K.T.O.G

Classical Mechanics Level 3

N N molecules each of mass m m of gas A and 2 N 2N molecules each of mass 2 m 2m of gas B are contained in the same vessel which is maintained at temperature T T . The mean square of the velocity of molecules of B B type is denoted by ( V R M S − B ) 2 { ({ V }_{ RMS-B } })^{ 2 } and the mean square of x-component of the velocity of A A type is denoted by ( V R M S − X − A ) 2 { ({ V }_{ RMS-X-A } })^{ 2 } . Find the value of ( V R M S − X − A ) 2 ( V R M S − B ) 2 . \frac { { ({ V }_{ RMS-X-A }) }^{ 2 } }{ { ({ V }_{ RMS-B }) }^{ 2 } }.

3 2 \frac { 3 }{ 2 } 2 3 \frac { 2 }{ 3 } 4 3 \frac { 4 }{ 3 } None of these choices

Jaswinder Singh by Jaswinder Singh , 23, India

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1 solution

Jaswinder Singh
Feb 22, 2016

for gas B ( V R M S − B ) 2 = 3 k T 2 m { ( }{ V }_{ RMS-B })^{ 2 }=\frac { 3kT }{ 2m } for gas A ( V R M S − X − A ) 2 = ( V R M S − Y − A ) 2 = ( V R M S − Z − A ) 2 { ({ V }_{ RMS-X-A }) }^{ 2 }={ ({ V }_{ RMS-Y-A }) }^{ 2 }={ ({ V }_{ RMS-Z-A }) }^{ 2 } ( V R M S − X − A ) 2 = ( V R M S − A ) 2 3 = 1 3 × 3 k T m = k T m { ({ V }_{ RMS-X-A }) }^{ 2 }=\frac { { ({ V }_{ RMS-A }) }^{ 2 } }{ 3 } =\frac { 1 }{ 3 } \times \frac { 3kT }{ m } =\frac { kT }{ m } ( V R M S − X − A ) 2 ( V R M S − B ) 2 = k T / m 3 k T / 2 m = 2 3 \frac { { ({ V }_{ RMS-X-A }) }^{ 2 } }{ { ({ V }_{ RMS-B }) }^{ 2 } } =\frac { kT/m }{ 3kT/2m } =\frac { 2 }{ 3 }

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