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` :

Options

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

A, C

Explanation

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)`.