wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The volume and temperature of air (assumed to be an ideal gas) in a closed vessel is 2.87m3 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
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
3.33
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
5.00
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
6.66
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 5.00
Pg = 0.5 bar = 50 kPa;
R = 287 J/kgK = 0.287 kJ/kgK
Ideas gas equation:
PV=mRT
Here P is absolute pressure:
P=Pabs=Pgauge+Patm
=50+101.325=151.325kPa
m=PVRT=151.325×2.870.287×300=5.04kg

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon