An ideal monoatomic gas undergoes a cyclic process `ABCA` as shown in the figure. The ratio of heat absorbed during `AB` to the work done on the gas during `BC` id

Options

A. `(5)/(2ln2)`
B. `(5)/(3)`
C. `(5)/(4ln2)`
D. `(5)/(6)`

Answer

C

Explanation

`W_(AB) = (2V_(0) - V_(0))P_(0) = P_(0)V_(0)`
[isobaric process]
`W_(BC) = |nRT(2T_(0))ln"(V_(0))/(2V_(0))| = 2P_(0)V_(0)ln2`
[Isothermal process]
`therefore (Q_(AB))/(W_(BC)) = ((3)/(2)P_(0)V_(0)+ P_(0)V_(0))/(2P_(0)V_(0)ln2) = (5)/(4ln2)`