Difficulty: Easy
Correct Answer: Density of dry air at N.T.P. is 1 g per litre.
Explanation:
Introduction / Context:
Gas property facts at reference conditions are frequent quick checks in chemical engineering calculations. N.T.P. is commonly taken as 0°C and 1 atm. Ideal gases at N.T.P. occupy a well known molar volume, and conversions between gram moles, kilogram moles, and pound moles are standard. The density of real air at N.T.P., however, is not exactly 1 g per litre. This item tests attention to canonical values and unit consistency.
Given Data / Assumptions:
Concept / Approach:
Use ideal gas molar volume to validate statements about gram moles, kilogram moles, and pound moles. Then compare the quoted dry air density to the accepted value. The statements giving 22.4 litres per gram mole and 22.4 m^3 per kilogram mole are consistent. For 1 lb-mol, the corresponding volume at 0°C is approximately 359 ft^3 (as the 379.5 ft^3 value corresponds to 60°F, not 32°F). The given air density of 1 g L^-1 is too low compared with the standard 1.293 g L^-1; therefore it is the incorrect statement.
Step-by-Step Solution:
Check molar volume: 22.414 L mol^-1 at N.T.P.Convert to kg-mol: 22.414 m^3 per kmol ≈ 22.4 m^3 per kg-mol.Convert to lb-mol at 0°C: ≈ 359 ft^3 per lb-mol (consistent).Compare dry air density: expected ≈ 1.293 g L^-1, not 1 g L^-1.
Verification / Alternative check:
Using PV = nRT at 273.15 K and 1 atm reproduces 22.414 L mol^-1. Standard air composition yields a molar mass ~28.97 g mol^-1, giving density 28.97 g per 22.414 L ≈ 1.293 g L^-1.
Why Other Options Are Wrong:
They are correct statements at N.T.P. and match accepted reference values for ideal gases.
Common Pitfalls:
Mixing N.T.P. with S.T.P. or 60°F base conditions; rounding aggressively; using 1 g L^-1 as a rough mnemonic instead of the accurate 1.293 g L^-1.
Final Answer:
Density of dry air at N.T.P. is 1 g per litre.
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