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

Temperature at which rms speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at 20C is
(Take molar mass of argon as 40 g and for helium as 4 g)

A
2530 K
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
3000 K
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2000 K
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
4000 K
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 2530 K
Given,
Temperature of helium gas atom, T1=20C=253 K
Let temperature of argon gas be T2

We know that,
Vrms=3RTM
For helium gas, (Vrms)He=3×8.3×2534×103 [molar mass of helium=4 g]

For Argon gas, (Vrms)Ar=3×8.3×T240×103 [molar mass of argon=40 g]

On equating both speeds,
3×8.3×2534×103=3×8.3×T240×1032531=T210T2=2530 K

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Using KTG
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon