0

Guest

Courses New

For two gases, A and B with molecular weight MA and MB, it is observed that at a certain temperature, T, the mean speed of A is equal to the root mean square speed of B. Thus the mean speed of A can be made equal to the mean speed of B, if. a) A is at temperature T and B at T ' , T T ' b) A is lowered to a temperature   2 8 3 T T while B is at T c) Both A and B are raised to a higher temperature d) Both A and B are placed at lower temperature

For two gases, A and B with molecular weight MA and MB, it is observed that at a certain temperature,
T, the mean speed of A is equal to the root mean square speed of B. Thus the mean speed of A can be
made equal to the mean speed of B, if.
a) A is at temperature T and B at T ' , T T '
b) A is lowered to a temperature

 2
8
3
T T while B is at T
c) Both A and B are raised to a higher temperature
d) Both A and B are placed at lower temperature

Grade:11

1 Answers

Bhavya
askIITians Faculty 1281 Points
7 years ago
Dear student
Average speed of A = root ( 8RT/\piMA)
RMS speed of B = root (3RT/MB)
Since, both are equal, we equate the two and squaring is done
we get,
8MB/3\pi= MA
In order to have equal RMS speeds just equate the two, and derive a relation (T and T’ relation is not clear in question) but I hope you have understood the method.

Think You Can Provide A Better Answer ?

Other Related Questions on Physical Chemistry

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More