Question

On a cold, 5°F day, the nosewheel tire of a Learjet, having an internal volume of 1000. in3 is inflated to 30 psi (i.e., 30 lb/in2) above the ambient pressure of 14.7 psi. Calculate the weight (in pounds) and specific volume of the air in the tire. (Be careful of units here--we usually use feet, not inches, with imperial units in this course, and you need to work with absolute temperatures.)

Answer 1

The weight in (pounds ) is
`m = 0.14994 \ lb`

The specific volume is
`V_s = 0.2403 \ m^3 /kg`

Explanation:

From the question we are told that

The temperature is
`T = 5^o F = (5^oF - 32) * (5)/(9) = -15 ^oC + 273 = 258 K`

The volume is
`V = 1000 \ in^3 = (1000)/(1728) = 0.5787 ft^3 = 0.0164\ m^3`

The initial absolute pressure is
`P = 30\ lb/in^2 = 30\ lb/in^2 * (1)/( (in^2)/(144ft^2) ) = 4320 \ lb/ft^2 = 308 .19 KPa`

Generally from ideal gas equation we have

`P * V = m RT`

Here m is the weight nose wheel tire in pounds

R is the gas constant of air with value
`R = 0.287 \ (KJ)/(kg\cdot K)`

So

`308 .19 * 0.0164 = m * 0.287* 258`

=>
`m = 0.068 \ kg`

Converting to pounds

`m = 0.068 * 2.205`

`m = 0.14994 \ lb`

Form this equation
`P * V = m RT` specific volume is

`(V)/(m) = (RT)/( P)`

=>
`V_s = (V)/(m) =  (0.287 * 258 )/(308 .19 )`

=>
`V_s = 0.2403 \ m^3 /kg`