In a cyclic process initially a gas is at `10^5` Pa pressure and its volume is `2 m^3`. First it undergoes an isobaric expansion to increase its volume to `2.5 m^3`. The in an isochoric process its pressure is doubled. Now the gas is brought back to its initial state by changing the pressure of gas linearly with its volume. Find the total amount of heat supplied to the gas in the process.

Answer

To solve the problem step by step, we need to analyze the cyclic process involving the gas and calculate the total amount of heat supplied to the gas during the process.

### Step 1: Understand the Process
The gas undergoes three main processes:
1. **Isobaric Expansion**: The gas expands at constant pressure from an initial volume of \( V_1 = 2 \, m^3 \) to a final volume of \( V_2 = 2.5 \, m^3 \) while the pressure remains constant at \( P_1 = 10^5 \, Pa \).
2. **Isochoric Process**: The pressure of the gas is doubled at constant volume. The new pressure becomes \( P_2 = 2 \times 10^5 \, Pa \) while the volume remains \( V_2 = 2.5 \, m^3 \).
3. **Linear Return to Initial State**: The gas returns to its initial state (pressure \( P_1 \) and volume \( V_1 \)) by changing pressure linearly with volume.

### Step 2: Calculate Work Done in Each Process
1. **Isobaric Expansion (1 to 2)**:
- Work done \( W_{12} = P \Delta V = P (V_2 - V_1) \)
- \( W_{12} = 10^5 \, Pa \times (2.5 \, m^3 - 2 \, m^3) = 10^5 \, Pa \times 0.5 \, m^3 = 50000 \, J \)

2. **Isochoric Process (2 to 3)**:
- Since the volume does not change, the work done \( W_{23} = 0 \).

3. **Linear Return to Initial State (3 to 1)**:
- The process can be represented as a straight line in the PV diagram. The area under the curve represents the work done.
- The pressure changes from \( P_2 = 2 \times 10^5 \, Pa \) at \( V_2 = 2.5 \, m^3 \) back to \( P_1 = 10^5 \, Pa \) at \( V_1 = 2 \, m^3 \).
- The work done \( W_{31} \) can be calculated as the area of the triangle formed in the PV diagram:
\[ W_{31} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times (2.5 - 2) \times (2 \times 10^5 - 10^5) \]
\[ W_{31} = \frac{1}{2} \times 0.5 \times 10^5 = 25000 \, J \]

### Step 3: Total Work Done in the Cycle
The total work done \( W \) over the entire cycle is:
\[ W = W_{12} + W_{23} + W_{31} = 50000 \, J + 0 + 25000 \, J = 75000 \, J \]

### Step 4: Apply the First Law of Thermodynamics
For a cyclic process, the change in internal energy \( \Delta U \) is zero. Thus, from the first law of thermodynamics:
\[ \Delta Q = \Delta U + W \implies \Delta Q = 0 + W \]
Since the work done is positive (work done by the gas), the heat supplied to the gas is equal to the work done:
\[ \Delta Q = 75000 \, J \]

### Final Answer
The total amount of heat supplied to the gas in the process is \( 75000 \, J \).

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