A quantity of a substance in a closed system is made to undergo a reversible process from an initial volume of `3 m^(3)` and initial pressure `10^(5) N//m^(2)` to a final volume of `5m^(3)`. If the pressure is proportional to the square of the volume `(i.e,P=AV^(2))`, the work done by the substance will be

Options

A. `3.6xx10^(2)J`
B. `7.4xx10^(3)J`
C. `2.2xx10^(4)J`
D. `3.6xx10^(5)J`

Answer

To solve the problem, we will calculate the work done by the substance during the reversible process from an initial volume of \(3 \, m^3\) and initial pressure \(10^5 \, N/m^2\) to a final volume of \(5 \, m^3\), given that the pressure is proportional to the square of the volume (\(P = A V^2\)).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial Volume, \(V_1 = 3 \, m^3\)
- Final Volume, \(V_2 = 5 \, m^3\)
- Initial Pressure, \(P_1 = 10^5 \, N/m^2\)

2. **Use the Relationship Between Pressure and Volume:**
- We know that \(P = A V^2\). We need to find the constant \(A\).
- Substitute the initial conditions into the equation:
\[ P_1 = A V_1^2 \implies 10^5 = A (3^2) \implies 10^5 = A \cdot 9 \]
- Solve for \(A\):
\[ A = \frac{10^5}{9} \]

3. **Substitute \(A\) Back into the Pressure Equation:**
- Now, we can express pressure \(P\) in terms of volume \(V\):
\[ P = \frac{10^5}{9} V^2 \]

4. **Set Up the Work Done Integral:**
- The work done \(W\) during the expansion from \(V_1\) to \(V_2\) is given by:
\[ W = \int_{V_1}^{V_2} P \, dV \]
- Substitute \(P\) into the integral:
\[ W = \int_{3}^{5} \frac{10^5}{9} V^2 \, dV \]

5. **Calculate the Integral:**
- Factor out the constant:
\[ W = \frac{10^5}{9} \int_{3}^{5} V^2 \, dV \]
- The integral of \(V^2\) is:
\[ \int V^2 \, dV = \frac{V^3}{3} \]
- Evaluate the integral from \(3\) to \(5\):
\[ W = \frac{10^5}{9} \left[ \frac{V^3}{3} \right]_{3}^{5} = \frac{10^5}{9} \left( \frac{5^3}{3} - \frac{3^3}{3} \right) \]

6. **Calculate the Values:**
- Calculate \(5^3\) and \(3^3\):
\[ 5^3 = 125, \quad 3^3 = 27 \]
- Substitute back:
\[ W = \frac{10^5}{9} \left( \frac{125 - 27}{3} \right) = \frac{10^5}{9} \left( \frac{98}{3} \right) \]
- Simplify:
\[ W = \frac{10^5 \cdot 98}{27} \approx 3.6296 \times 10^5 \, J \]

7. **Final Result:**
- The work done by the substance is approximately:
\[ W \approx 3.6 \times 10^5 \, J \]