Question

The volume and temperature of air (assumed to be an ideal gas) in a closed vessel is 2.87$m^3$ and 300K, 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/kgK and the atmospheric pressure is 1 bar, the mass of air (in kg) in the vessel is

• A. 1.67
• B. 3.33
• C. 5.00
• D. 6.66

Answer 1

The correct option is C 5.00

$P_g$ = 0.5 bar = 50 kPa;
R = 287 J/kgK = 0.287 kJ/kgK
Ideas gas equation:
$PV=mRT$
Here P is absolute pressure:
$P=P_{abs}=P_{gauge}+P_{atm}$
$=50+101.325=151.325kPa$
$m=\frac{PV}{RT}=\frac{151.325\times 2.87}{0.287\times 300}=5.04kg$