help with gas perfect question ?

Question

The photosphere of the Sun is a gaseous layer about 550 km thick. The temperature in the photosphere falls from 6500 K at the bottom to 4300 Kat the top. The pressure at the bottom of the photosphere is estimated to be p0 =1.2×10^4 Pa. A model of the photosphere assumes that:

• the gas is static;

• the perfect gas law applies, with gas constant R =6.4×10^3 in SI units;

• the magnitude g of the acceleration due to the Sun’s gravity is constant, with g =274ms−2;

• if z (in metres) is the height above the bottom of the photosphere, then the temperature distribution Θ(z)(in K) is a linear function between the two temperature values given above, within the range 0 ≤ z ≤ 5.5×10^5.

(i)Find the appropriate linear function Θ(z)=a−b z, where a and b are positive constants to be evaluated

(ii) Find the corresponding pressure distribution p(z)within the photosphere.

 (iii) Hence show that according to this model, the gas pressure drops by a factor of roughly 80 between the bottom and top of the photosphere

Answer 1

This question involves quite a bit of interesting physics and some calculus.

 

A place to start is the stated value for R  ,  R =  6.4 E3  .   Most text books quote R =  8.31 .    This shows that the

person who set this problem intends R to be what is usually called the Specific value of R.   The Specific value of R is

the 8.31  divided by the molar mass (in kg) of particles.  Plugging in the numbers shows that the molar mass of the particles must be 0.0013 kg.   This is about the right value for a mixture of protons and deuterium. 

 

Next we consults some text books to find two equations for the lapse rate  (dT/dz)  ≡  Γ .

 

  (1)  Γ = - g / C_p      where g = 274  and C_p  is the specific heat at constant pressure.   An important fact is that C_p does not depend on the height above the bottom (z) .   This means that Γ is also a constant with respect to z.  In fact Γ = - .004.  Using the given facts about the temperature at the top and bottom and the difference in height, this constancy of Γ lets us write an expression for the temperature     Θ  = 6500 - .004 z     ( the units of the slope .004  are deg per meter).  This analysis answers part (i)  with a = 6500 and b = .004

 

  Using formula (1) we can work out that C_p = 68500  J/deg/kg

 

 The second text book expression for Γ is      (2)     Γ = dP/dz /( C_p ρ )    this ρ is a mass density.

 

  The ideal gas law can be written as P = ρ R T   where R is the Specific value of R mentioned above.

   Rearranging this gives  (1/ρ) =  R T / P   substituting into   equation (2) and rearranging gives

 

   (1/P) dP/dz  = - 0428/ (6500 - .004 z)   

 

This differential equation can be solved by separation of variables followed by integration of both sides.

 

The result is   ln(P(z) / P(z = 0)) = 10.7 ln[ (6500 - .004z)/ 6500]   .

 

  Exponentiation of both side provides the answer for part (ii)

 

Plugging in  z =  5.5E5  yields   P(z = 5.5E5) /P(z =0) = exp(-4.42) = 0.012.

 

Indeed   0.012 ~ 1/80    which address part (iii)