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` `V_(A) = 2 xx 10^(-3) m^(3)`
`P_(B) = 8 xx 10^(4) Pa` `V_(D) = 5 xx 10^(-3) m^(3)`
In the process `AB` `600 J` of heat is added to the system. The change in internal energy of the system in the process `AB` would be

Options

A. `560 J`
B. `800 J`
C. `60 J`
D. `640 J`

Answer

A

Explanation

a. No work is done along path `AB` because the process is isochoric.
`:.` Work done `= P_(B) (V_(D) - V_(A))`
`= 8 xx 10^(4) (5 xx 10^(-3) - 2 xx 10^(-3)) = 240 J`
`(Delta Q)_(AC) = (Delta Q)_(AB) + (Delta Q)_(BC) = 600 + 200 = 800 J`
`:. (Delta U)_(AC) = (Delta Q) - (Delta W) = 800 - 240 = 560 J`