An ideal gas undergoes a cyclic process as shown in Figure. ΔUBC = -9 kJ/mol, qAB = 2 kJ/mol, WAB = -9 kJ/mol, WCA = 3 kJ/mol. Heat absorbed by the system during process CA is:

ΔUAB = qAB + WAB = 2 + (-9) = -7 kJ/mol
For a cyclic process, the total change in internal energy is zero. Therefore, ΔUtotal = ΔUAB + ΔUBC + ΔUCA = 0.
ΔUCA = -ΔUAB - ΔUBC = -(-7) - (-9) = 16 kJ/mol
Since ΔUCA = qCA + WCA, we can find qCA:
qCA = ΔUCA - WCA = 16 - 3 = 13 kJ/mol
However, this value is not among the options. Let's reconsider the given data.
ΔUBC = -9 kJ/mol
qAB = 2 kJ/mol
WAB = -9 kJ/mol
WCA = 3 kJ/mol
For a cyclic process, the total work done is the sum of the work done in each step. The total change in internal energy is zero.
Wtotal = WAB + WBC + WCA = 0
ΔUtotal = ΔUAB + ΔUBC + ΔUCA = 0
We know that ΔU = q + W. Let's look at process AB:
ΔUAB = qAB + WAB = 2 kJ/mol + (-9 kJ/mol) = -7 kJ/mol
For the entire cycle, ΔUtotal = 0. Therefore:
ΔUAB + ΔUBC + ΔUCA = 0
-7 kJ/mol + (-9 kJ/mol) + ΔUCA = 0
ΔUCA = 16 kJ/mol
Now we use the first law of thermodynamics for process CA:
ΔUCA = qCA + WCA
16 kJ/mol = qCA + 3 kJ/mol
qCA = 13 kJ/mol
Since 13 kJ/mol is not in the options, there might be an error in the provided data or options.