At which of the following temperature would the molecules of a gas have twice the average kinetic energy they have at $20^{\circ} C$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
$E \infty T ∴ \frac{E_{2}}{E_{1}} = \frac{T_{2}}{T_{1}} \Rightarrow \frac{2 E_{1}}{E_{1}} = \frac{T_{2}}{(20 + 273)} \Rightarrow T_{2} = 293 \times 2 = 586 K = 313^{\circ} C .$

Suggest Corrections

Similar questions

20^{o} C

20^{o} C

At what temperature is the kinetic energy of a gas molecule half its value at $327^{\circ} C$ ?

At what temperature would the root mean square speed of a gas molecule have twice its value at $100^{\circ}$ C?

O_{2}

- 73^{\circ}

Join BYJU'S Learning Program