Question

A vertical piston-cylinder device initially contains \(0.12 \mathrm{m}^{3}\) of helium at \(20^{\circ} \mathrm{C}\). The mass of the piston is such that it maintains a constant pressure of \(200 \mathrm{kPa}\) inside. \(\mathrm{A}\) valve is now opened, and helium is allowed to escape until the volume inside the cylinder is decreased by one-half. Heat transfer takes place between the helium and its surroundings at \(20^{\circ} \mathrm{C}\) and 95 kPa so that the temperature of helium in the cylinder remains constant. Determine ( \(a\) ) the maximum work potential of the helium at the initial state and \((b)\) the exergy destroyed during this process.

Short answer

Answer: The maximum work potential at the initial state is 22,964 J, and the exergy destroyed during the process is -22,964 J.

Step-by-step solution

Calculate the initial moles of helium

To find the initial moles of helium, we use the ideal gas equation: $$PV = nRT$$ where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature. We are given \(P=200\,\text{kPa}\), \(V=0.12\,\text{m}^3\), and \(T=20^{\circ} \mathrm{C}\). We first need to convert the temperature to Kelvin: $$T_K = 20 + 273.15 = 293.15\,\mathrm{K}$$ Now, we can find the initial moles of helium: $$n = \frac{PV}{RT} = \frac{(200\times10^3\,\mathrm{Pa})(0.12\,\mathrm{m^3})}{(8.314\,\mathrm{J/(mol\cdot K)})(293.15\,\mathrm{K})} = 9.69\,\mathrm{mol}$$

Calculate the initial and final volume

The initial volume of helium is given as \(V_1 = 0.12\,\mathrm{m^3}\). Since the volume decreases by one-half, the final volume of helium is: $$V_2 = \frac{1}{2}V_1 = 0.06\,\mathrm{m^3}$$

Find the maximum work potential

The process is isothermal, meaning the temperature remains constant. We can use the equation for isothermal work on an ideal gas to find the maximum work potential: $$W_{\text{max}} = nRT \ln\frac{V_1}{V_2}$$ Substitute the values we found in Steps 1 and 2: $$W_{\text{max}} = (9.69\,\mathrm{mol})(8.314\,\mathrm{J/(mol\cdot K)})(293.15\,\mathrm{K})\ln\frac{0.12\,\mathrm{m^3}}{0.06\,\mathrm{m^3}} = 22964.0\,\mathrm{J}$$ The maximum work potential at the initial state is \(22,964\,\mathrm{J}\).

Calculate the exergy destroyed

For an isothermal process with heat transfer, we can write the energy and exergy balance equations: $$Q - W = \Delta{U} = 0 \quad\text{and}\quad \Phi = -T_0 \cdot \Delta{S} = W - W_{\text{reversible}}$$ Since the process is isothermal, \(\Delta{U} = 0\), and the work done on the system equal to the heat transfer (\(W = Q\)). We can solve for the exergy destroyed, \(\Phi\): $$\Phi = -T_0 \cdot \Delta{S} = (293.15\,\mathrm{K})(-nR\ln\frac{V_1}{V_2}) = -22964.0\,\mathrm{J}$$ The exergy destroyed during the process is \(-22,964\,\mathrm{J}\) (note the negative sign, indicating that exergy is destroyed and not created).