An ideal gas undergoes a process consisting of adiabatic and isothermal process only as shown in the figure. Then :

Options

A. `T_(1) lt T_(2)`
B. `V_(A) V_(C)=V_(B)V_(D)`
C. work done by the gas in process will be negative
D. efficiency `=(W_(AB)+W_(BC))/(Q_(AB))`

Answer

B

Explanation

`P_(A)V_(A)=P_(B)V_(B)`.....(1) (isothermal process)
`P_(B)V_(B)^(gamma)=P_(C)V_(C)^(gamma)`.......(2) (adiabatic process)
`P_(C)V_(C)=P_(D)V_(D)`......(3) (isothermal process)
`P_(D)V_(D)^(gamma) =P_(A)V_(A)^(gamma)` .......(4) (adiabatic process)
`:.` multiplying all the above equations
`P_(A)P_(B)P_(C)V_(A)V_(B)^(gamma)V_(C)V_(D)^(gamma)=P_(A)P_(B)P_(C)V_(A)^(gamma)V_(B)V_(C)^(gamma)V_(D)`
`V_(B)^(gamma-1)V_(D)^(gamma-1)=V_(A)^(gamma-1)V_(C)^(gamma-1)` or `V_(B)V_(D)=V_(A)V_(C)`