Titan, the largest moon of Saturn, has an atmosphere that is about 94% nitrogen (N2) and 6% methane (CH4). The molecular mass of nitrogen is 4.651e-26kg and the molecular mass of methane is 2.663e-26kg. The acceleration due to gravity at the surface is about 1.35 m/s^2. For this problem, assume that Titan's atmosphere is in thermal equilibrium at 100 K. What fraction of the atmosphere is methane at altitude 2 km

9.493 e-01

Explanation:

Given that Titan's temperature at thermal equilibrium = 100 K

Determine the fraction of the atmosphere that is methane

at an altitude of 2 km

applying Ideal gas law

P = Po e^ -Z/k ( 1 )

In this question Z  = M₀ g z  ( replace -Z in  equation 1 with M₀gz )

P = Po e^- M₀gz / RT ( 2 )

where : Mo = 16 * 10^ -3 kg/mol , g = 1.35 m/s^2 , z = 2 * 10^3 m , R = 8.3 J/mol.k , T = 100 k

insert values Back to equation 2

P = Po e^-0.052

P = 0.9493 Po

hence the fraction of the atmosphere that is methane

= P / Po = 9.493 * 10^-1

= 9.493 e-01