A gas is initially at a pressure of `100 kPa` and its volume is `2.0 m^(3)`. Its pressure is kept constant and the volume is changed from `2.0 m^(3)` to `2.5 m^(3)`. Its volume is now kept constant and the pressure is increased from `100 kPa` to `200 kPa`. The gas is brought back to its initial state, the pressure varying linearly with its volume.
(a) Whether the heat is supplied to or exerted from the gas in the complete cycle ?
(b) How much heat was supplied or extracted ?

Answer

To solve the problem step by step, we will analyze the processes involved and calculate the heat exchanged during the complete cycle.

### Step 1: Identify the Processes
1. The gas starts at an initial state with:
- Pressure \( P_1 = 100 \, \text{kPa} = 100 \times 10^3 \, \text{Pa} \)
- Volume \( V_1 = 2.0 \, \text{m}^3 \)

2. The first process is isobaric (constant pressure) where the volume changes from \( V_1 = 2.0 \, \text{m}^3 \) to \( V_2 = 2.5 \, \text{m}^3 \).

3. The second process is isochoric (constant volume) where the pressure increases from \( P_2 = 100 \, \text{kPa} \) to \( P_3 = 200 \, \text{kPa} \) at \( V_2 = 2.5 \, \text{m}^3 \).

4. Finally, the gas returns to its initial state, which involves a linear change in pressure and volume.

### Step 2: Draw the P-V Diagram
- On the P-V diagram, plot the points:
- Point 1: \( (V_1, P_1) = (2.0, 100) \)
- Point 2: \( (V_2, P_1) = (2.5, 100) \)
- Point 3: \( (V_2, P_3) = (2.5, 200) \)
- Point 4: Back to \( (V_1, P_1) = (2.0, 100) \)

### Step 3: Calculate Work Done in the Cycle
- The work done during the isobaric process (1 to 2) can be calculated using:
\[ W_{12} = P \Delta V = P_1 (V_2 - V_1) = 100 \times 10^3 \times (2.5 - 2.0) = 100 \times 10^3 \times 0.5 = 50,000 \, \text{J} \]

- The work done during the isochoric process (2 to 3) is zero since volume does not change.

- The work done during the process from point 3 back to point 1 (3 to 4) can be calculated using the area of the triangle formed in the P-V diagram:
\[ W_{34} = -\frac{1}{2} \times \text{Base} \times \text{Height} = -\frac{1}{2} \times (V_2 - V_1) \times (P_3 - P_1) \]
\[ = -\frac{1}{2} \times (2.5 - 2.0) \times (200 - 100) \times 10^3 = -\frac{1}{2} \times 0.5 \times 100 \times 10^3 = -25,000 \, \text{J} \]

- Total work done in the cycle:
\[ W_{\text{cycle}} = W_{12} + W_{34} = 50,000 - 25,000 = 25,000 \, \text{J} \]

### Step 4: Apply the First Law of Thermodynamics
- According to the first law of thermodynamics:
\[ \Delta U = Q - W \]
For a complete cycle, \( \Delta U = 0 \), thus:
\[ Q = W \]
Since we calculated \( W_{\text{cycle}} = 25,000 \, \text{J} \), the heat exchanged is:
\[ Q = 25,000 \, \text{J} \]

### Step 5: Determine if Heat is Supplied or Extracted
- Since the work done is positive, it indicates that heat is supplied to the gas during the cycle.

### Final Answers
(a) Heat is supplied to the gas in the complete cycle.

(b) The amount of heat supplied is \( 25,000 \, \text{J} \).