Gas Expansion Process $pv^2=c$ Explained

This problem involves calculating the work done by a gas during an expansion process where the relationship between pressure ($p$) and volume ($V$) is given by $pv^2 = c$. This type of process is known as a polytropic process, and in this specific case, the polytropic index ($n$) is 2.

The general formula for work done ($W$) during a polytropic process for an ideal gas is:

$$W = \frac{p_1V_1 - p_2V_2}{n-1}$$

where:

• $p_1$ is the initial pressure
• $V_1$ is the initial volume
• $p_2$ is the final pressure
• $V_2$ is the final volume
• $n$ is the polytropic index

For the case where $n=2$, the formula simplifies to:

$$W = \frac{p_1V_1 - p_2V_2}{2-1} = p_1V_1 - p_2V_2$$

Thermodynamics Parameters Given

The problem provides the following values:

• Initial Pressure, $p_1 = 350 \text{ kPa}$
• Initial Volume, $V_1 = 0.049 \text{ m}^3$
• Final Pressure, $p_2 = 150 \text{ kPa}$
• Polytropic Index, $n = 2$

Calculating Final Volume ($V_2$) for Expansion

To calculate the work done, we first need to determine the final volume ($V_2$). We use the process equation $pv^2 = c$, which implies $p_1V_1^2 = p_2V_2^2$.

Rearranging the equation to solve for $V_2$:

$$V_2^2 = V_1^2 \frac{p_1}{p_2}$$

Taking the square root of both sides:

$$V_2 = V_1 \sqrt{\frac{p_1}{p_2}}$$

Now, substitute the given values:

$$V_2 = 0.049 \text{ m}^3 \times \sqrt{\frac{350 \text{ kPa}}{150 \text{ kPa}}}$$

$$V_2 = 0.049 \text{ m}^3 \times \sqrt{\frac{7}{3}}$$

$$V_2 \approx 0.049 \text{ m}^3 \times 1.5275$$

$$V_2 \approx 0.07485 \text{ m}^3$$

For consistency with the options, we can use a slightly rounded value for $V_2$, such as $V_2 \approx 0.0749 \text{ m}^3$, for the next step.

Calculating Work Delivered in kJ

Now we can calculate the work done ($W$) using the simplified formula for $n=2$:

$$W = p_1V_1 - p_2V_2$$

Substitute the known values:

$$W = (350 \text{ kPa} \times 0.049 \text{ m}^3) - (150 \text{ kPa} \times 0.0749 \text{ m}^3)$$

First, calculate the initial $p_1V_1$ term:

$$p_1V_1 = 350 \times 0.049 = 17.15 \text{ kPa} \cdot \text{m}^3$$

Next, calculate the final $p_2V_2$ term:

$$p_2V_2 = 150 \times 0.0749 = 11.235 \text{ kPa} \cdot \text{m}^3$$

Now, subtract the final term from the initial term to find the work done:

$$W = 17.15 \text{ kPa} \cdot \text{m}^3 - 11.235 \text{ kPa} \cdot \text{m}^3$$

$$W = 5.915 \text{ kPa} \cdot \text{m}^3$$

Since $1 \text{ kPa} \cdot \text{m}^3$ is equivalent to $1 \text{ kJ}$, the work delivered is:

$$W = 5.915 \text{ kJ}$$

Final Work Done Calculation

The work delivered during the expansion process, calculated using the given parameters and the process equation $pv^2 = c$, is 5.915 kJ.