3-G) SHORT ANSWER: No GOAL write-up needed. But clearly show SKETCH, PROPERTY TA

Question

3-G) SHORT ANSWER: No GOAL write-up needed.
But clearly show SKETCH, PROPERTY TABLE, MASS & ENERGY BALANCE EQUATIONS FOR EACH.
a) A rigid tank containing 20 kg of NITROGEN GAS (N2) at 1.0 bars and -26.43 oC is heated until the
temperature reaches 200 oC. The final pressure and volume of the N2 are _______ kPa &_______ m3.
b) A rigid tank with a volume of 5.0 m3 contains ARGON GAS at -16 oC. The ARGON GAS absorbs
2500 kJ of external heat transfer causing the temperature to reach 160 oC. The total mass of the
ARGON is ____________ kg. The initial and final pressures are _________ kPa and ________ kPa.
c) A rigid tank containing 20 kg of AIR at 6.0 bars and 140 oC is cooled causing the pressure to fall
to 4.0 bars. The final temperature is _____ oC. The total heat transfer out of the system = _______ kJ
d) Twenty kilograms (20 kg) of NEON GAS contained in a piston-cylinder device initially at 6.0 bars
and 140 oC is cooled at constant pressure to 273 K. The system work = ________ kJ and total
transfer = ____________ kJ.

Explanation / Answer

a)

P2/P1 = T2/T1

P2/1 = (273+200)/(273-26.43)

P2 = 1.9183 bar = 191.83 kPa

P2*V2 = mR*T2

191.83*10^3 *V2 = 20*(8314/28)*(273+200)

V2 = 14.642 m^3

b)

Q = m*Cv*(T2 - T1)

2500 = m*0.312*(160-(-16))

m = 45.527 kg

P1*V1 = mR*T1

P1*5 = 45.527*0.208*(273-16)

P1 = 486.74 kPa

P2*V2 = mR*T2

P2*5 = 45.527*0.208*(273+160)

P2 = 820.06 kPa

c)

T2/T1 = P2/P1

T2 / (140+273) = 4/6

T2 = 275.3 K = 275.3-273 deg C = 2.3 deg C

Q = m*Cv*(T2 - T1)

Q = 20*0.718*(2.3 - 140) = 1976.89 kJ

d)

W = P*(V2 - V1) = mR*(T2 - T1)

W = 20*0.412*(273 - (140+273)) = -1153.6 kJ

Q = m*Cp*(T2 - T1)

Q = 20*1.03*(273-(140+273)) = -2884 kJ

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