How to Use This Page
This page provides complete exercise solutions for Chapter 6 — Gaseous State of the Samacheer Kalvi Class 11 Chemistry textbook. Work through each section methodically:
1. Key Concepts — quick revision of formulas and definitions before attempting questions. 2. MCQs through Match the Following — test your recall and understanding. 3. Brief and Detailed Answers — practice writing structured responses for the board exam. 4. Worked Numericals — follow each step carefully; the method matters more than the final answer. 5. Common Mistakes & Exam Checklist — review these before your exam to avoid losing marks.
Key Concepts (Quick Recap)
| Concept | Formula / Statement |
|---|
| Boyle's Law | At constant T, P₁V₁ = P₂V₂ (V is proportional to 1/P) |
| Charles's Law | At constant P, V₁/T₁ = V₂/T₂ (V is proportional to T) |
| Gay-Lussac's Law | At constant V, P₁/T₁ = P₂/T₂ (P is proportional to T) |
| Avogadro's Law | At same T and P, equal volumes contain equal moles (V is proportional to n) |
| Ideal Gas Equation | PV = nRT; R = 0.0821 L·atm mol⁻¹ K⁻¹ = 8.314 J mol⁻¹ K⁻¹ |
| Dalton's Law | P_total = P₁ + P₂ + P₃ + ... ; P_i = x_i times P_total |
| Graham's Law | r₁/r₂ = square root of (M₂/M₁) ; rate of effusion is proportional to 1/square root of M |
| Kinetic Molecular Theory | KE = (3/2)nRT per mole; v_rms = square root of (3RT/M) |
| Van der Waals Equation | (P + an²/V²)(V minus nb) = nRT |
| Compressibility Factor | Z = PV/nRT; Z = 1 for ideal gas |
Important Values:
R = 0.0821 L·atm mol⁻¹ K⁻¹
R = 8.314 J mol⁻¹ K⁻¹
R = 1.987 cal mol⁻¹ K⁻¹
Standard temperature = 273.15 K (0 °C)
Standard pressure = 1 atm = 760 mm Hg = 101.325 kPa
Molar volume at STP = 22.4 L
I. Choose the Best Answer (MCQs)
1. Which of the following graphs correctly represents Boyle's law?
Answer
(a) P vs V — straight line through origin (b) P vs 1/V — straight line through origin (c) P vs V — parabola (d) P vs V — hyperbola at constant T
Answer(b) Boyle's law states PV = constant at fixed T and n. Rearranging, P = constant/V, so P is directly proportional to 1/V. A plot of P vs 1/V gives a straight line through the origin.
2. A gas occupies 2.5 L at 1.2 atm. What volume will it occupy at 3.6 atm at the same temperature?
Answer
(a) 0.83 L (b) 1.0 L (c) 7.5 L (d) 0.5 L
Answer(a)
Given: P₁ = 1.2 atm, V₁ = 2.5 L, P₂ = 3.6 atm
By Boyle's law: P₁V₁ = P₂V₂
V₂ = P₁V₁ / P₂ = (1.2 × 2.5) / 3.6 = 3.0 / 3.6 = 0.833 L ≈ 0.83 L
3. According to Graham's law, the rate of diffusion of O₂ (M = 32) compared to H₂ (M = 2) is:
Answer
(a) 4 times faster (b) 4 times slower (c) 16 times faster (d) 16 times slower
Answer(b)
r(O₂)/r(H₂) = √(M(H₂)/M(O₂)) = √(2/32) = √(1/16) = 1/4
O₂ diffuses 4 times slower than H₂.
4. The compressibility factor Z for a real gas at high pressure is:
Answer
(a) Z < 1 (b) Z = 1 (c) Z > 1 (d) Z = 0
Answer(c) At high pressure, the finite volume of gas molecules becomes significant. The excluded volume makes V_real > V_ideal, so Z = PV/nRT > 1. At low pressure, intermolecular attractions dominate and Z < 1.
5. The van der Waals constant 'a' accounts for:
Answer
(a) Finite molecular volume (b) Intermolecular forces of attraction (c) Kinetic energy of molecules (d) Number of molecules
Answer(b) The constant 'a' corrects for the attractive intermolecular forces that reduce the effective pressure. The constant 'b' corrects for the finite volume occupied by gas molecules.
6. At what temperature will the RMS velocity of SO₂ (M = 64) be the same as that of O₂ (M = 32) at 300 K?
Answer
(a) 300 K (b) 600 K (c) 150 K (d) 1200 K
Answer(b)
v_rms = √(3RT/M)
For RMS velocities to be equal:
√(3RT₁/M₁) = √(3RT₂/M₂)
T₁/M₁ = T₂/M₂
T(SO₂)/64 = 300/32
T(SO₂) = (300 × 64)/32 = 600 K
7. According to Dalton's law, if a gas mixture contains 2 moles of N₂ and 3 moles of O₂ at 10 atm total pressure, the partial pressure of N₂ is:
Answer
(a) 6 atm (b) 4 atm (c) 2 atm (d) 5 atm
Answer(b)
Mole fraction of N₂ = n(N₂)/n(total) = 2/5
P(N₂) = x(N₂) × P_total = (2/5) × 10 = 4 atm
II. Fill in the Blanks
1. At constant temperature, the product of pressure and volume of a fixed amount of gas is constant. 2. The value of gas constant R in L·atm mol⁻¹ K⁻¹ is 0.0821. 3. According to Charles's law, V₁/T₁ = V₂/T₂. 4. The volume of one mole of any gas at STP is 22.4 litres. 5. Graham's law of diffusion states that the rate of diffusion is inversely proportional to the square root of its molar mass. 6. The van der Waals constant 'b' accounts for the finite volume of gas molecules. 7. The average kinetic energy of gas molecules is directly proportional to absolute temperature. 8. If Z > 1, the real gas shows positive deviation from ideal behaviour. 9. The total pressure of a gas mixture is equal to the sum of partial pressures of individual gases. 10. The RMS velocity of a gas molecule is given by √(3RT/M). 11. At low temperature and high pressure, real gases deviate most from ideal behaviour. 12. The SI unit of pressure is pascal (Pa). 13. Boyle's law is valid only for ideal gases under isothermal conditions. 14. The ratio of rates of diffusion of two gases is equal to the inverse ratio of the square roots of their molar masses. 15. The liquefaction of a gas becomes easier when the temperature is below its critical temperature.
III. True or False (with Corrections)
1. Boyle's law states that at constant volume, pressure is proportional to temperature.
False. Boyle's law states that at constant temperature, volume is inversely proportional to pressure. The statement describes Gay-Lussac's law, not Boyle's law.
2. An ideal gas obeys PV = nRT at all temperatures and pressures.
True. By definition, an ideal gas follows the ideal gas equation at all conditions. Real gases approximate this behaviour only at low pressure and high temperature.
3. The average kinetic energy of a gas depends on the nature of the gas.
False. The average kinetic energy per mole of any gas equals (3/2)RT, which depends only on temperature, not on the nature (molar mass or identity) of the gas.
4. The van der Waals constant 'a' is larger for gases that are more easily liquefied.
True. Gases with stronger intermolecular forces (which make them easier to liquefy) have larger 'a' values because 'a' measures the magnitude of attractive forces.
5. The rate of effusion of a gas is directly proportional to the square root of its molar mass.
False. The rate of effusion is inversely proportional to the square root of the molar mass. Graham's law: r is proportional to 1/√M.
IV. Match the Following
| Column A | Column B |
|---|
| 1. Boyle's law | (a) V is proportional to T at constant P |
| 2. Charles's law | (b) PV = nRT |
| 3. Dalton's law | (c) V is proportional to 1/P at constant T |
| 4. Graham's law | (d) P_total = P₁ + P₂ + P₃ + ... |
| 5. Ideal gas equation | (e) r is proportional to 1/√M |
| 6. Gay-Lussac's law | (f) P is proportional to T at constant V |
Answer
1 → (c) Boyle's law: V is proportional to 1/P at constant T
2 → (a) Charles's law: V is proportional to T at constant P
3 → (d) Dalton's law: P_total = P₁ + P₂ + P₃ + ...
4 → (e) Graham's law: r is proportional to 1/√M
5 → (b) Ideal gas equation: PV = nRT
6 → (f) Gay-Lussac's law: P is proportional to T at constant V
V. Answer Briefly (2-mark)
1. State Boyle's law. Give its mathematical expression.
Answer
Boyle's law states that at constant temperature, the volume of a fixed amount of gas is inversely proportional to its pressure.
V is proportional to 1/P (at constant T and n)
PV = constant
P₁V₁ = P₂V₂
2. What is the significance of the van der Waals constants 'a' and 'b'?
Answer
The van der Waals constant 'a' measures the magnitude of intermolecular attractive forces. Gases with stronger attractions have larger 'a' values and liquefy more easily. The constant 'b' represents the excluded volume per mole of gas molecules — the volume that is unavailable due to the finite size of the molecules themselves.
3. Define compressibility factor Z. What is its value for an ideal gas?
Answer
The compressibility factor is defined as Z = PV/(nRT). It measures how much a real gas deviates from ideal behaviour. For an ideal gas, Z = 1 at all conditions. When Z > 1, the gas shows positive deviation (repulsive forces dominate); when Z < 1, it shows negative deviation (attractive forces dominate).
4. State Dalton's law of partial pressures. Write its mathematical expression.
Answer
Dalton's law states that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of the individual gases.
P_total = P₁ + P₂ + P₃ + ...
where Pᵢ = xᵢ × P_total
xᵢ = mole fraction of gas i = nᵢ / n_total
VI. Answer in Detail (5-mark)
1. Derive the ideal gas equation from the gas laws. Explain the significance of R and calculate the volume of 2 moles of gas at 300 K and 5 atm.
Answer
The ideal gas equation is derived by combining the three empirical gas laws:
Boyle's law: V is proportional to 1/P (at constant T and n)
Charles's law: V is proportional to T (at constant P and n)
Avogadro's law: V is proportional to n (at constant P and T)
Combining:
V is proportional to nT/P
V = R × nT/P (where R is the proportionality constant)
PV = nRT
Where:
P = pressure (atm)
V = volume (L)
n = number of moles
R = gas constant = 0.0821 L·atm mol⁻¹ K⁻¹
T = temperature (K)
Significance of R: R (universal gas constant) relates the energy scale to the temperature scale for one mole of an ideal gas. Its value depends on the units of P and V. R = PV/nT = 8.314 J mol⁻¹ K⁻¹ in SI units.
Numerical example:
Given: n = 2 mol, T = 300 K, P = 5 atm
PV = nRT
V = nRT/P
V = (2 × 0.0821 × 300) / 5
V = 49.26 / 5
V = 9.852 L
The gas occupies 9.85 L under these conditions.
2. State and explain Graham's law of diffusion. A gas X diffuses 3 times faster than gas Y. If the molar mass of Y is 36 g/mol, find the molar mass of X.
Answer
Graham's law states that at constant temperature and pressure, the rate of diffusion (or effusion) of a gas is inversely proportional to the square root of its molar mass.
r is proportional to 1/√M
r₁/r₂ = √(M₂/M₁)
Where:
r₁, r₂ = rates of diffusion of gases 1 and 2
M₁, M₂ = molar masses of gases 1 and 2
This law has important applications:
- Separating isotopes (e.g., enrichment of U-235 using UF₆)
- Comparing molecular masses of unknown gases
- Explaining why lighter gases (H₂, He) diffuse faster than heavier ones (CO₂, SO₂)
Numerical example:
Given: r(X)/r(Y) = 3, M(Y) = 36 g/mol
By Graham's law:
r(X)/r(Y) = √(M(Y)/M(X))
3 = √(36/M(X))
Squaring both sides:
9 = 36/M(X)
M(X) = 36/9 = 4 g/mol
The molar mass of gas X is 4 g/mol (this is helium).
3. Explain deviation of real gases from ideal behaviour. Discuss the van der Waals equation with a worked example.
Answer
Real gases deviate from ideal behaviour because of two assumptions that are not valid under all conditions:
1. Intermolecular forces of attraction: Ideal gas molecules are assumed to have no attractive forces. Real gas molecules attract each other, reducing the effective pressure. This causes Z < 1 at moderate pressures.
2. Finite molecular volume: Ideal gas molecules are assumed to have zero volume. Real gas molecules occupy space, reducing the available volume. This causes Z > 1 at very high pressures.
At low temperature and high pressure, deviations are most pronounced because molecules are closer together, making intermolecular forces significant.
The van der Waals equation corrects for both:
(P + an²/V²)(V − nb) = nRT
Where:
an²/V² = correction for intermolecular attraction
(increases effective pressure needed)
nb = correction for finite molecular volume
(reduces the available volume)
For 1 mole: (P + a/V²)(V − b) = RT
Numerical example: Calculate the pressure exerted by 2 moles of CO₂ in a 5 L vessel at 300 K using the van der Waals equation. Given a = 3.59 L²·atm/mol², b = 0.0427 L/mol.
Given: n = 2 mol, V = 5 L, T = 300 K
a = 3.59 L²·atm/mol², b = 0.0427 L/mol
Van der Waals equation:
(P + an²/V²)(V − nb) = nRT
Step 1: Calculate an²/V²
an²/V² = (3.59 × 4) / 25 = 14.36 / 25 = 0.5744 atm
Step 2: Calculate (V − nb)
V − nb = 5 − (2 × 0.0427) = 5 − 0.0854 = 4.9146 L
Step 3: Calculate nRT
nRT = 2 × 0.0821 × 300 = 49.26 L·atm
Step 4: Solve for P
(P + 0.5744) × 4.9146 = 49.26
P + 0.5744 = 49.26 / 4.9146 = 10.023
P = 10.023 − 0.5744 = 9.45 atm
Ideal gas pressure would be:
P_ideal = nRT/V = 49.26/5 = 9.85 atm
The real gas pressure (9.45 atm) is less than ideal (9.85 atm)
because intermolecular attractions reduce the effective pressure.
Common Mistakes to Avoid
1. Forgetting to convert temperature to Kelvin — always convert °C to K using T(K) = T(°C) + 273.15 before using any gas law formula.
2. Using inconsistent units — ensure P is in atm, V in litres, and T in Kelvin when using R = 0.0821. Convert mmHg to atm (divide by 760) before calculation.
3. Confusing Boyle's law with Charles's law — Boyle's law relates P and V (at constant T); Charles's law relates V and T (at constant P). Mix them up and your entire calculation is wrong.
4. Ignoring the square root in Graham's law — the rate ratio equals the square root of the inverse mass ratio, not the mass ratio itself. Always write: r₁/r₂ = √(M₂/M₁).
5. Mixing up 'a' and 'b' in van der Waals equation — 'a' corrects for intermolecular attraction (appears in the pressure term), 'b' corrects for molecular volume (appears in the volume term).
6. Using n instead of n² in the pressure correction — the pressure correction term is an²/V², not a/V² (for n moles). Remember to square the number of moles.
7. Confusing RMS velocity with average velocity — v_rms = √(3RT/M), v_avg = √(8RT/πM), v_mp = √(2RT/M). These are different and cannot be used interchangeably.
8. Not applying Dalton's law correctly — partial pressure Pᵢ = xᵢ × P_total, where xᵢ is the mole fraction, not the mass fraction.
Exam Checklist
- [ ] Can you state Boyle's, Charles's, and Gay-Lussac's laws with their mathematical expressions?
- [ ] Can you derive the ideal gas equation PV = nRT from the gas laws?
- [ ] Do you know the value of R in different units (0.0821 L·atm, 8.314 J, 1.987 cal)?
- [ ] Can you convert between atm, mmHg, kPa, and bar for pressure?
- [ ] Can you calculate partial pressure using mole fraction (Dalton's law)?
- [ ] Can you apply Graham's law to find molar mass from diffusion rates?
- [ ] Can you calculate v_rms, v_avg, and v_mp and distinguish between them?
- [ ] Do you understand why real gases deviate from ideal behaviour (two reasons)?
- [ ] Can you write and apply the van der Waals equation to solve numericals?
- [ ] Can you interpret the compressibility factor (Z > 1, Z < 1, Z = 1)?
