During the expansion process the volume of the gas changes form `4m^(3)` to `6m^(3)` while the pressure change according to `p = 30 V +100` where pressure is in `Pa` and volume is in `m^(3)`. The work done by gas in `N xx 10^(2)J`. Find `N`.

Answer

To solve the problem, we need to calculate the work done by the gas during its expansion from a volume of \(4 \, m^3\) to \(6 \, m^3\) given the pressure-volume relationship \(p = 30V + 100\).

### Step-by-Step Solution:

1. **Identify the Work Done Formula**:
The work done \(W\) by the gas during expansion can be calculated using the formula:
\[ W = \int_{V_1}^{V_2} P \, dV \]
where \(P\) is the pressure and \(V_1\) and \(V_2\) are the initial and final volumes, respectively.

2. **Substitute the Pressure Equation**:
From the problem, we have the pressure as a function of volume:
\[ P = 30V + 100 \]
So, we can substitute this into the work done formula:
\[ W = \int_{4}^{6} (30V + 100) \, dV \]

3. **Split the Integral**:
We can split the integral into two parts:
\[ W = \int_{4}^{6} 30V \, dV + \int_{4}^{6} 100 \, dV \]

4. **Calculate Each Integral**:
- For the first integral:
\[ \int 30V \, dV = 30 \cdot \frac{V^2}{2} = 15V^2 \]
Evaluating from 4 to 6:
\[ 15(6^2) - 15(4^2) = 15(36) - 15(16) = 540 - 240 = 300 \]

- For the second integral:
\[ \int 100 \, dV = 100V \]
Evaluating from 4 to 6:
\[ 100(6) - 100(4) = 600 - 400 = 200 \]

5. **Combine the Results**:
Now, we can combine the results from both integrals to find the total work done:
\[ W = 300 + 200 = 500 \, J \]

6. **Convert to Required Format**:
The problem states that the work done is expressed as \(N \times 10^2 \, J\). We have:
\[ 500 \, J = 5 \times 10^2 \, J \]
Therefore, \(N = 5\).

### Final Answer:
The value of \(N\) is \(5\).