An ideal gas undergoes a thermodynamic process as shown in Fig. The process consists of two isobaric and two isothermal steps. Show that the network done during the whole process is W _((Net)) = P _(1) (V _(2) - V _(1)) log _(e)""(P_(2))/( P _(1))

An ideal gas undergoes a thermodynamic process as shown in Fig. The pr

1.	An ideal gas undergoes a thermodynamic process as shown in Fig. The process consists of two isobaric and two isothermal steps. Show that the network done during the whole process is W _((Net)) = P _(1) (V _(2) - V _(1)) log _(e)""(P_(2))/( P _(1))
Answer» <P> Solution :`W _(nct) =P_(2)(V _(B) -V _(A)) + P _(1) P_(2)LOG _(e) ( V _(2) // V _(B)) + P _(1) (V _(1) - V _(2)) + P _(1) V _(1) log (V _(A) //V_(1)) , P _(2) V _(B) = P _(1) V _(2), P _(2) V _(A) = P _(1) V _(1) W _(net) = P _(2) V _(B) - P _(2) V _(A) + P _(1) V _(2) log _(e) (P _(2) // P _(1)) + P _(1) V _(1) - P _(1) V _(2) - P _(1) V _(1) log ( P _(1) // P _(1)) = P _(1) (V _(2) - V _(1)) log _(e) ( P _(2) // P _(1))`