One mole of ideal monoatmic gas is carried throught the reversible cyclic process as shown in figure. Calculate net heat absorbed by the gas in the path BC

Options

A. `(1)/(2)P^(@)V^(@)`
B. `(7)/(2)P^(@)V^(@)`
C. `2P^(@)V^(@)`
D. `(5)/(2)P^(@)V^(@)`

Answer

A

Explanation

`DeltaE=q+w`
`W_(BC) =(1)/(2) (2V^(@) -V^(@))(P^(@)-3P^(@))(0-P^(@))=-2P^(@)V^(@)`
`DeltaE=nC_(v)DeltaT=1xx(3)/(2)R((P^(@)2V^(@))/(R)-(3P^(@)V^(@))/(R)) =-(3)/(2) P^(@)V^(@)`
`q_("BC")=DeltaE-W=-(3)/(2) P^(@)V^(@)+2P^(@)V^(@)=(1)/(2)P^(@)V^(@)`