A monatomic idea gas of `2 mol` is taken through a cyclic process starting from `A` as shown in figure. The volume ratio are `V_(B)//V_(A)=2` and `V_(D)//V_(A)=4`. If the temperature `T_(A)` at `A` is `27^(@)C`, and gas constant is `R`. Calculate heat absorbed or released by the gas in process `D rarr A`.

Options

A. `1500, ` added
B. `1200R ln (2)`, added
C. `900 R,` rejected
D. `1200 R ln (2)`, rejected

Answer

D

Explanation

Gas is monatomic, hence
`C_(v)=(3)/(2)R` and `C_(p)=(5)/(2)R`
Number of moles, `n=2`
`T_(A)=27^(@)C=300K`
Process `A rarr B` is a striaght line passing through origin, hence it is an isobaric process.
`(V_(A))/(T_(A))=(V_(B))/(T_(B))`
`T_(B)=((V_(B))/(V_(A)))T_(A)`
`=2 xx 300 K=600K` ltbr. Porcess `A rarr B ` is isobaric.
`Q_(p)=nC_(p)(T_(B)-T_(A))`
`Q_(p)=2((5)/( 2)R)(600-300)=1500R`
Heat is added to the system.
Process` B rarr C` is isothermal.
From the first law of thermodynamics.
`Q_(BC)=W_(BC)`
`(` as `Delta U=0)`
` W=nRT_(B)1n((V_(C))/(V_(B)))`
`=(2)(R)(600)1n((4V_(A))/(2V_(A)))`
`=(1200R)1n (2)`
Heat is added to the system.
Process `C rarr D ` is isochoric.
From the first law of thermodynamics,
`Q_(CD)=nC_(v)(T_(D)-T_(C))=n((3)/(2)R)(T_(A)-T_(B))`
`=2((3)/(2) R)(300-600)=-900R`
Negative sign implies that heat is rejected by the system.
Process `D rarr A` is isothermal.
`Q_(DA)=W_(DA)=nRT_(0)1n((V_(A))/(V_(D)))`
`=(2)(R)3001n((V_(A))/(4V_(A)))=600R 1n((1)/(4))`
Negative sign implies that heat is rejected by the system.
In a cyclic process, `Delta U=0`
`Q_(n et)=W_( n et)`
`W_(n et)=Q_(AB)+Q_(BC)+Q_(CD)+Q_(DA)`
`=1500 R+8.316R-900R-831.6R=600R`