91Ó°ÊÓ

In thermodynamics, a piston-cylinder process involves a gas confined within a cylinder that has a movable piston. The piston can move up and down, changing the volume of the cylinder and thereby affecting the pressure of the gas inside. Here, the process for carbon dioxide (CO2) transitions from a state with a volume of 2.5 cubic feet to 0.5 cubic feet. The piston moves, compressing the gas and increasing its pressure from 5 to 20 lbf/in². This piston-cylinder setup is a useful model for understanding many thermodynamic processes in engines and machinery.

The pressure-volume relationship in a thermodynamic process is crucial for determining the work done. This relationship is often non-linear but can sometimes be represented linearly as in our exercise. The equation given here is: \[ p = 23.75 - 7.5 V \].This equation tells us how the pressure (p) varies with volume (V). As the volume decreases, the pressure increases, which reflects the gas being compressed in the piston-cylinder assembly. This type of relationship can be experimentally determined or derived from gas laws and helps in calculating the work done during the process.

The work done by a gas during a thermodynamic process can be calculated using an integral. For a piston-cylinder process, the work is represented as:\[ W = \int_{V_1}^{V_2} p dV \]. This integral sums up the infinitesimal pressure-volume changes from the initial volume (V1) to the final volume (V2). By substituting the provided pressure-volume relationship into the integral, one obtains:\[ W = \int_{2.5}^{0.5} (23.75 - 7.5V) dV \].Evaluating this integral involves breaking it into simpler parts and integrating each term separately.

Unit conversion is key in ensuring that all the quantities are consistent for meaningful calculations. In this exercise, pressure is initially given in lbf/in² and volume in ft³. To calculate work, a conversion to a common unit system is necessary. We first keep pressure in lbf/in² until finishing the work integral. The final value of work is then converted from lbf-ft to Btu using the conversion factor: 1 Btu = 778.17 ft-lbf.The work calculated as -24.0625 lbf-ft is converted as:\[ -24.0625 \times \frac{1}{778.17} \approx -0.0309 \text{ Btu} \. \]

Integrating a linear function, as seen with our pressure-volume relationship, simplifies calculations significantly. The equation simplifies the integral into manageable parts:\[ W = 23.75 \int_{2.5}^{0.5} dV - 7.5 \int_{2.5}^{0.5} V dV \].Each part is then integrated separately. For a constant term like 23.75, the integral is straightforward:\[ 23.75 (V \bigg|_{2.5}^{0.5}) = 23.75 (0.5 - 2.5) = 23.75 \times (-2) = -47.5 \.\]For the term involving V, the integration process is:\[ -7.5 \times \frac{V^2}{2} \bigg|_{2.5}^{0.5} = -7.5 \times ( \frac{0.5^2}{2} - \frac{2.5^2}{2} ) = -7.5 \times (-3.125) = 23.4375 \.\]Summing these results gives us the total work done during the process.