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^(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 exterted from the gas in the complete cycle ? (b) how much heat was supplied or extracted ?

Explanation

To solve the problem step by step, we will analyze the processes involved in the gas cycle and apply the first law of thermodynamics.

### Step 1: Identify the Initial State
The gas is initially at:
- Pressure (P1) = 100 kPa = 100,000 Pa
- Volume (V1) = 2.0 m³

### Step 2: Process 1 - Isobaric Expansion
The gas undergoes an isobaric (constant pressure) expansion from:
- Initial Volume (V1) = 2.0 m³
- Final Volume (V2) = 2.5 m³

Since the pressure remains constant at 100 kPa, we can calculate the work done (W1) during this process using the formula:
\[ W_1 = P \Delta V = P (V_2 - V_1) \]
\[ W_1 = 100,000 \, \text{Pa} \times (2.5 - 2.0) \, \text{m}^3 \]
\[ W_1 = 100,000 \, \text{Pa} \times 0.5 \, \text{m}^3 = 50,000 \, \text{J} \]

### Step 3: Process 2 - Isovolumetric Heating
Next, the volume is kept constant at V2 = 2.5 m³, and the pressure is increased from 100 kPa to 200 kPa. Since the volume does not change, the work done (W2) during this process is:
\[ W_2 = 0 \, \text{J} \]

### Step 4: Process 3 - Isobaric Compression
Finally, the gas returns to its initial state (P1, V1) by varying the pressure linearly with volume. The pressure decreases from 200 kPa back to 100 kPa while the volume decreases from 2.5 m³ back to 2.0 m³.

The work done (W3) during this process can be calculated as the area of the triangle formed in the PV diagram:
\[ W_3 = \frac{1}{2} \times \text{Base} \times \text{Height} \]
Where:
- Base = \( V_2 - V_1 = 2.5 - 2.0 = 0.5 \, \text{m}^3 \)
- Height = \( 200 \, \text{kPa} - 100 \, \text{kPa} = 100 \, \text{kPa} = 100,000 \, \text{Pa} \)

Thus,
\[ W_3 = \frac{1}{2} \times 0.5 \, \text{m}^3 \times 100,000 \, \text{Pa} \]
\[ W_3 = \frac{1}{2} \times 0.5 \times 100,000 = 25,000 \, \text{J} \]

### Step 5: Total Work Done
The total work done in the entire cycle (W_total) is:
\[ W_{\text{total}} = W_1 + W_2 + W_3 \]
\[ W_{\text{total}} = 50,000 \, \text{J} + 0 + (-25,000 \, \text{J}) \]
\[ W_{\text{total}} = 25,000 \, \text{J} \]

### Step 6: Apply the First Law of Thermodynamics
According to the first law of thermodynamics:
\[ \Delta U = Q - W \]
Where:
- \( \Delta U \) = change in internal energy (0 for a complete cycle)
- \( Q \) = heat added to the system
- \( W \) = work done by the system

Since \( \Delta U = 0 \):
\[ 0 = Q - W_{\text{total}} \]
Thus,
\[ Q = W_{\text{total}} = 25,000 \, \text{J} \]

### Conclusion
(a) Heat is supplied to the gas in the complete cycle.
(b) The amount of heat supplied is 25,000 J.