One mole of an ideal monoatomatic gas is taken round the cylic process `ABCA` as shown Then

Options

A. Work done by the gas `P_(0)V_(0)`
B. Heat rejected in the process `CA` is `(5P_(0)V_(0))/(2)` and absorbed in the process `AB` is `3P_(0)V_(0)`
C. Heat absorbed in the process `BC` is `(P_(0)V_(0))/(2)`
D. `(25)/(8)(P_(0)V_(0))/(R)` is the maximu temperature attained by the gas during the cycle

Answer

A, B, C

Explanation

(a). The work done by the gas is equal to the area inder the closed curve. Thus work done in cycle is `W=P_(0)V_(0)`
(b) Heat rejected in path `CA` is given as
`Q_(CA)=nC_(P)DeltaT=(5)/(2)P_(0)V_(0)`
Heat absored in path `AC` is
`Q_(AC)=nC_(P)(T_(B)-T_(A))=3P_(0)V_(0)`
(c) For cycle `ABC`, we have heat supplied = work done
by the gas `-((5)/(2))P_(0)V_(0)+3P_(0)V_(0)+Q_(BC)=P_(0)V_(0)`
`implies Q_(BC)=P_(0)V_(0)//2` .