Refer to figure. Let `DeltaU_(1)` and `DeltaU_(2)` be the changes in internal energy of the system in the processes A and B. then

Options

A. `DeltaU_(1)gtDeltaU_(2)`
B. `DeltaU_(1)=DeltaU_(2)`
C. `DeltaU_(1)ltDeltaU_(2)`
D. `DeltaU_(1)!=DeltaU_(2)`

Explanation

To solve the problem, we need to analyze the changes in internal energy (ΔU) during two processes (A and B) that form a cyclic process. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Cyclic Process
The system starts from state 1, undergoes process A to reach state 2, and then undergoes process B to return to state 1. Since the system returns to its initial state, this is a cyclic process.

**Hint:** Remember that in a cyclic process, the initial and final states are the same.

### Step 2: Apply the First Law of Thermodynamics
According to the first law of thermodynamics, the net heat added to the system over a complete cycle is equal to the net work done by the system. Mathematically, this can be expressed as:
\[ \oint dQ = \oint dW \]
where the integral is taken over the entire cycle.

**Hint:** The first law relates heat, work, and internal energy changes.

### Step 3: Relate Heat, Work, and Internal Energy
For any thermodynamic process, the first law can be expressed as:
\[ dQ = dU + dW \]
Integrating this over the entire cycle gives:
\[ \oint dQ = \oint dU + \oint dW \]

**Hint:** This equation shows how changes in internal energy relate to heat and work.

### Step 4: Analyze the Cyclic Integral
Since the system returns to its initial state, the change in internal energy over the entire cycle is zero:
\[ \oint dU = 0 \]
This implies that the sum of the changes in internal energy for the two processes (A and B) must also equal zero:
\[ \Delta U_A + \Delta U_B = 0 \]

**Hint:** The internal energy change for the entire cycle is zero because the system returns to its original state.

### Step 5: Relate Changes in Internal Energy
From the above equation, we can express the change in internal energy for process A (ΔU1) and process B (ΔU2):
\[ \Delta U_1 + \Delta U_2 = 0 \implies \Delta U_1 = -\Delta U_2 \]
This means that the change in internal energy during process A is equal in magnitude but opposite in sign to the change in internal energy during process B.

**Hint:** The negative sign indicates that if one process increases internal energy, the other decreases it by the same amount.

### Step 6: Conclusion
Since ΔU1 is equal to -ΔU2, we can conclude that:
\[ \Delta U_1 = \Delta U_2 \]
This corresponds to option 2 from the question.

**Final Answer:** The correct option is that ΔU1 is equal to ΔU2.