An ideal diatomic gas undergoes a thermodynamic process as shown in the P - V diagram. The process AB is isochoric while the process BC is isothermal. The total heat given to the gas in the process is nearly (use l n 2 = 0.7 )

Step-by-step solution

Q A B = ∆ U A B + W A B ⇒ W A B = 0 ∆ U A B = f 2 n R ∆ T ⇒ f 2 ∆ P V ∆ U A B = 5 2 ∆ P V ⇒ Q A B = 2.5 P 0 V 0 Process BC Q B C = ∆ U B C + W B C ⇒ Q B C = 0 + 2 P 0 V 0 l n 2 = 1.4 P 0 V 0 Q n e t = Q A B + Q B C = 3.9 P 0 V 0