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The volume and temperature of air (assumed to be an ideal gas) in a closed vessel is 2.87m3 and 300 K, respectively. The gauge pressure indicated by a manometer fitted to the wall of the vessel is 0.5 bar. If the gas constant of air is R = 287 J/kg K and the atmospheric pressure is 1 bar, the mass of air (in kg) in the vessel is
1. 1.67
2. 3.33
3. 5.00
4. 6.66

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Correct Answer - Option 3 : 5.00

V = 2.87 m3; T = 300 K

Pgauge = 0.5 bar; Patm = 1 bar; R = 287 j/Kg-K

PV = mRT (P is absolute pressure)

Pabs = Patm + Pgauge = 0.5 + 1 = 1.5 bar = 150 kPa

\(PV = mRT \Rightarrow m = \frac{{PV}}{{RT}} = \frac{{150*2.87}}{{0.287*300}}\)

m = 5 Kg

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