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 find the work done by the gas during the expansion process, we can follow these steps:

### Step 1: Understand the Work Done Formula
The work done by the gas during an expansion or compression is given by the integral of pressure with respect to volume:
\[ W = \int_{V_i}^{V_f} P \, dV \]
where \( V_i \) is the initial volume and \( V_f \) is the final volume.

### Step 2: Substitute the Pressure Equation
Given the pressure as a function of volume:
\[ P = 30V + 100 \]
We can substitute this into the work done formula:
\[ W = \int_{4}^{6} (30V + 100) \, dV \]

### Step 3: Break Down the Integral
We can separate the integral into two parts:
\[ W = \int_{4}^{6} 30V \, dV + \int_{4}^{6} 100 \, dV \]

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

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

### Step 5: Combine the Results
Now, add the results of both integrals to find the total work done:
\[ W = 300 + 200 = 500 \, \text{J} \]

### Step 6: Express the Work Done in the Required Format
The problem states that the work done can be expressed as \( N \times 10^2 \, \text{J} \). We have:
\[ 500 \, \text{J} = 5 \times 10^2 \, \text{J} \]
Thus, by comparing, we find:
\[ N = 5 \]

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

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