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) ?

Answer

To find the work done by the gas during its expansion from \(1 \, m^3\) to \(3 \, m^3\) under the given conditions, we can follow these steps:

### Step 1: Understand 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 \(V_1\) and \(V_2\) are the initial and final volumes, respectively.

### Step 2: Substitute the Pressure Equation
We are given that the pressure \(P\) is related to the volume \(V\) by the equation:
\[ P = K V^2 \]
where \(K = 6 \, \text{bar/m}^6\). Substituting this into the work done formula gives:
\[ W = \int_{1}^{3} K V^2 \, dV \]

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

### Step 4: Evaluate the Integral
Now we need to evaluate the integral \(\int V^2 \, dV\):
\[ \int V^2 \, dV = \frac{V^3}{3} \]
Now we can apply the limits from \(1\) to \(3\):
\[ W = K \left[ \frac{V^3}{3} \right]_{1}^{3} = K \left( \frac{3^3}{3} - \frac{1^3}{3} \right) \]
Calculating this gives:
\[ W = K \left( \frac{27}{3} - \frac{1}{3} \right) = K \left( 9 - \frac{1}{3} \right) = K \left( \frac{27 - 1}{3} \right) = K \left( \frac{26}{3} \right) \]

### Step 5: Substitute the Value of K
Now substituting \(K = 6 \, \text{bar/m}^6\):
\[ W = 6 \left( \frac{26}{3} \right) = \frac{156}{3} = 52 \, \text{bar} \cdot m^3 \]

### Step 6: Convert Units
Since \(1 \, \text{bar} \cdot m^3 = 100 \, \text{kJ}\), we convert the work done:
\[ W = 52 \, \text{bar} \cdot m^3 \times 100 \, \text{kJ/bar} = 5200 \, \text{kJ} \]

### Final Answer
Thus, the work done by the gas is:
\[ \boxed{5200 \, \text{kJ}} \]