Answer:
9.493 e-01
Step-by-step 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