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Chapter 6: Gaseous State — Online MCQ Test

CHEMISTRY · CLASS 11th · Tamil Nadu State Board

Practice Chapter 6: Gaseous State with a free chapter-wise online MCQ test for Tamil Nadu State Board CLASS 11th CHEMISTRY. This chapter covers: This chapter details gas laws including Boyle Law Charles Law Avogadro Law and ideal gas equations. Students study Dalton law of partial pressures Graham law of diffusion kinetic t.... AI-generated questions from basic to board-exam level, with instant results and explanations.

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Chapter 6: Gaseous State — Important Questions & Answers (FAQ)

Frequently asked questions from Tamil Nadu State Board CLASS 11th CHEMISTRY — Chapter 6: Gaseous State, with answers and explanations. These are sample questions; the exam has its own separate question set.

According to Boyle's law, for a fixed amount of gas at constant temperature, the pressure is __________ to its volume.
  • A. Directly proportional
  • B. Inversely proportional ✓
  • C. Independent
  • D. Equal
Answer: B. Inversely proportional
Boyle's law states that P is inversely proportional to V (P ∝ 1/V) at constant temperature.
Which of the following is the correct mathematical expression for the Ideal Gas Equation?
  • A. PV = nRT ✓
  • B. PT = nRV
  • C. PV = n/RT
  • D. P/V = RT
Answer: A. PV = nRT
The ideal gas equation represents the relationship between pressure, volume, temperature, and moles of a gas as PV = nRT.
Under what conditions do real gases behave most like an ideal gas?
  • A. High pressure and high temperature
  • B. Low pressure and low temperature
  • C. High pressure and low temperature
  • D. Low pressure and high temperature ✓
Answer: D. Low pressure and high temperature
At low pressure and high temperature, the intermolecular forces are negligible and volume is significant, approaching ideality.
The compressibility factor (Z) for an ideal gas is:
  • A. Z < 1
  • B. Z > 1
  • C. Z = 1 ✓
  • D. Z = 0
Answer: C. Z = 1
For an ideal gas, PV = nRT, so the ratio PV/nRT = 1.
If the pressure of a gas is doubled and its absolute temperature is halved, the volume becomes:
  • A. Double the initial volume
  • B. Four times the initial volume
  • C. One-fourth of the initial volume ✓
  • D. Remains unchanged
Answer: C. One-fourth of the initial volume
Using V2 = V1 * (P1/P2) * (T2/T1), if P2 = 2P1 and T2 = 0.5T1, V2 = V1 * (1/2) * (1/2) = V1/4.

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