A rigid tank contains 20lbm of air at 20psia and 70F (state 1). More air is adde

Question

A rigid tank contains 20lbm of air at 20psia and 70F (state 1). More air is added to the tank until the pressure and temperature rise to 35psia and 90F, respectively (state 2). (a) Show why you can (or why you can not) assume that air behaves as an ideal gas during this process. [The critical pressure and temperature of air are 547psia and 238.5R, respectively.] (b) Determine the mass of the air in the tank after the additional air is added. (Please give all the solutions and steps, if you give the answer by picture please post a clear picture. Please don't answer the question if you can't do all the questions, THANKS)

Explanation / Answer

Ideal gas law applied here taking air is an ideal gas since 80% of air is mixed with vapour pressure.

We have the formula m = PV/RT

State 1 : m1 = P1V1/RT1

m1 : 20 lbm; P1 : 20 psia , T1 : 70oF = 529.67 oR ( OoF = 459.67 oR

m2 = P2V2/RT2

m2 :?

P2: 35 psia ; T2: 90 degree F = 549.67 oR

Change of mass delta m = m2-m1 = (P2V2/ RT2 ) - m1 = m1 ( P2T1/ P1T2) - m1

= m1 [ P2T1/ P1T2 -1]

Delta m = 20 lbm [ (35 psia * 529.67 R / 20 psia * 549.67 R) -1] = 13.73 lbm

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