One mole of an ideal gas in initial state A undergoes a cyclic process ABCA, as shown in the figure. Its pressure at A is `P_0`. Choose the correct option (s) from the following

Options

A. Internal energies at `A` and `B` are the same
B. Work done by the gas is process `AB` is `P_(0)V_(0)ln4`
C. Pressure at `C` is `(P_(0))/(4)`
D. Temperature at C is `(T_(0))/(4)`

Answer

A, B, C, D

Explanation

`:. U = (f)/(2) nRT`, where `f, n,R` are constants. Also
temperature `T` is same at `A & B`.
`:. U_(A) = U_(B)`
Also, `DeltaW_(AB) = nRT_(0) ln ((V_(f))/(V_(i)))`
`= nRT_(0) ln4 = P_(0)V_(0)ln4`
Process `BA` is isothermal so
`V_(A) P_(A) = V_(B) P_(B), V_(0)P_(0) = 4 V_(0) P_(B)`
`P_(B) = P_(0)//4`
For process `BC, V prop T` So this process is isobaric
`:. T_(C) = T_(0)//4`
So, answer are `(A)(B)(C)(D)`.