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A certain quantity of ideal gas takes up `56J` of heat in the process AB and 360 J in the process AC. What is the number of degrees of freedom of the gas. |
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Answer» `DeltaQ_(AB)=nC_(p)DeltaT=(gamma)/(gamma-1)nRDeltaT` `=(gamma)/(gamma-1)[3P_(0)V_(0)-P_(0)V_(0)]=2PV_(0)xx(gamma)/(gamma-1)` `DeltaQ_(AC)=DeltaU+Deltaw` `=(nR)/(gamma-1)DeltaT+(1)/(2)xx3V_(0)[P_(0)+4P_(0)]` `=([16P_(0)V_(0)-P_(0)V_(0)])/(gamma-1)+(15P_(0)V_(0))/(2)` `56=2P_(0)V_(0)xx(gamma)/(gamma-1)` ,brgt `360=15P_(0)V_(0)[(gamma+1)/(2(gamma-1))]` `(360)/(56)=(15)/(4)((gamma+1))/(gamma)` `12gamma=7gamma+7` `gamma=(7)/(5)=1+(2)/(f)impliesf=5` |
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