The ratio of average kinetic energy of $H_{2}$ and $C H_{4}$ at $27^{\circ} C$ will be:

1 : 1

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

1 : 8

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

8 : 1

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

4 : 1

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

Open in App

Solution

The correct option is A 1 : 1
We know,
$A v g . K E = \frac{3}{2} k T$
Where $k$ = boltzmann constant.
Temperature is the same in both cases here.
Since the average kinetic energy depends only on temperature so both $H_{2}$ and $C H_{4}$ will have the same average kinetic energy and the required ratio will be $1 : 1$ .

Suggest Corrections

Similar questions

H_{2}

C H_{4}

27^{\circ} C

H_{2}

{C H}_{4}

K E

C O_{2}

27^{\circ} C

E

N_{2}

C O_{2}

27^{0}

N_{2}

The average kinetic energy of a gas molecule at $27^{\circ}$ is $6.21 \times 10^{- 21} J .$ The average kinetic energy at $227^{\circ}$ will be

Join BYJU'S Learning Program