Air has a pressure of 200 kPa, and occupies a volume of 0.25 m3. Determine the mass of the air present, assuming ideal gas behavior, for air temperatures of
(a) 25°C
(b) 100°C
(c) 250°C
(d) 500°C
Given: P = 200 kPa; V = 0.25 m3
What will be an ideal response?
The ideal gas law gives m = PV/RT
For air, R = 0.287 kJ/kg-K
(a) m = (200 kPa)(0.25m3)/(0.287 kJ/kg-K)(25 + 273)K = 0.585 kg
(b) m = (200 kPa)(0.25m3)/(0.287 kJ/kg-K)(100 + 273)K = 0.467 kg
(c) m = (200 kPa)(0.25m3)/(0.287 kJ/kg-K)(250 + 273)K = 0.333 kg
(d) m = (200 kPa)(0.25m3)/(0.287 kJ/kg-K)(500 + 273)K = 0.225 kg
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