Basic gas law reference: The molar volume occupied by one gm·mol (one mole) of an ideal gas at standard temperature and pressure (S.T.P.) is approximately equal to which of the following?

Chemical Engineering Stoichiometry Difficulty: Easy
Choose an option
  • A
    22.4 litres
  • B
    22400 litres
  • C
    22.4 c.c.
  • D
    359 litres

Answer

Correct Answer: 22.4 litres

Explanation

Introduction / Context:The molar volume at S.T.P. is a benchmark figure for gas calculations, quick estimates of moles from volumes, and stoichiometry in reaction engineering. Though different standards exist (e.g., IUPAC’s 0°C, 1 bar), many textbooks still reference 22.414 L/mol at 0°C and 1 atm.

Given Data / Assumptions:

  • Ideal-gas behaviour assumed.
  • Classical S.T.P. convention: 0°C (273.15 K) and 1 atm.

Concept / Approach:Using PV = nRT with n = 1 mol, P = 1 atm, T = 273.15 K, and R = 0.082057 L·atm·mol^-1·K^-1 gives V ≈ 22.414 L. This is typically rounded to 22.4 L for quick work.

Step-by-Step Solution:

Step 1: Write PV = nRT.Step 2: Substitute P = 1 atm, n = 1 mol, R = 0.082057 L·atm·mol^-1·K^-1, T = 273.15 K.Step 3: Compute V ≈ 22.414 L.Step 4: Round appropriately to ~22.4 L.

Verification / Alternative check:If using 1 bar instead of 1 atm, the molar volume is ~22.71 L/mol. The answer here follows the 1 atm convention consistent with many problem sets.

Why Other Options Are Wrong:

  • 22400 litres: Off by a factor of 1000 (confuses litres with millilitres).
  • 22.4 c.c.: Off by 1000 (1 L = 1000 c.c.).
  • 359 litres: Approximates 12.7 moles at STP, not one mole.

Common Pitfalls:Mixing definitions of S.T.P. or mixing units (L, mL, c.c.). Always check the stated standard before using the molar volume constant.

Final Answer:22.4 litres

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