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

Explanation

To solve the problem, we need to find the amount of heat supplied to the system during a two-step process that takes the gas from its initial state to its final state. The two steps are an isobaric expansion followed by an isochoric process.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial pressure, \( P_i = 10^5 \, \text{Pa} \)
- Initial volume, \( V_i = 10^{-3} \, \text{m}^3 \)
- Final pressure, \( P_f = \frac{1}{32} \times 10^5 \, \text{Pa} = 3.125 \times 10^4 \, \text{Pa} \)
- Final volume, \( V_f = 8 \times 10^{-3} \, \text{m}^3 \)

2. **Calculate Work Done in the Isobaric Process:**
- The work done during an isobaric process (constant pressure) is given by:
\[ W_{isobaric} = P_i (V_f - V_i) \]
- Substituting the values:
\[ W_{isobaric} = 10^5 \, \text{Pa} \times (8 \times 10^{-3} \, \text{m}^3 - 10^{-3} \, \text{m}^3) = 10^5 \times 7 \times 10^{-3} = 700 \, \text{J} \]

3. **Calculate Change in Internal Energy (\( \Delta U \)):**
- For an ideal gas, the change in internal energy during an isochoric process is given by:
\[ \Delta U = n C_v \Delta T \]
- However, we can also use the first law of thermodynamics:
\[ Q = \Delta U + W \]
- Since we are considering the entire process, we need to find the total change in internal energy from initial to final state.

4. **Calculate the Change in Internal Energy:**
- The internal energy change during the adiabatic process can be calculated using:
\[ \Delta U = -W_{adiabatic} \]
- From the video transcript, we found that the work done during the adiabatic process is \( W_{adiabatic} = 112.5 \, \text{J} \). Thus:
\[ \Delta U = -112.5 \, \text{J} \]

5. **Calculate the Total Heat Supplied:**
- Using the first law of thermodynamics:
\[ Q_{total} = \Delta U + W_{isobaric} \]
- Substituting the values:
\[ Q_{total} = -112.5 \, \text{J} + 700 \, \text{J} = 587.5 \, \text{J} \]

### Final Answer:
The amount of heat supplied to the system in the two-step process is approximately \( Q \approx 587.5 \, \text{J} \).