. P = 1 bar. Step 2: Apply the Ideal Gas Equation (PV = nRT). We need to find the volume (V), so we rearrange the equation: V = nRT / P. Step 3: Substitute the values and calculate. V = (0.2 mol) × (0.083 L bar K⁻¹ mol⁻¹) × (304.25 K) / (1 bar) V = 5.05075 L Final Answer: The volume occupied by 8.8 g of CO₂ under these conditions is approximately 5.05 L.
• Example 2: Calculating Pressure A 2 L flask contains 1.6 g of methane (CH₄) at 27°C. What is the pressure of the gas? (R = 0.0821 L atm K⁻¹ mol⁻¹) Step 1: Convert all units to match the gas constant 'R'. Moles (n): Molar mass of CH₄ = 12 + 4(1) = 16 g/mol. n = given mass / molar mass = 1.6 g / 16 g/mol = 0.1 mol. Temperature (T): Convert Celsius to Kelvin. T = 27°C + 273 = 300 K. Volume (V): The volume is already in Liters. V = 2 L. Step 2: Apply the Ideal Gas Equation (PV = nRT). We need to find the pressure (P), so we rearrange the equation: P = nRT / V. Step 3: Substitute the values and calculate. P = (0.1 mol) × (0.0821 L atm K⁻¹ mol⁻¹) × (300 K) / (2 L) P = (0.1 × 0.0821 × 150) atm * P = 1.2315 atm Final Answer: The pressure of the methane gas in the flask is approximately 1.23 atm.

Exam Trap: Ideal Gases are an Idealization

A common question in exams involves the difference between ideal and real gases. Remember, the Ideal Gas Law assumes two things that aren't true for real gases:

1. Gas particles have no volume.
2. There are no intermolecular attractive forces between gas particles.

Real gases deviate most from ideal behavior under high pressure and low temperature.

• High Pressure: The molecules are forced closer together, so their individual volume becomes significant compared to the total volume of the container. The assumption of zero particle volume fails.
• Low Temperature: The molecules move slower, giving intermolecular forces more time to act. The assumption of no attractive forces fails.

To account for this, the van der Waals equation introduces correction factors: 'a' for intermolecular attraction and 'b' for molecular volume. A gas with a higher 'a' value is more easily liquefied.

Practice Questions with Solutions

• Q: A gas occupies 300 mL at 27°C and 730 mm Hg pressure. What would be its volume at Standard Temperature and Pressure (STP: 273 K and 760 mm Hg)? A: Step 1: Identify the initial and final conditions and use the combined gas law: (P₁V₁)/T₁ = (P₂V₂)/T₂. Initial state (1): P₁ = 730 mm Hg, V₁ = 300 mL, T₁ = 27°C + 273 = 300 K. Final state (2) at STP: P₂ = 760 mm Hg, V₂ = ?, T₂ = 273 K. Step 2: Rearrange the formula to solve for V₂. V₂ = (P₁V₁T₂) / (T₁P₂) Step 3: Substitute the values and calculate. V₂ = (730 mm Hg × 300 mL × 273 K) / (300 K × 760 mm Hg) * V₂ = (730 × 273) / 760 mL = 198990 / 760 mL = 261.8 mL. Final answer: The volume at STP would be 261.8 mL.
• Q: At 25°C and 760 mm of Hg pressure a gas occupies 600 mL volume. What will be its pressure at a height where temperature is 10°C and volume of the gas is 640 mL? A: Step 1: Use the combined gas law, (P₁V₁)/T₁ = (P₂V₂)/T₂. Initial state (1): P₁ = 760 mm Hg, V₁ = 600 mL, T₁ = 25°C + 273 = 298 K. Final state (2): P₂ = ?, V₂ = 640 mL, T₂ = 10°C + 273 = 283 K. Step 2: Rearrange the formula to solve for P₂. P₂ = (P₁V₁T₂) / (T₁V₂) Step 3: Substitute the values and calculate. P₂ = (760 mm Hg × 600 mL × 283 K) / (298 K × 640 mL) * P₂ = (129168000) / (190720) mm Hg = 677.26 mm Hg. Final answer: The pressure will be approximately 677.26 mm Hg.
• Q: 38 mL of moist nitrogen gas were collected at 27°C and 746.5 mm pressure. Calculate the volume of the gas at 0°C and 760 mm pressure. (Aqueous tension at 27°C is 26.5 mm). A: Step 1: Calculate the pressure of the dry gas. The total pressure is the sum of the partial pressure of the gas and the aqueous tension (vapor pressure of water). P(total) = P(dry gas) + P(water vapour) P(dry gas) = 746.5 mm - 26.5 mm = 720 mm. Step 2: Apply the combined gas law using the pressure of the dry gas. Initial state (1): P₁ = 720 mm, V₁ = 38 mL, T₁ = 27°C + 273 = 300 K. Final state (2): P₂ = 760 mm, V₂ = ?, T₂ = 0°C + 273 = 273 K. Step 3: Solve for V₂. V₂ = (P₁V₁T₂) / (T₁P₂) V₂ = (720 mm × 38 mL × 273 K) / (300 K × 760 mm) * V₂ = 7464960 / 228000 mL = 32.74 mL. Final answer: The volume of the dry nitrogen gas at STP is 32.74 mL.
• Q: What will be the minimum pressure required to compress 500 dm³ of air at 1 bar to 200 dm³ at 30°C? A: Step 1: Identify the relevant variables. The temperature is constant (30°C), so we can use Boyle's Law: P₁V₁ = P₂V₂. Initial state (1): P₁ = 1 bar, V₁ = 500 dm³. Final state (2): P₂ = ?, V₂ = 200 dm³. Step 2: Rearrange the formula to solve for P₂. P₂ = (P₁V₁) / V₂ Step 3: Substitute the values and calculate. P₂ = (1 bar × 500 dm³) / 200 dm³ * P₂ = 2.5 bar. Final answer: The minimum pressure required is 2.5 bar.

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