At a particular sea level location at a particular time the atmospheric pressure is 14.8 lbs/in^2 and the temperature is 80 degrees F. Estimate the temperature, pressure, and air density at an altitude of 8,000 ft.

30.52 °F

1572 lbs/ft²

0.0019 slug/ft³

Explanation:

Given that

Pressure at sea level, P₁ = 14.8 PSI

Temperature, T₁ = 80 °F

At 8,000 ft (m) altitude

for altitude h, less than 36000 ft, the model used to modify temperature and pressure is written below

T = 59 - 0.00356h, where h is the altitude

T = 59 - 0.00356(8,000)

T = 59 - 28.48

T = 30.52 °F

Next, we find the pressure which is calculated using the formula

P = 2116 [(T + 459.7)/518.6]^5.256

P = 2116 [(30.52 + 459.7) / 518.6]^5.256

P = 2116 [490.22/518.6]^5.256

P = 2116 (0.945)^5.256

P = 2116 * 0.743

P = 1572 lbs/ft²

The density is gotten using the formula

ρ = P / [1718 (T + 459.7)]

ρ = 1572 / [1718 (30.52 + 459.7)]

ρ = 1572 / (1718 * 490.22)

ρ = 1572 / 842198

ρ = 0.0019 slug/ft³

hello, i am papaguy.

i have found an answer to your question but use it as information.

information: because weak acids/bases have low concentration of hydrogen ions, the addition of water has a large impact on the ph. 2. a difference in 1 unit is a 10x difference in concentration for example; a liquid with ph of 3 is 10x more acidic than a liquid with a ph of 4. therefore, a liquid with a ph of 3 is stronger.

best of luck!

c. newton's third law of motion