One mole of a monoatomic gas is enclosed in a cylinder and occupies a volume of `4` liter at a pressure `100N//m^(2)` . It is subjected to process `T = = alphaV^(2)`, where `alpha` is a positive constant , `V` is volume of the gas and `T` is kelvin temperature. Find the work done by gas (in joule) in increasing the volume of gas to six times initial volume .

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Information
We have:
- One mole of a monoatomic gas.
- Initial volume \( V_0 = 4 \) liters.
- Initial pressure \( P_0 = 100 \, \text{N/m}^2 \).
- The process is defined by the equation \( T = \alpha V^2 \), where \( \alpha \) is a constant.

### Step 2: Convert Volume to SI Units
Convert the volume from liters to cubic meters:
\[ V_0 = 4 \, \text{liters} = 4 \times 10^{-3} \, \text{m}^3 \]

### Step 3: Determine the Final Volume
The final volume \( V_f \) is six times the initial volume:
\[ V_f = 6 V_0 = 6 \times 4 \times 10^{-3} = 24 \times 10^{-3} \, \text{m}^3 \]

### Step 4: Use the Ideal Gas Law
From the ideal gas law, we know:
\[ PV = nRT \]
Rearranging gives us:
\[ P = \frac{nRT}{V} \]
Substituting \( T = \alpha V^2 \) into the equation:
\[ P = \frac{nR(\alpha V^2)}{V} = nR\alpha V \]

### Step 5: Calculate Work Done
The work done \( W \) by the gas during the expansion is given by:
\[ W = \int_{V_0}^{V_f} P \, dV \]
Substituting for \( P \):
\[ W = \int_{V_0}^{V_f} nR\alpha V \, dV \]
Since \( n \), \( R \), and \( \alpha \) are constants, we can factor them out of the integral:
\[ W = nR\alpha \int_{V_0}^{V_f} V \, dV \]

### Step 6: Evaluate the Integral
The integral of \( V \) is:
\[ \int V \, dV = \frac{V^2}{2} \]
Thus:
\[ W = nR\alpha \left[ \frac{V^2}{2} \right]_{V_0}^{V_f} = nR\alpha \left( \frac{(V_f)^2}{2} - \frac{(V_0)^2}{2} \right) \]
Substituting \( V_f = 24 \times 10^{-3} \, \text{m}^3 \) and \( V_0 = 4 \times 10^{-3} \, \text{m}^3 \):
\[ W = nR\alpha \left( \frac{(24 \times 10^{-3})^2}{2} - \frac{(4 \times 10^{-3})^2}{2} \right) \]

### Step 7: Calculate the Values
Calculating the squares:
\[ (24 \times 10^{-3})^2 = 576 \times 10^{-6} \, \text{m}^6 \]
\[ (4 \times 10^{-3})^2 = 16 \times 10^{-6} \, \text{m}^6 \]
Thus:
\[ W = nR\alpha \left( \frac{576 \times 10^{-6} - 16 \times 10^{-6}}{2} \right) = nR\alpha \left( \frac{560 \times 10^{-6}}{2} \right) = nR\alpha \cdot 280 \times 10^{-6} \]

### Step 8: Substitute Known Values
We know:
- \( n = 1 \) mole,
- \( R = 8.314 \, \text{J/(mol K)} \),
- To find \( \alpha \), we can use the initial conditions. From the ideal gas law, we can find \( T \) at \( V_0 \):
\[ T_0 = \frac{P_0 V_0}{nR} = \frac{100 \times 4 \times 10^{-3}}{1 \times 8.314} \approx 4.8 \, \text{K} \]
Using \( T_0 = \alpha V_0^2 \):
\[ \alpha = \frac{T_0}{V_0^2} = \frac{4.8}{(4 \times 10^{-3})^2} = \frac{4.8}{16 \times 10^{-6}} = 300000 \, \text{K/m}^6 \]

### Step 9: Final Calculation of Work Done
Now substituting \( \alpha \) into the work done formula:
\[ W = 1 \times 8.314 \times 300000 \times 280 \times 10^{-6} \]
Calculating this gives:
\[ W \approx 7 \, \text{J} \]

### Final Answer
The work done by the gas in increasing the volume to six times the initial volume is approximately:
\[ \boxed{7 \, \text{J}} \]