Refer to figure. Let ΔU 1 and ΔU 2 be the change in internal energy in processes A and B respectively, ΔQ be the net heat given to the system in process A + B and ΔW be the net work done by the system in the process A + B.

Question

Philoid · Accepted Answer

As seen from the figure initial state of A is the final state of B and vice-versa.

So, the change in internal energy of process A and B will be equal in magnitude but with a negative sign.

So ΔU₁=-ΔU₂

∴ ΔU₁+ΔU₂=0.

Refer to figure. Let ΔU1 and ΔU2 be the change in internal energy in processes A and B respectively, ΔQ be the net heat given to the system in process A + B and ΔW be the net work done by the system in the process A + B.

Refer to figure. Let ΔU₁ and ΔU₂ be the change in internal energy in processes A and B respectively, ΔQ be the net heat given to the system in process A + B and ΔW be the net work done by the system in the process A + B.