A thermodynamic process is as shown in figure such that
`P_(A) = 3 xx 10^(4)Pa`,
`P_(B) = 8 xx 10^(4) Pa`,
`V_(A) = 2 xx 10^(-3) m^(3)`,
`V_(D) = 5 xx 10^(-3) m^(3)`

In procss 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 system in proess AC is

Options

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

Answer

A

Explanation

In going from A to C via, B, `Delta Q = 600 + 200 = 200 800 J`
`Delta W = W_(AB) + W_(BC) = 0 + P_(B)(V_(C )-V_(B)) = 8 xx 10^(4) xx (V_(D) - V_(A))`
`= 8 xx 10^(4)(5 xx 10^(-3) - 2 xx 10^(-3)) = 240 J`
`Delta U = Delta Q - Delta W = 800 - 240 = 560 J`