W17.The pV-diagram of figure below shows a series of thermodynamic processes.
In process ab, 150J of heat is added to the system; in process bd, 600J of heat
is added. Find
(b) the internal energy change in process ab;
(c) the internal energy change in process abd; and
(d) the total heat added in process acd.
To find the internal energy change in each process, we can use the first law of thermodynamics:
ΔU = Q - W
Where:
ΔU = change in internal energy
Q = heat added to the system
W = work done by the system
(a) Process ab:
Given:
Q_ab = 150J (heat added)
W_ab = 0 (since there is no area enclosed by the process ab on the PV diagram)
Therefore:
ΔU_ab = Q_ab - W_ab
ΔU_ab = 150J - 0
ΔU_ab = 150J
(b) The internal energy change in process ab is 150J.
(c) Process abd:
To find the internal energy change in process abd, we need to add the internal energy changes of processes ab and bd:
ΔU_abd = ΔU_ab + ΔU_bd
ΔU_abd = 150J + ΔU_bd
Given:
Q_bd = 600J (heat added in process bd)
W_bd = 300J (work done by the system in process bd)
Therefore:
ΔU_bd = Q_bd - W_bd
ΔU_bd = 600J - 300J
ΔU_bd = 300J
Therefore:
ΔU_abd = 150J + 300J
ΔU_abd = 450J
(c) The internal energy change in process abd is 450J.
(d) Total heat added in process acd:
To find the total heat added in process acd, we need to calculate the area enclosed by the process acd on the PV diagram. Since the work done in process acd is negative, the total heat added is equal to the change in internal energy in this process.
ΔU_acd = -W_acd
ΔU_acd = -300J
Therefore, the total heat added in process acd is 300J.