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A monoatomic ideal gas of two moles is t...

A monoatomic ideal gas of two moles is taken through a cyclic process starting from A as shown in the figure. The volume ratio are `V_B/V_A =2` and `V_D/V_A = 4`. If the temperature `T_A` at `27^@C` is Calculate
the total work done by the gas during the complete cycle (in J).

Text Solution

Verified by Experts

Taking `Vk_(A)=V_(0), V_(B)=2V_(0),V_(D)=4V_(0)`
Precess `AtoB`. Isobaric process as V/T=constant
`V_(0)//(273+27)=(2V_(0))//T_(B)impliesT_(B)=600K=327^(@)C`
`Q_(1)=nC_(P)DeltaT=2(5//2)R(600-300)=1500R` (absorbed)
Process `BtoC` (isothermal)
`Q_(2)=W_(2)=nRT_(B)"In"(V_(2))//V_(1)0=831.78R` (absorbed)
Process `CtoD` (Isochoric)
`Q_(3)=nC_(V) DeltaT=-900R` (released).

Process `DtoA` (Isothermal)
`Q_(4)=W_(4)=nRT_9a)"In"(V_(2)//V_(1))=-831.78R`
Total work done, `W=Q_(1)+Q_(2)+Q_(3)+Q_(4)=600R`
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