MyQuestionIcon

1

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

Relating Translational KE and T

The temperatu...

Question

The temperature at which most probable speed of $O_{2}$ becomes equal to the root mean square speed of $N_{2}$ is:
[Given: $N_{2}$ at $427^{0} C$ ]

Open in App

Solution

$V_{most probable} = \sqrt{\frac{2 R T}{M}}; V_{r m s} = \sqrt{\frac{3 R T}{M}} V_{m p O_{2}} = \sqrt{\frac{2 R T_{1}}{32}}; V_{m p N_{2}} = \sqrt{\frac{3 R T_{2}}{28}} Given : T_{2} = 427 + 273 = 700 K \Rightarrow \sqrt{\frac{2 R T_{1}}{32}} = \sqrt{\frac{3 R T_{2}}{28}} \Rightarrow \frac{2 T_{1}}{32} = \frac{3 T_{2}}{28} T_{1} = \frac{16 \times 3 T_{2}}{28} T_{1} = 1200 K$

Suggest Corrections

thumbs-up

0

Similar questions

Q. Temperature at which most probable speed of

O_{2}

becomes equal to root mean square speed of

N_{2}

N_{2}

^{\circ}

Q. At what temperature most probable speed of

O_{2}

molecules have the samevalue of root mean square speed of

O_{2}

molecules at 300 K?

Q. At what temperature is the root mean square speed of

N_{2}

gas equal to that of propane gas at STP?

Q. At what temperature is the root mean square speed of

N_{2}

gas equal to that of propane gas at

440 K

Q. At what temperature most probable speed of

S O_{2}

molecules have the same value of root mean square speed of

O_{2}

300 K

?

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