A 28 liter rigid enclosure contains air at 140 kPa and 20oC. Heat is added to the container until the pressure reaches 345 kPa. Calculate the heat added.
Formulae:
For an ideal gas, PV = nRT
From first law of thermodynamics,
Q + W =
ΔUFor an ideal gas ΔU is a function of temperature only, and work done at constant volume = 0
Therefore, Q = ΔU.
And, ΔU = mCV(T2 - T1)
m = n x Molecular weight
Calculations:
n = PV / (RT) = 140 x 1000 x 0.028 / (8314 x 293) = 1.609 x 10-3 kmol
m = 1.609 x 10-3 x 29 = 0.0467 kg.
For constant volume system, P ∝ T.
Therefore, P1/T1 = P2/T2
140/293 = 345/T2
T2 = 722 K.
CV = 0.718 kJ/kg.oC (data, for air at 20oC)
Q = ΔU = 0.0467 x 0.718 x (722 - 293) = 14.385 kJ
i.e., Heat added = 14.385 kJ
Last Modified on: 04-Feb-2022
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