A thermodynamical process is shown in the figure with `p_A=3xxp_(atm)`, `V_A=2xx10^-4m^3`, `p_B=8xxp_(atm)`, `V_C=5xx10^-4m^3`. In the process AB and BC, 600J and 200J heat are added to the system. Find the change in internal energy of the system in the process CA. `[1 p_(atm)=10^5N//m^2]`

Options

A. (a) `560J`
B. (b) `-560J`
C. (c) `-240J`
D. (d) `+240J`

Answer

B

Explanation

In a cyclic process,
`Q_(n et)=W_(n et)`
`:.` `Q_(AB)+Q_(BC)+Q_(CA)=` area under the graph
`:.` `600+200+Q_(CA)=1/2(3xx10^-4)(5xx10^5)`
`=75J`
`:.` `Q_(CA)=-725J=W_(CA)+DeltaU_(CA)`
`=-725=` (-Area under the graph)`+DeltaU_(CA)`
`=-(1/2xx11xx10^5)(3xx10^-4)+DeltaU_(CA)`
Solving we get, `DeltaU_(CA)=-560J`