Answer:
a) P = 1240 lb/ft^2
b) P = 1040 lb/ft^2
c) P = 1270 lb/ft^2
Explanation:
Given:
- P_a = 2216.2 lb/ft^2
- β = 0.00357 R/ft
- g = 32.174 ft/s^2
- T_a = 518.7 R
- R = 1716 ft-lb / slug-R
- γ = 0.07647 lb/ft^3
- h = 14,110 ft
Find:
(a) Determine the pressure at this elevation using the standard atmosphere equation.
(b) Determine the pressure assuming the air has a constant specific weight of 0.07647 lb/ft3.
(c) Determine the pressure if the air is assumed to have a constant temperature of 59 oF.
Solution:
- The standard atmospheric equation is expressed as:
P = P_a* ( 1 - βh/T_a)^(g / R*β)
(g / R*β) = 32.174 / 1716*0.0035 = 5.252
P = 2116.2*(1 - 0.0035*14,110/518.7)^5.252
P = 1240 lb/ft^2
- The air density method which is expressed as:
P = P_a - γ*h
P = 2116.2 - 0.07647*14,110
P = 1040 lb/ft^2
- Using constant temperature ideal gas approximation:
P = P_a* e^ ( -g*h / R*T_a )
P = 2116.2* e^ ( -32.174*14110 / 1716*518.7 )
P = 1270 lb/ft^2
Answer:
B
Explanation:
Heat increase molecular motion
Because the Moon has a very small surface area compared to other spacial geo-bodies, it has cooled down much faster than Earth. Any water on the moon would freeze.
Answer: 888.45 K or 615.3 °c
Explanation:
According to Gay Lussacs law which states that at constant volume, pressure of an ideal gas is directly proportional to it's absolute temperature.
P/T = Constant
Therefore, P1/T1 = P2/T2
P1 = 6.7 atm
T1= 23°c = 273.15 + 23 = 296.15K
Since P2 is tripled, then,
P2 = 6.7 x 3= 20.1 atm
T2 = (20.1 x 296.15) ÷ 6.7
T2 = 888.45 K
Or in celcius 615.3°c
