InterviewSolution
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InterviewSolution
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A gaseous mixture enclosed in a vessel consists of one gram mole of a gas A with `gamma=(5/3)` and some amount of gas B with `gamma=7/5` at a temperature T. The gases A and B do not react with each other and are assumed to be ideal. Find the number of gram moles of the gas B if `gamma` for the gaseous mixture is `(19/13)`. |
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Answer» Correct Answer - B As for ideal gas `C_(P)-C_(V)=R andgamma=(C_(P)//C_(V))` , `gamma-1=(R)/(C_(V)) orC_(V)=(R)/((gamma-1))` `(C_(V))_(1)=(R)/((5//3)-1)=(3)/(2)R` `(C_(V))_(2)=(R)/((7//5)-1)=(5R)/(2)` and `(C_(V))_(mix)=(R)/((19//13)-1)=(13)/(6)R` Now from the conservation of energy,i.e., `DeltaU=DeltaU_(1)+DeltaU_(2)` , we get `(mu_(1)+mu_(2))(C_(V))_(mix)DeltaT=[mu_(1)(C_(V))_(1)+mu_(2)(C_(V))_(2)]DeltaT` `(V_(V))_(mix)=(mu_(1)(C_(V))_(1)+mu_(2)(C_(V))_(2))/(mu_(1)+mu_(2))` `(13)/(6)R=(1xx(3)/(2)R+mu_(2)xx(5)/(2)R)/(1+mu_(2))=((3+5mu_(2))R)/(2(1+mu_(2)))` or `13+13mu_(2)=9+15mu_(2)` ,i.e.,`mu_(2),=2g "mole"` |
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