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.
For an ideal gas, PV = nRT
From first law of thermodynamics,
Q + W =DU
For an ideal gas DU is a function of temperature only, and work done at constant volume = 0
Therefore, Q = DU.
And, DU = mCV(T2 - T1)
m = n x Molecular weight
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 a 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 = DU = 0.0467 x 0.718 x (722 - 293) = 14.385 kJ
i.e., Heat added = 14.385 kJ
Last Modified on: 11-Sep-2014
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