Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
59 views
in Chemistry by (62.9k points)
closed by
Pay load is defined as the difference between the mass of displaced air and the mass of the ballon Calculate the pay-load when a balloon of radius `10m` mass `100 kg` is filled with helium at 1.66 bar at `27^(@)C` (Density of air ` = 1.2 kg m^(-3)` and `R = 0.083` nar `dm^(-3) K^(-1) mo1^(-1)`) .

1 Answer

0 votes
by (66.1k points)
selected by
 
Best answer
Volume of balloon, `V=4/3pi r^(3)`
`r=10 m`
`V=4/3xx3.14xx(10)^(3)=4186.6 m^(3)=4186.6xx10^(3) dm^(3)`
Mass of displaced air is
`4186.6 m^(3)xx1.2 kg m^(-3)=5024 kg`
`T=27+273=300 K`
Moles of gas present,
`n=(PV)/(RT)`
`=(1.66xx4186.6xx10^(3) dm^(3))/(0.083xx300)=279.1xx10^(3) mol`
Mass of He present `=279.1xx4xx10^(3) mol`
`=1116.4xx10^(3) g`
`=1116.4 kg`
Mass of filled ballon `=100+1116.4=1216.4 kg`
Pay load=Mass of displaced air- Mass of ballon
`=5024-1216.4=3807.6 kg`

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...