With the assumption that the air temperature is a uniform 0.0°C (as in Example 18.4), what is the density of the air at an altitude of 1.00 km as a percentage of the density at the surface?
Solution 20E We need to first find the expression for density with height. This can be derived from the expression of pressure as shown below. pM We have density = RT …..(1) RTy The variation of pressure with height is given by p = p e 0 …..(2) Here, p =0tmospheric pressure , M =molar mass of air, g =acceleration due to gravity y =height, R =universal gas constant, T =absolute temperature 0 Given, temperature T = 0 C = 273.15 K Altitude y = 1.00 km = 1000 m M = 0.029 kg/mol R = 8.314 J/mol.K g = 9.8 m/s 2 Now, substituting the value of p from equation (2) in equation (1), p 0 Mgy = RT e RT …..(3) p 0 The density of air on surface = s RT …..(4) Now, dividing equation (3) by (4), Mgy = e RT …..(5) s Now, substituting the given values in equation (5), we can calculate the density of air at the given altitude. From equation (5), 0.029×9.8×1000 = e 8.314×273.15 s = e 0.125 s = 0.125 s e 1 = s 1.13 = 0.88 s = 88% s Therefore, the density of air at the given altitude is 88% of the density at the surface.