Key idea: both processes connect the same initial and final states

From the $P$-$V$ diagram, processes $A$ and $B$ start from the same initial state and end at the same final state.

Therefore, the change in internal energy depends only on the initial and final states, not on the path.

$\Delta U_A = \Delta U_B$

Use the first law of thermodynamics

$\Delta Q = \Delta U + W$

where $W$ is the work done by the gas.

Compare work done in the two paths

Work done is the area under the curve in a $P$-$V$ diagram:

$W = \int P\, dV$

Since both processes involve the same change in volume, the path with higher pressure at intermediate points gives larger area under the curve, hence larger work.

From the figure, path $A$ lies above path $B$, so

$W_A > W_B$

Compare heat absorbed

Since

$\Delta Q_A = \Delta U_A + W_A$ $\Delta Q_B = \Delta U_B + W_B$

and we already have

$\Delta U_A = \Delta U_B$ $W_A > W_B$

therefore,

$\Delta Q_A > \Delta Q_B$

Final conclusion

$\Delta Q_A > \Delta Q_B \quad \text{and} \quad \Delta U_A = \Delta U_B$

So the correct option is C.