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A process 1rightarrow2 using monoatomic gas is shown on the P-V diagram on the right P_(1)=2P_(2)=10^(6)N//m^(2),V_(2)=4V_(1)=0.4m^(3)The heat ansorbed by the gas |
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Answer» 350kJ `W=3/2P_(0)v_(0)+3v_(0)P_(0)` `W=a/2P_(0)v_(0)` `DeltaU=nC_(v)DeltaT=n((3R)/2)(T_(F)-T_(i))` `DeltaU= 3/2nR (T_(f)-T_(i))=3/2[P_(f)V_(f)-P_(i)V_(i)]` `DeltaU-3/2[4P_(0)V_(0)-2P_(0)V_(0)]` `DeltaU=3P_(0)V_(0)` `DeltaQ=DeltaU+W=9/2P_(0)V_(0)+3P_(0)V_(0)=15/2P_(0)V_(0)` `P_(0)=10^(6)/2 , V_(0)=0.1` `DeltaQ=15/2xx10^(6)/2xx0.1=375000J` `DeltaQ=375kJ` 
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