Question

Calculate the work (in kJ) when substance initially at 99 kPa, 16 m³ is expanded against a linear spring to 208 kPa, 70 m³.

Answer 1

For linear spring;

In other words, pressure is proportional to the volume.

Average pressure during the process is given as;

Hence, work done is given as;

...(Answer)

Answer 2

To calculate the work done when a substance initially at 99 kPa and 16 m³ is expanded against a linear spring to 208 kPa and 70 m³, we can follow these steps:

Given:

• Initial pressure, $P_1 = 99 \, \text{kPa}$

• Initial volume, $V_1 = 16 \, \text{m}^3$

• Final pressure, $P_2 = 208 \, \text{kPa}$

• Final volume, $V_2 = 70 \, \text{m}^3$

Assumption:

• The pressure-volume relationship is linear due to the linear spring, meaning pressure varies linearly with volume.

Calculation:

1. Determine the linear relationship between pressure and volume:

   Since the relationship is linear, we can express pressure as:

   $$P = mV + c$$

   where $m$ is the slope and $c$ is the intercept.

   Using the initial and final conditions:

   $$P_1 = mV_1 + c \implies 99 = m \times 16 + c \quad \text{(Equation 1)}$$$$P_2 = mV_2 + c \implies 208 = m \times 70 + c \quad \text{(Equation 2)}$$

   Subtract Equation 1 from Equation 2 to solve for $m$:

   $$208 - 99 = m \times (70 - 16)$$$$109 = m \times 54$$$$m = \frac{109}{54} \approx 2.0185 \, \text{kPa/m}^3$$

   Substitute $m$ back into Equation 1 to find $c$:

   $$99 = 2.0185 \times 16 + c$$$$99 = 32.296 + c$$$$c = 99 - 32.296$$$$c \approx 66.704 \, \text{kPa}$$

   Therefore, the pressure-volume relationship is:

   $$P = 2.0185V + 66.704$$

2. Calculate the work done:

   The work done during expansion is the area under the $P$-$V$ curve, which can be calculated as:

   $$W = \int_{V_1}^{V_2} P \, dV$$

   Substitute the expression for $P$:

   $$W = \int_{16}^{70} (2.0185V + 66.704) \, dV$$

   Evaluate the integral:

   $$W = \left[ 2.0185 \frac{V^2}{2} + 66.704V \right]_{16}^{70}$$

   Calculate the definite integral:

   $$W = \left( 2.0185 \times \frac{70^2}{2} + 66.704 \times 70 \right) - \left( 2.0185 \times \frac{16^2}{2} + 66.704 \times 16 \right)$$$$W = \left( 2.0185 \times 2450 + 4669.28 \right) - \left( 2.0185 \times 128 + 1067.264 \right)$$$$W = \left( 4945.825 + 4669.28 \right) - \left( 258.368 + 1067.264 \right)$$$$W = 9615.105 - 1325.632$$$$W = 8289.473 \, \text{kPa} \cdot \text{m}^3$$

   Since $1 \, \text{kPa} \cdot \text{m}^3 = 1 \, \text{kJ}$:

   $$W \approx 8289.473 \, \text{kJ}$$