A certain balloon maintains an internal gas pressure of `P_(0) = 100 k Pa` until the volume reaches `V_(0) = 20 m^(3)`. Beyond a volume of `20 m^(3)`, the internal pressure varies as `P = P_(0) + 2 k (V - V_(0))^(2)` where `P` is in kPa. `V` is in `m^(3)` and `k` is a constant `(k = 1 kPa//m^(3))`. Initially the balloon contains helium gas at `20^(@)C`, 100 kPa with a `15 m^(3)` volume. The balloon is then heated until the volume becomes `25 m^(3)` and the pressure is 150 kPa. Assume ideal gas behavior for helium. The work done by the balloon for the entire process in kJ is

Options

A. 1256
B. 1414
C. 1083
D. 1512

Explanation

To solve the problem, we need to calculate the work done by the balloon during the heating process as it expands from an initial volume of 15 m³ to a final volume of 25 m³. The work done can be calculated in two parts: from 15 m³ to 20 m³ where the pressure is constant, and from 20 m³ to 25 m³ where the pressure varies according to the given equation.

### Step 1: Calculate Work Done from 15 m³ to 20 m³

The pressure \( P_0 \) is constant at 100 kPa during this segment. The work done \( W_1 \) can be calculated using the formula:

\[ W_1 = P_0 \int_{V_1}^{V_2} dV = P_0 (V_2 - V_1) \]

Where:
- \( P_0 = 100 \, \text{kPa} = 100 \times 10^3 \, \text{Pa} \)
- \( V_1 = 15 \, \text{m}^3 \)
- \( V_2 = 20 \, \text{m}^3 \)

Substituting the values:

\[ W_1 = 100 \times 10^3 \times (20 - 15) = 100 \times 10^3 \times 5 = 500 \times 10^3 \, \text{J} = 500 \, \text{kJ} \]

### Step 2: Calculate Work Done from 20 m³ to 25 m³

In this segment, the pressure varies according to the equation:

\[ P = P_0 + 2k(V - V_0)^2 \]

Where:
- \( P_0 = 100 \, \text{kPa} \)
- \( k = 1 \, \text{kPa/m}^3 \)
- \( V_0 = 20 \, \text{m}^3 \)

The work done \( W_2 \) can be calculated using:

\[ W_2 = \int_{V_2}^{V_3} P \, dV = \int_{20}^{25} \left(100 + 2 \times 1 \times (V - 20)^2\right) dV \]

Calculating the integral:

\[ W_2 = \int_{20}^{25} \left(100 + 2(V - 20)^2\right) dV \]

This can be split into two integrals:

\[ W_2 = \int_{20}^{25} 100 \, dV + \int_{20}^{25} 2(V - 20)^2 \, dV \]

Calculating the first integral:

\[ \int_{20}^{25} 100 \, dV = 100 \times (25 - 20) = 500 \, \text{J} \]

Calculating the second integral:

Let \( u = V - 20 \), then \( dV = du \) and the limits change from \( 0 \) to \( 5 \):

\[ \int_{0}^{5} 2u^2 \, du = 2 \left[\frac{u^3}{3}\right]_{0}^{5} = 2 \left[\frac{5^3}{3} - 0\right] = 2 \left[\frac{125}{3}\right] = \frac{250}{3} \, \text{J} \]

Now, substituting back into \( W_2 \):

\[ W_2 = 500 + \frac{250}{3} \, \text{J} = \frac{1500 + 250}{3} = \frac{1750}{3} \, \text{J} \]

### Step 3: Total Work Done

Now, we can find the total work done \( W \):

\[ W = W_1 + W_2 = 500 \, \text{kJ} + \frac{1750}{3} \, \text{J} \times \frac{1 \, \text{kJ}}{1000 \, \text{J}} = 500 + \frac{1.75}{3} \, \text{kJ} \]

Calculating \( \frac{1.75}{3} \):

\[ \frac{1.75}{3} \approx 0.5833 \, \text{kJ} \]

So,

\[ W \approx 500 + 0.5833 \approx 500.5833 \, \text{kJ} \]

### Final Answer

The total work done by the balloon for the entire process is approximately:

\[ \boxed{500.58 \, \text{kJ}} \]