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Grade: 11
        
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
3 years ago

Answers : (1)

Bhavya
askIITians Faculty
1284 Points
							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.
3 years ago
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