Question

Using the ideal gas law, find the density of air at the following elevations. Also list it as a percentage of the density at sea level. Assume that the temperature is constant at 273K. HINT: Lookup the standard pressure at these elevations online to calculate this. (3 Points) (a.) Sea level. O ft. (b.) 5000 ft. (c.) 10,000 ft. (d.) 14,000 ft What temperature would be needed to change the density of sea level air to that of 14,000 ft.?

Answer 1

Answer:

a) p = 1.2891 kg / m^3

b) p = 1.0777 kg / m^3

c) p = 0.8898 kg / m^3

d) p = 0.7611 kg / m^3

e) T = 462.4 K

Explanation:

Given:

R-air = 0.287 KJ / kgK

T = 273 K

From Look up tables:

Sea Level: P_o = 101 KPa

5000 ft = 1524 m : P_1 = 84.44 KPa

10000 ft = 3048 m : P_2 = 69.714 KPa

14000 ft = 4267.2 m : P_2 = 59.63 KPa

Ideal Gas Law:

p = P / R*T

p: Density

P: absolute pressure

R: Gas constant

T: Absolute Temperature

part a

Sea Level: P_o = 101 KPa

p = 101 KPa / 0.287 *273

p = 1.2891 kg / m^3

part b

5000 ft = 1524 m : P_1 = 84.44 KPa

p = 84.44 KPa / 0.287 *273

p = 1.0777 kg / m^3

part c

10000 ft = 3048 m : P_2 = 69.714 KPa

p = 69.714 KPa / 0.287 *273

p = 0.8898 kg / m^3

part d

14000 ft = 4267.2 m : P_2 = 59.63 KPa

p = 59.63 KPa / 0.287 *273

p = 0.7611 kg / m^3

part e

T = P / R*p

T @ p=0.7611 kg / m^3 and Patm = 101 KPa

T = 101 KPa / 0.287*0.7611

T = 462.4 K