To solve this problem we will apply the definition of the ideal gas equation, where we will clear the density variable. In turn, the specific volume is the inverse of the density, so once the first term has been completed, we will simply proceed to divide it by 1. According to the definition of 1 atmosphere, this is equivalent in the English system to
[tex]1atm = 2116lb/ft^2[/tex]
The ideal gas equation said us that,
PV = nRT
Here,
P = pressure
V = Volume
R = Gas ideal constant
T = Temperature
n = Amount of substance (at this case the mass)
Then
[tex]\frac{n}{V} = \frac{P}{RT}[/tex]
The amount of substance per volume is the density, then
[tex]\rho = \frac{P}{RT}[/tex]
Replacing with our values,
[tex]\rho = \frac{5.6*2116}{1716*850}[/tex]
[tex]\rho = 0.00812slug/ft^3[/tex]
Finally the specific volume would be
[tex]v = \frac{1}{\rho}[/tex]
[tex]v = 123ft^3/slug[/tex]