A sample of an ideal gas is expanded `1 m^(3)` to `3 m^(3)` in a reversible process for which `P=KV^(2)` , with `K=6 "bar"//m^(6).` What is work done by the gas (in kJ) ?

Options

A. 5200 KJ
B. 15600 kJ
C. 52 kJ
D. 5267 .6 kJ

Explanation

To solve the problem of calculating the work done by an ideal gas during a reversible expansion from \(1 \, m^3\) to \(3 \, m^3\) with the pressure given by \(P = KV^2\) (where \(K = 6 \, \text{bar/m}^6\)), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Work Done Formula**:
The work done \(W\) by the gas during an expansion is given by the integral:
\[ W = \int_{V_1}^{V_2} P \, dV \]
where \(V_1\) is the initial volume and \(V_2\) is the final volume.

2. **Substitute the Pressure Equation**:
Given that \(P = KV^2\), we can substitute this into the work formula:
\[ W = \int_{1}^{3} KV^2 \, dV \]

3. **Factor Out the Constant**:
Since \(K\) is a constant, it can be factored out of the integral:
\[ W = K \int_{1}^{3} V^2 \, dV \]

4. **Integrate \(V^2\)**:
The integral of \(V^2\) is:
\[ \int V^2 \, dV = \frac{V^3}{3} \]
Therefore, we evaluate this from \(1\) to \(3\):
\[ W = K \left[ \frac{V^3}{3} \right]_{1}^{3} = K \left( \frac{3^3}{3} - \frac{1^3}{3} \right) \]

5. **Calculate the Values**:
Now calculate the values:
\[ W = K \left( \frac{27}{3} - \frac{1}{3} \right) = K \left( 9 - \frac{1}{3} \right) = K \left( \frac{27 - 1}{3} \right) = K \left( \frac{26}{3} \right) \]

6. **Substitute the Value of \(K\)**:
Substitute \(K = 6 \, \text{bar/m}^6\):
\[ W = 6 \left( \frac{26}{3} \right) = \frac{156}{3} = 52 \, \text{bar} \cdot m^3 \]

7. **Convert to Kilojoules**:
Since \(1 \, \text{bar} \cdot m^3 = 100 \, \text{kJ}\):
\[ W = 52 \, \text{bar} \cdot m^3 \times 100 \, \text{kJ/bar} = 5200 \, \text{kJ} \]

### Final Answer:
The work done by the gas is \(5200 \, \text{kJ}\).