Question

Determine the percentage of mass of the atmosphere that resides between sea level and a height of 25.7 km. Assume an average pressure of 1.00 atm at sea level and a temperature of the atmosphere of 15 °C. The average molar mass of air is 28.96 g/mol. You may or may not need Earth's radius, which is 6,371 km.

Answer 1

Given

Pressure P = 1atm = 101.325 kPa = 1.01325 * 10⁵ pa or Kg/m.s2

Temperature T = 15 C = 288 K

Radius of earth = 6371 km

outer radius of atmosphere = 6371 + 25.7 km = 6396.7 km

volume of atmoshpere present between sea level and height = 4**(R³-r³)

Volume V= 4 * 3.14 * ( 6396.7³ - 6371³ ) = 39.465 * 10⁹ km³ = 39.465 * 10¹⁸ m³  

PV = nRT

1.0325 * 10⁵ Pa *  39.465 * 10¹⁸ m³ = n* 8.314 J/mol.K * 288 K

n = 16.65 * 10²⁰ moles

Mass = n * molar mass = 16.65 * 10²⁰ moles * 28.96 g/mol = 482.13  * 10²⁰ g = 4.82 * 10¹⁹ Kg Answer