To solve the problem of calculating the work done by the gas during its expansion through a quasi-static process described by the equation \( P = \alpha V^2 \), we will follow these steps:
### Step-by-Step Solution:
1. **Identify the Given Values:**
- The constant \( \alpha = 5.00 \, \text{atm/m}^6 \).
- The initial volume \( V_1 = 1.00 \, \text{m}^3 \).
- The final volume \( V_2 = 2 \times V_1 = 2.00 \, \text{m}^3 \).
2. **Convert Pressure Units:**
- Convert \( \alpha \) from atm to Pascals:
\[
1 \, \text{atm} = 101325 \, \text{Pa}
\]
Therefore,
\[
\alpha = 5.00 \, \text{atm/m}^6 = 5.00 \times 101325 \, \text{Pa/m}^6 = 506625 \, \text{Pa/m}^6
\]
3. **Set Up the Work Integral:**
- The work done \( W \) by the gas during expansion is given by the integral:
\[
W = \int_{V_1}^{V_2} P \, dV
\]
- Substitute \( P = \alpha V^2 \) into the integral:
\[
W = \int_{V_1}^{V_2} \alpha V^2 \, dV
\]
4. **Factor Out the Constant:**
- Since \( \alpha \) is a constant, we can take it out of the integral:
\[
W = \alpha \int_{V_1}^{V_2} V^2 \, dV
\]
5. **Evaluate the Integral:**
- The integral of \( V^2 \) is:
\[
\int V^2 \, dV = \frac{V^3}{3}
\]
- Thus, we evaluate from \( V_1 \) to \( V_2 \):
\[
W = \alpha \left[ \frac{V^3}{3} \right]_{V_1}^{V_2} = \alpha \left( \frac{(V_2)^3}{3} - \frac{(V_1)^3}{3} \right)
\]
6. **Substitute the Limits:**
- Substitute \( V_1 = 1 \, \text{m}^3 \) and \( V_2 = 2 \, \text{m}^3 \):
\[
W = \alpha \left( \frac{(2)^3}{3} - \frac{(1)^3}{3} \right) = \alpha \left( \frac{8}{3} - \frac{1}{3} \right) = \alpha \left( \frac{7}{3} \right)
\]
7. **Calculate the Work Done:**
- Substitute the value of \( \alpha \):
\[
W = 506625 \, \text{Pa/m}^6 \times \frac{7}{3} = \frac{7 \times 506625}{3} \, \text{J}
\]
- Calculate:
\[
W = \frac{3546375}{3} \, \text{J} \approx 1182125 \, \text{J}
\]
8. **Convert to MegaJoules:**
- Finally, convert the work done into MegaJoules:
\[
W \approx 1.18 \, \text{MJ}
\]
### Final Answer:
The work done by the gas in expanding is approximately \( 1.18 \, \text{MJ} \).
