The average translational energy and the rms speed of molecules of a sample of oxygen gas at 300 K are 6.21×10−21J and 484 ms-1 respectively. The corresponding values at 600 K are nearly (assuming ideal gas behaviour)

8.78 × 10 − 21 J , 684 ms − 1

The average translational energy and the rms speed of molecules of a sample of oxygen gas at 300 K are 6.21×10−21J and 484 ms-1 respectively. The corresponding values at 600 K are nearly (assuming ideal gas behaviour)

8.78 × 10 − 21 J , 684 ms − 1

12.42 × 10 − 21 J , 968 ms − 1

6.21 × 10 − 21 J , 968 ms − 1

12.42 × 10 − 21 J , 684 ms − 1

8.78 × 10 − 21 J , 684 ms − 1

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

The average translational energy and the rms speed of molecules of a sample of oxygen gas at 300 K are $6.21 \times 10^{- 21} J$ and 484 ms^-1 respectively. The corresponding values at 600 K are nearly (assuming ideal gas behaviour)

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$12.42 \times 10^{- 21} J, 968 {ms}^{- 1}$

$6.21 \times 10^{- 21} J, 968 {ms}^{- 1}$

$12.42 \times 10^{- 21} J, 684 {ms}^{- 1}$

$8.78 \times 10^{- 21} J, 684 {ms}^{- 1}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$E \propto T and v_{rms} \propto \sqrt{T}$ . Hence at 600 K,

$\begin{array}{l} \frac{E_{2}}{E_{1}} = \frac{T_{2}}{T_{1}} \\ \Rightarrow E = 6 .21 \times 10^{- 21} \times \frac{600}{300} \\ = 12 .42 \times 10^{- 21} J \\ and \frac{v_{2}}{v_{1}} = \sqrt{\frac{T_{2}}{T_{1}}} \\ v_{rms} = 484 \times \sqrt{\frac{600}{300}} ≃ 684 {ms}^{- 1} \end{array}$