The above p-v diagram represents the thermodynamic cycle of an engine, operating with an ideal monoatomic gas. The amount of heat, extracted from the source in a single cycle is

Options

A. `P_(0)V_(0)`
B. `((13)/(2))p_(0)v_(0)`
C. `((11)/(2))p_(0)v_(0)`
D. `4p_(0)v_(0)`

Answer

B

Explanation

`eta=(1-(T_(2))/(T_(1)))xx 100 implies (40)/(100) = 1-(T_(2))/(500) implies T_(2) = 300K` Again `(60)/(100) = 1-(300)/(T_(1)) implies T_(1) = 750K`
heat supplied =`nC_(v)(2T_(0) - T_(0)) + nC_(P)(4T_(0) -2T_(0)) = (n.3RT_(0))/(2) + (D5R)/(2)(2T_(0)) = (13)/(2)nRT_(0) = (13)/(2)P_(0)V_(0)`