Verified Solution • 2025–26 Session moderate

Q12.9: NCERT Physics - Part 2 KINETIC THEORY 12.9 Solution: At what temperature is the root mean square speed of an atom in …

At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at � 20 �C? (atomic mass of Ar = 39.9 u, of He = 4.0 u).

Step-by-Step Verified Solution

**Answer:** TAr=2.52×103T_{Ar} = 2.52 \times 10^3 K **Step-by-Step Explanation:** For helium: THe=−20∘C=253T_{He} = -20^\circ\text{C} = 253 K, MHe=4.0M_{He} = 4.0 u. For argon: MAr=39.9M_{Ar} = 39.9 u. The rms speed is:
vrms=3RTMv_{rms} = \sqrt{\frac{3RT}{M}}
Setting vrms,Ar=vrms,Hev_{rms,Ar} = v_{rms,He}:
3RTArMAr=3RTHeMHe\frac{3RT_{Ar}}{M_{Ar}} = \frac{3RT_{He}}{M_{He}}
TAr=THe×MArMHe=253×39.94.0=2523.7 K≈2.52×103 KT_{Ar} = T_{He} \times \frac{M_{Ar}}{M_{He}} = 253 \times \frac{39.9}{4.0} = 2523.7 \text{ K} \approx 2.52 \times 10^3 \text{ K}
[citation:19]

All Questions in this Chapter

Q12.1 Q12.2 Q12.3 Q12.4 Q12.5 Q12.6 Q12.7 Q12.8 Q12.9 Q12.10

Related Learning Resources & Exam Preparation

Chapter Practice Bank
Practice all topic-wise questions from KINETIC THEORY
Reference Textbook Solutions
Explore complete chapter exercises for NCERT Physics - Part 2
Competitive PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring