Question

Determine the specific weight of air when the temperature is 100∘F and the absolute pressure is 60 psi . The gas constant for the air is R = 1716 ft⋅lb/[slug⋅R]. Express your answer using three significant figures.

Answer 1

Answer:

so specific weight is 009 lb-f/ft³

Explanation:

given data

temperature = 100°F = 559.67°R

absolute pressure = 60 psi = 8640 lb/ft²

gas constant for the air R = 1716 ft⋅lb/[slug⋅R]

to find out

specific weight of air

solution

we know specific weight is express as

specific weight = $\frac{wight}{volume}$  

specific weight = $\frac{mg}{v}$              ...........1

m is mass and v is volume and g is acceleration due to gravity = 32.174 ft/s²

and by ideal gas equation

Pv = nRT

$\frac{m}{v} = \frac{P}{R*T}$

so specific weight

$\frac{mg}{v} = \frac{P*g}{R*T}$

specific weight = $\frac{8640*32.174}{1716*559.67}$

specific weight = 0.2894472 slug/ft²-s²

specific weight = 9.312677 lbm/ft²-s²

so specific weight is 009 lb-f/ft³