MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Relating Translational KE and T

At certain te...

Question

At certain temperature rms velocity of ethane molecules is $5.33 \times 10^{4} c m$ find the rms velocity of $O_{2}$ at temperature.

Open in App

Solution

We know,
$U_{R M S} = \sqrt{\frac{3 R T}{M}}$
So,
$\frac{U_{a v g} (C_{2} H_{6})}{U_{a v g} (O_{2})} = \sqrt{\frac{T_{C_{2} H_{6}} \times M_{O_{2}}}{T_{O_{2}} \times M_{C_{2} H_{6}}}}$
$M_{C_{2} H_{6}} = 30$ , $M_{O_{2}} = 32$
Here, $T_{C_{2} H_{6}} = T_{O_{2}}$
Hence,
$\frac{U_{a v g} (C_{2} H_{6})}{U_{a v g} (O_{2})} = 1.038$
Hence,
$U_{R M S} (O_{2}) = \frac{1.038}{5.33 \times 10^{2}} = 0.0019 \frac{m}{s}$

Suggest Corrections

thumbs-up

0

Similar questions

Q. At a certain temperature, the

r m s

O_{2}

400 m / s e c

. At the same temperature, the

r m s

H_{2}

molecules will be

Q. At some temperature, the rms velocity of

H_{2}

2 \times 10^{3} m / s

. what will be the rms velocity of

O_{2}

molecules at the same temperature:

Q. At what temperature, the

r m s

O_{2}

r m s

S O_{2}

127^{o} C

:

Q. The temperature at which

R M S

O_{2}

is equal to the

R M S

S O_{2}

327^{o} C

Q. Initially, the root mean square (rms) velocity of

N_{2}

molecules at certain temperature is

u

. If this temperature is doubled and all the nitrogen molecules dissociate into nitrogen atoms, then the new rms velocity will be :

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Using KTG

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Relating Translational KE and T

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon