Standard tropospheric lapse rate relation In the troposphere (the weather-dominant layer), the air temperature t at a height h (in metres) above mean sea level follows which relation, given sea-level temperature is 15°C and t is in °C?

Chemical Engineering Environmental Engineering Difficulty: Easy
Choose an option
Answer

Correct Answer: t = 15 - 0.0065h

Explanation

Introduction / Context: Weather, clouds, and most air pollution dispersion occur in the troposphere. A widely used engineering approximation for temperature variation with altitude is the standard tropospheric lapse rate. This question checks your ability to recall the canonical linear relationship between temperature and height in the lower atmosphere.

Given Data / Assumptions:

  • Layer: Troposphere (surface up to roughly 8–18 km).
  • Sea-level temperature: 15°C.
  • Height h is in metres; temperature t is in °C.
  • Use the International Standard Atmosphere (ISA) average lapse rate.

Concept / Approach: The standard environmental lapse rate for the troposphere is about 6.5°C per 1000 m (i.e., 0.0065°C per metre). Temperature decreases approximately linearly with altitude: t(h) = t0 - L*h, where t0 is sea-level temperature and L is the lapse rate.

Step-by-Step Solution:

Identify lapse rate L = 0.0065 °C/m (equivalently, 6.5 °C/km).Set sea-level temperature t0 = 15 °C.Write linear relation: t = 15 - 0.0065 h.Compare with the options and select the matching expression.

Verification / Alternative check: For h = 1000 m, t = 15 - 6.5 = 8.5 °C, which is within typical tropospheric profiles and aligns with standard atmosphere tables.

Why Other Options Are Wrong:

t = 15 + 0.0065h: Implies temperature increases with height, which contradicts tropospheric behavior.t = 0.0035h - 15 / t = 15 - 0.0035h: Use an incorrect magnitude (3.5 °C/km), not the standard 6.5 °C/km rate.

Common Pitfalls: Mixing up units (metres vs. kilometres) or confusing troposphere with stratosphere, where temperature can increase with height due to ozone absorption.

Final Answer: t = 15 - 0.0065h

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