The heat absorbed in process II will be 40 J.
In this problem, we are given that the work done in process I is 80 J and the work done in process II is 60 J. We need to find the heat absorbed in process II.
From the first law of thermodynamics, we know that the change in internal energy of a system is equal to the heat absorbed by the system minus the work done on the system. Mathematically, this can be represented as:
ΔU = Q - W
Since process I and process II combined constitute the entire process from state a to state b, we can equate the total change in internal energy to the sum of the individual changes:
ΔU = ΔU1 + ΔU2
Since the internal energy of an ideal gas depends only on its temperature, we can write the change in internal energy as:
ΔU = ΔT × nCv
where ΔT is the change in temperature, n is the number of moles of gas, and Cv is the molar specific heat at constant volume.
For an ideal gas, Cv is given by the equation Cv = (f/2)R, where f is the degrees of freedom of the gas molecule and R is the molar gas constant.
Since the process is adiabatic (no heat is transferred), the change in internal energy is equal to 0. Therefore, we have:
0 = Q1 - W1 + Q2 - W2
Substituting the known values, we have:
0 = 100 J - 80 J + Q2 - 60 J
Simplifying the equation, we find:
Q2 = 40 J