The average translational energy and the rms speed of molecules in a sample of oxygen gas at 300 K are 6.21 × 1 0 − 21 J \& 484 m / s respectively. The corresponding values at 600K are nearly (assuming ideal gas behaviour)

K E = 2 3 k T ⇒ K E ∝ T , so KE doubles to 12.42 × 1 0 − 21 J . v r m s ∝ T ⇒ v r m s = 2 × 484 = 684 m / s .

Start Free Mock LoginAlready registered? Login here

Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoKTGMedium2 minQB

PhysicsMediumsingle choice

The average translational energy and the rms speed of molecules in a sample of oxygen gas at 300 K are $6.21 \times 1 0^{- 21} J$ \& $484 m / s$ respectively. The corresponding values at 600K are nearly (assuming ideal gas behaviour)

Options:

Answer:

Solution:

$K E = \frac{3}{2} k T \Rightarrow K E \propto T$ , so KE doubles to $12.42 \times 1 0^{- 21} J . v_{r m s} \propto T \Rightarrow v_{r m s} =$ $2 \times 484 = 684 m / s$ .

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:KTG

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...