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A thermodynamical process is shown in th...

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]`

A

560 J

B

800 J

C

600 J

D

640 J

Text Solution

Verified by Experts

Since internal energy is a state function so the change in internal energy during the process AC will be same as that during the total process AB plue BC. Calculate the total wark `DeltaW` during AB plus BC and total heat `Delta`Q during these processes. Then calculate `DeltaU = DeltaQ -DeltaW`
During the processes AB the volume doesn.t change so work done is zero.
During BC the pressure is constant, so total work done
`P_B(V_C -V_B) =P_B (V_C -V_A) :.V_B=V_A`
`=B xx 10^(4) [N//m^2](5 xx 10^(-3)m^3 - 2 xx 10^(-3)m^3) = 240J`
`:. DeltaW` during process AB plus BC = 240 J
Now given `DeltaQ` during the same total process `= 600 + 200 = 800J`.
`:. DeltaU = DeltaQ - DeltaQ = 800 - 240 = 560 J`
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