Following figure shows two processes A and B for a gas. If $$\Delta Q_A$$ and $$\Delta Q_B$$ are the amount of heat absorbed by the system in two cases, and $$\Delta U_A$$ and $$\Delta U_B$$ are changes in internal energies, respectively, then:

Based on the First Law of Thermodynamics and the properties of state functions, the solution is as follows:

1. Internal Energy Change ($$\Delta U$$):

Internal energy is a state function, meaning it only depends on the initial and final states of the system, not the path taken. Since both processes A and B start and end at the same points on the P-V diagram:

$$\Delta U_A = \Delta U_B$$

2. Work Done (W):

Work done in a thermodynamic process is equal to the area under the curve on a P-V diagram.

• Process A stays "above" process B.
• Therefore, the area under curve A is greater than the area under curve B.$$W_A > W_B$$

3. Heat Absorbed ($$\Delta Q$$):

According to the First Law of Thermodynamics:

$$\Delta Q = \Delta U + W$$

Applying this to both processes:

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

Since $$\Delta U$$ is the same for both but $$W_A > W_B$$, it follows that:

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