Part 1).The piston (.8 ft2) is being supported by 20 psig of air pressure at 100
ID: 1859648 • Letter: P
Question
Part 1).The piston (.8 ft2) is being supported by 20 psig of air pressure at 1000degF. If the gas is now cooled so that the piston slowly falls to rest against the 'stops' in the figure, what will the pressure then be (psi, gauge)?
a.20
b.10
c.40
d.70
Part 2) the gas is now cooled so that the piston slowly falls to rest against the 'stops' in the figure, what will the temperature then be degF?
a.270
b.400
c.730
d.1000
Part 3)How much WORK (ft-lb) is done by the gas during cooling?
a.5.1
b.28
c.7994
d.3997
Part 4)What is the mass of air captured in the container AFTER the 'compression' (lbm)?
a. .008
b. .074
c. .13
d. .19
Explanation / Answer
Part 1.
Pressure will remain constant during cooling. So pressure will be 20 psi. Hence, option a is correct.
Part 2.
For constant pressure process, V1/T1 = V2/T2
T2 = (V2/V1)*T1
By looking at figure, V2/V1 = 0.5
Thus, T2 = 0.5*(1000+460) = 730 R = 730-460 F = 270 deg F. Option a is corrrect.
Part 3.
Work done = P*(V1 - V2)
Pressure = 20 psig = 20 + 14.7 psia = 34.7 psia (14.7 psi = atmospheric pressure)
W = 34.7*12^2 *(0.8*(2 - 1)) = 3997 ft-lb
Hence, option d is correct.
Part 4.
Mass captured = PV/RT
For air, R = 1716 ft-lb / slug-R
m = (34.7*12^2)*(0.8*1)/(1716*(270+460)) = 0.0032 slug = 0.0032*32.2 lbm = 0.103 lbm
Hence, option c is correct.