A thermodynamic process is shown in Fig. The pressures and volumes corresponding to some points in the figure are : `P_(A) = 3 xx 10^(4) Pa, P_(B) = 8 xx 10^(4) Pa` and `V_(A) = 2 xx 10^(-3) m^(3) , V_(D) = 5 xx 10^(-3) m`.
In process `AB, 600 J` of heat is added to the system and in process `BC, 200 J` of heat is added to the system. The change in internal energy of the energy of the process `AC` would be.

Options

A. 560 J
B. 800 J
C. 600 J
D. 640 J

Answer

A

Explanation

Since, `W_(AB) = W_(DC) = 0`
`W_(BC) =P(V_2-V_1) =8xx10^4(5-2)xx10^(-3) = 240 J`
`W_(AC)= W_(AB)+W_(BC)=240J`
Now, `DeltaQ_(AC)=DeltaQ_(AB)+DeltaQ_(BC) = 600+200 = 800J`
`:. 800=DeltaU_(AC) +240 implies DeltaU_(AC) = 560 J`
[as `DeltaQ_(AC) = DeltaU_(AC) + DeltaW_(AC)`].