One mole of an ideal monoatomaic gas is taken from `A` to `C` along the path `ABC` . The temperature of the gas at `A` is `T_(0)` . For the process `ABC` : A. work done by the gas is `RT_(0)`B. change in internal energy of the gas is `(11)/(2) RT_(0)`C. heat absorbed by the gas is `(11)/(2) RT_(0)` D. heat sbsorbed by the gas is `(13)/(2) R T_(0)` . (`R`=uniyersal gas constant)

One mole of an ideal monoatomaic gas is taken from `A` to `C` along th

1.	One mole of an ideal monoatomaic gas is taken from `A` to `C` along the path `ABC` . The temperature of the gas at `A` is `T_(0)` . For the process `ABC` : A. work done by the gas is `RT_(0)`B. change in internal energy of the gas is `(11)/(2) RT_(0)`C. heat absorbed by the gas is `(11)/(2) RT_(0)` D. heat sbsorbed by the gas is `(13)/(2) R T_(0)` . (`R`=uniyersal gas constant)
Answer» Correct Answer - A::C Process `AB:P` =constant `:. Vprop P V_(B)=2V_(A)` `T_(B)=2T_(A)=2T_(0) DeltaW_(AB)=P_(0)V_(0)=RT_(0) DeltaQ_(AB)=C_(p)DeltaT=((5)/(3)R)(2T_(0)-T_(0))=(5)/(2)RT_(0)` `:. DeltaU_(AB)=DeltaQ_(AB)-DeltaW_(AB)=(3)/(2)RT_(0)` . Process `BC:V`=constant `:. P propT , P_(C)=2P_(B)` `:. T_(C)=2T_(B)=4T_(0) , DeltaW_(BC)=0` `:. DeltaQ_(BC)=DeltaU_(BC)=C_(v)DeltaT` `=((3)/(2)R)(rT_(0)-2T_(0))=3RT_(0)` `:. DeltaW_("net") =RT_(0),DeltaQ_("net")=(11)/(2)RT_(0)` and `DeltaU_("net")=(9)/(2)RT_(0)`.

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