1.

One mole of monoatomic gas is taken through cyclic process shown below T_(A)=300 K.Process AB is defined as PT=constant.

Answer»

Work done in process AB is `-400` R.
Change in internal ENERGY in process CA is 900 R.
Heat transferred in the process BC is 2000R.
Change in internal energy in process CA is -900 R.

Solution :(A) Process AB: `PT=CONST. RIGHTARROW nRT^(2)=const`
`W=undersetAoversetBintPdV=undersetAoversetBint(const)/T dv Rightarrow(dV)/(dT)=(2nRT)/(const.)`
`W=underset300overset100int(const.)/T.(2nRT)/(const.) dt Rightarrow (P_(A))/P_(B)=T_(B)/T_(A) Rightarrow 1/3=T_(B)/T_(A)`
`Rightarrow T_(B)=300/3=100 RightarrowW=2nR(100-300) RightarrowW_(AB)=-400` nR
(B)Process CA: Isochoric P//Tconst:`T_(A)//T_(C)=P_(A)//P_(C)`
`T_(A)//T_(C)=P_(A)//P_(B) RightarrowT_(A)//T_(C)=1//3RightarrowT_(C)=3T_(A)`
`T_(C)=900R RightarrowDeltaU=nC_(v)DeltaT`
`=(1) 3/2Rxx(T_(A)-T_(C))=3/2Rxx(300-900)=3/2Rxx-600=-900R Rightarrow|DeltaU|=(900R)`
(C) Process BC: Isobaric
`Q=nC_(P)DeltaT RightarrowQ=(1) 5/2Rxx(T_(C)-T_(B))`
`Q=5/2Rxx(900-100)Rightarrow Q=5/2Rxx800 RightarrowQ=2000R`


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