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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 bemade 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 283T T while B is at Tc) Both A and B are raised to a higher temperatured) Both A and B are placed at lower temperature


4 years ago

Bhavya
1281 Points
							Dear studentAverage speed of A = root ( 8RT/MA)RMS speed of B   = root (3RT/MB)Since, both are equal, we equate the two and squaring is donewe get,8MB/3= MAIn 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.

4 years ago
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