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

A balloon contains 1500 m3 of helium gas at 27 C and 4 atm . The volume of helium gas at 3 C and 2 atm will be

A
2700 m3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
1900 m3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
1700 m3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1500 m3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 2700 m3
Given:
Initial volume, V1=1500 m3
Initial temperature, T1=27 C=300 K
Initial pressure, P1=4 atm
Final temperature, T2=3 C=270 K
Final pressure, P2=2 atm
To find:
Final volume, V2=?

We know that, ideal gas equation is given by
PV=nRT
As number of moles will remain constant, so
P1V1T1=P2V2T2
V2=P1V1T2P2T1
V2=4×1500×2702×300
V2=2700 m3

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon