Pressure & Temperature in Saturns Atmosphere

[CaptainEvil]

Homework Statement

At the top of the NH₃ clouds in Saturn’s atmosphere, the temperature To = 110 K and the pressure Po is about 0.5 bar = 0.5 atm. Below this level at a radius of r_o=60268 km, the atmosphere is convective and the temperature changes with radius at a constant rate of dT/dr = -7 x 10^-4 K m^-1 (which you may find convenient to call by a name, say L, in the question below). The gravity at Saturn’s cloud tops is 1.06 times that on the surface of the Earth.

a. Write down and explicit expression for the value of the temperature as a function of
depth T(r) and the expression for the equation of hydrostatic equilibrium including
this explicit variation of T(r). Eliminate the density from the equation of hydrostatic
equilibrium using the ideal gas law and assuming that Saturn’s atmosphere is pure H2.
State any assumptions you use.

b. Integrate your equation of hydrostatic equilibrium to find an explicit expression for
the pressure P(r) as a function of depth below the NH₃ cloud tops.

c. Use the ideal gas law and your expressions for P(r) and T(r) to find an expression for
the density ρ(r) as a function of depth.

d. The separation of the two nuclei in H₂ is about 7.4 x 10_-10m , so two molecules will essentially touch when their separation is of order 10^-10 m. When the density is high enough for the average separation between molecules to be of this order, H₂ will sure be essentially liquid rather than gaseous. Imagine for counting purposes that the molecules in a liquid are arranged in a simple lattice of rows, columns and layers with all the spacing between adjacent molecules being 10^-10 m. What is the density of H₂ in this state, in kg m^-3?

e. Use your estimate from parts c) and d) to estimate the depth below the NH₃ cloud tops at which Saturn’s interior changes from gas to a molecular liquid. What is the
temperature there?

Homework Equations

The Attempt at a Solution

To = 110k
Po = 0.5 atm
r_o = 60268 km
g_s = 1.06g = 10.388 m/s²
L = -7x10^-4 K/m

a) T(r) = To + Lr

hydrostatic quilibrium = dP/dr = -ρg

density ρ can be eliminated via I.G.L => ρ = $$\mu$$P/RT(r) where $$\mu$$ is the molecular weight, R is the gas constant and P pressure.

dP/dR = -$$\mu$$Pg/RT(r)

I can then separate these and integrate to solve for Pressure, but I am having a lot of trouble.

Could someone help me with the integration (if I'm even on the right track) and part d (where I don't even know where to start)

Thanks.

P.S. my teacher gave me a hint for the integration that I should get something like dP/dr = P*f(r) and so I will have to integrate 1/P dP = f(r) dr

Thanks

 

[Jhero]

for your question! Let's break it down step by step.

a) To find the temperature as a function of depth, we can use the given information that the temperature changes at a constant rate of dT/dr = L. This means that the temperature at any depth r can be expressed as T(r) = To + Lr.

Now, let's look at the equation of hydrostatic equilibrium. This equation states that the change in pressure with depth is equal to the product of the density and the acceleration due to gravity. In this case, we can write it as dP/dr = -ρgs, where gs is the gravity at Saturn's cloud tops.

Next, we can eliminate the density from this equation by using the ideal gas law, which states that PV = nRT. We can rearrange this to solve for density, ρ = n/V = (molecular weight of gas)P/RT. Substituting this into the equation of hydrostatic equilibrium, we get dP/dr = -\muPg/RT(r).

Now, we can separate the variables and integrate to solve for P(r). We can write this as 1/P dP = -\muPg/RT(r) dr. Integrating both sides, we get ln(P) = -\muPg/(R L) ln(T(r)) + C, where C is a constant of integration. Solving for P, we get P(r) = P0(T(r)/T0)^{-\muPg/(R L)}, where P0 and T0 are the initial pressure and temperature at the top of the NH3 clouds.

b) To find the pressure as a function of depth, we can substitute our expression for T(r) into the equation we found in part a, P(r) = P0(T(r)/T0)^{-\muPg/(R L)}. This gives us P(r) = P0(To + Lr)^{-\muPg/(R L)}.

c) To find the density as a function of depth, we can use the ideal gas law again, ρ = (molecular weight of gas)P/RT(r). Substituting our expression for P(r) from part b, we get ρ(r) = \muP0(To + Lr)^{-\muPg/(R L)}/RT(r).

d) To find the density of H2 in its liquid state,