A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure `P_i=10^5` Pa and volume `V_i=10^-3m^3` changes to a final state at `P_f=(1//32)xx10^5Pa and V_f=8xx10^-3m^3` in an adiabatic quasi-static process, such that `P^3V^3=constant.` Consider another thermodynamic process that brings the system form the same initial state to the same final state in two steps: an isobaric expansion at `P_i` followed by an isochoric (isovolumetric ) process at volume `V_r.` The amount of heat supplied to the system i the two-step process is approximately

Options

A. `112 J`
B. `294 J`
C. `588 J`
D. `813 J`

Explanation

To solve the problem step-by-step, we will follow the thermodynamic processes described in the question.

### Step 1: Understand the Initial and Final States
- **Initial State**:
- Pressure, \( P_i = 10^5 \, \text{Pa} \)
- Volume, \( V_i = 10^{-3} \, \text{m}^3 \)

- **Final State**:
- Pressure, \( P_f = \frac{1}{32} \times 10^5 \, \text{Pa} = 3.125 \times 10^4 \, \text{Pa} \)
- Volume, \( V_f = 8 \times 10^{-3} \, \text{m}^3 \)

### Step 2: Identify the Processes
We need to analyze two processes:
1. An **adiabatic process** where \( P^3 V^5 = \text{constant} \).
2. A **two-step process** consisting of:
- An **isobaric expansion** at \( P_i \).
- An **isochoric process** at volume \( V_r \).

### Step 3: Calculate Work Done in the Adiabatic Process
Using the formula for work done in an adiabatic process:
\[ W = \frac{P_f V_f - P_i V_i}{1 - \gamma} \]
where \( \gamma = \frac{5}{3} \).

Substituting the values:
- \( P_f = 3.125 \times 10^4 \, \text{Pa} \)
- \( V_f = 8 \times 10^{-3} \, \text{m}^3 \)
- \( P_i = 10^5 \, \text{Pa} \)
- \( V_i = 10^{-3} \, \text{m}^3 \)

Calculating:
\[ W = \frac{(3.125 \times 10^4 \times 8 \times 10^{-3}) - (10^5 \times 10^{-3})}{1 - \frac{5}{3}} \]
\[ W = \frac{(250 - 100)}{-\frac{2}{3}} = \frac{150}{-\frac{2}{3}} = -112.5 \, \text{J} \]

### Step 4: Change in Internal Energy
For an adiabatic process, the change in internal energy \( \Delta U \) is equal to the negative of the work done:
\[ \Delta U = -W = 112.5 \, \text{J} \]

### Step 5: Analyze the Two-Step Process
1. **Isobaric Process**:
- Work done \( W_1 = P_i \Delta V \)
- \( \Delta V = V_f - V_i = 8 \times 10^{-3} - 10^{-3} = 7 \times 10^{-3} \, \text{m}^3 \)
- \( W_1 = 10^5 \times 7 \times 10^{-3} = 700 \, \text{J} \)

2. **Isochoric Process**:
- Work done \( W_2 = 0 \) (since volume does not change).

### Step 6: Apply the First Law of Thermodynamics
Using the first law of thermodynamics:
\[ Q_{\text{net}} = W_{\text{net}} + \Delta U \]
Where:
- \( W_{\text{net}} = W_1 + W_2 = 700 + 0 = 700 \, \text{J} \)
- \( \Delta U = 112.5 \, \text{J} \)

Thus:
\[ Q_{\text{net}} = 700 + 112.5 = 812.5 \, \text{J} \]

### Step 7: Final Calculation
The total amount of heat supplied to the system in the two-step process is approximately:
\[ Q_{\text{net}} \approx 812.5 \, \text{J} \]

### Conclusion
The closest option to our calculated value is \( 588 \, \text{J} \), which indicates a possible error in the interpretation of the problem or the options provided.