MyQuestionIcon

1

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

Graham's Law of Effusion

A and B are t...

Question

A and B are two gases. $\frac{T_{A}}{M_{A}} = 4 \frac{T_{B}}{M_{B}}$ , where T is the temperature and $M$ is the molar mass. If $C_{A}$ and $C_{B}$ are rms speeds, then $\frac{C_{A}}{C_{B}}$ will be

A

2

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

B

4

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

C

0.5

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

D

0.25

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

Open in App

Solution

The correct option is A 2
$\frac{C_{A}}{C_{B}} = \frac{\sqrt{\frac{3 R T_{A}}{M_{A}}}}{\sqrt{\frac{3 R T_{B}}{M_{B}}}} = 2$

Suggest Corrections

thumbs-up

1

Similar questions

Q. A and B are two gases.

\frac{T_{A}}{M_{A}} = 4 \frac{T_{B}}{M_{B}}

where T is the temperature and M is the molecular mass. If

C_{A}

C_{B}

are rms speeds, then

\frac{C_{A}}{C_{B}}

Q. Let A and B be two gases and given

\frac{T_{A}}{M_{A}} = 4. \frac{T_{B}}{M_{B}};

where T is the temperature and M is the molecular mass. If

C_{A}

C_{B}

are the rms speed, then the ration

\frac{C_{A}}{C_{B}}

will be equal to

Q. Let A and B the two gases and given :

\frac{T_{A}}{M_{A}} = 4. \frac{T_{B}}{M_{B}}

; where T is the temperature and M is the molecular mass. If

C_{A}

C_{B}

are the rms speed, then the ratio

\frac{C_{A}}{C_{B}}

will be equal to

Q. If for two gases of molar masses

M_{A}

M_{B}

T_{A}

T_{B}, T_{A} M_{B} = T_{B} M_{A}

, then which property has the same magnitude for both the gases?

Q. In the right triangle shown the sum of the distances

B M

M A

is equal to the distances

B C

C A

M B = x, C B = h

C A = d

x

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

The ideal gas model

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Graham's Law of Effusion

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon