What is the normal pressure of the atmosphere at the summit of Mt. Everest, 8850m above sea level? (Assume g is constant and that the density of air, ρ, is proportional to the pressure, P, at any height. Use the values at sea level ρ0 = 1.29kg/m3 and P0 = 1.013 × 105N/m2.)
I just need the answer- what is it? Too many numbers for any normal human being to understand.
To find the normal pressure of the atmosphere at the summit of Mt. Everest, we can use the barometric formula, which relates the pressure and altitude in a standard atmosphere.
The barometric formula is given by:
P = P0 * (1 - (L * h) / T0)^(g * M / (R * L))
Where:
P0 = Pressure at sea level (1.013 × 10^5 N/m^2)
L = Temperature lapse rate (0.0065 K/m)
h = Altitude above sea level (8850 m)
T0 = Standard temperature at sea level (288.15 K)
g = Acceleration due to gravity (9.81 m/s^2)
M = Molar mass of Earth's air (0.02896 kg/mol)
R = Ideal gas constant (8.314 J/(mol K))
Plugging in the given values:
P = 1.013 × 10^5 * (1 - (0.0065 * 8850) / 288.15)^(9.81 * 0.02896 / (8.314 * 0.0065))
Simplifying the equation:
P = 1.013 × 10^5 * (1 - 0.0212)^(0.00499)
P = 1.013 × 10^5 * (0.9788)^(0.00499)
P ≈ 1.013 × 10^5 * 0.9999
P ≈ 1.012 × 10^5 N/m^2
Therefore, the normal pressure of the atmosphere at the summit of Mt. Everest is approximately 1.012 × 10^5 N/m^2.
To find the normal pressure of the atmosphere at the summit of Mt. Everest, we can use the given information about the density of air, ρ, being proportional to the pressure, P, at any height. We can use the equation P = ρgh, where P is the pressure, ρ is the density, g is the gravitational acceleration (approximately 9.8 m/s²), and h is the height above sea level.
First, let's find the density of air at the summit of Mt. Everest. We know that the density at sea level is ρ0 = 1.29 kg/m³, and we can use the proportionality relation to find the density at the summit. Since the pressure decreases with height, we can use the ratio of heights to calculate the density at the summit.
The height of Mt. Everest is given as 8850 m above sea level, so the ratio of heights is (h - 0)/h = 1 - 0/h = 1. From the given information, we are told that density is proportional to pressure, which means that the ratio of densities is equal to the ratio of pressures. Therefore, the density at the summit of Mt. Everest is:
ρ = ρ0 * (h - 0)/h = 1.29 kg/m³ * (1 - 0/8850) = 1.29 kg/m³
Now that we have found the density at the summit of Mt. Everest, we can use the equation P = ρgh to find the pressure. Substituting the given values, we have:
P = ρ * g * h = 1.29 kg/m³ * 9.8 m/s² * 8850 m = 111,705 N/m²
Therefore, the normal pressure of the atmosphere at the summit of Mt. Everest, 8850 m above sea level, is approximately 111,705 N/m².
Well, to figure the proportional constant, you have to assume how thick is the atmosphere (ie, where ia density zero).
So assume a zero density at h=100km (the edge of space).
so density= m:(100km- h)+ b
and we quickly deduce that b=0 at h=100km; and
at sealevel
1.29=m*100km or m=1.29/1E5
so the density at h=8850
density=1.29E-5*8.850E3 kg/m^3
Use a similar argument to find pressure at 8850.