One mole of ideal monatomic gas is carried through the reversible cyclic process as shown in figure. Calculate the max temperature attained by the gas during the cycle.

Options

A. `(25)/(8)((P^(@)V^(@))/(R))`
B. `-(25)/(8)((P^(@)V^(@))/(R))`
C. `(35)/(8)((P^(@)V^(@))/(R))`
D. `-(35)/(8)((P^(@)V^(@))/(R))`

Answer

A

Explanation

BC is a straight line & equation can be obtained by using
`(y-y_(1)) = (Y_(2)-Y_(1))/(x_(2)-x_(1))(x-x_(1))`
`(P-3P^(@))=(P^(@)-3P^(@))/(2V^(@)-V_(@))(V-V^(@)) implies (P-3P^(@)) =-(2P^(@))/(V^(@))(V-V^(@)) " "…..(1)`
Replacing P by using `P=(RT)/(V)` in equation (1)
`((RT)/(V)-3P^(@))=-(2P^(@))/(V^(@)) (V-V^(@)) implies T=(2P^(@)V(V-V^(@)))/(V^(@)R) + (3P^(@)V)/(R)`
For T to be maximum
`(dT)/(dV)=0`
`(dT)/(dV) =-(2P^(@)(2V-V^(@)))/(V^(@)R)+ (3P^(@))/(R) =0 implies V=(5)/(4)V^(@)`
`T_("max")=-(2P^(@)xx(5)/(4)V^(@)((5)/(4)V^(@)-V^(@)))/(V^(@)R)+(3P^(@)(5)/(4)V^(@))/(R)=(25)/(8)((P^(@)V^(@))/(R))`