Question

A mass of 8 kg of helium undergoes a process from an initial state of \(3 \mathrm{m}^{3} / \mathrm{kg}\) and \(15^{\circ} \mathrm{C}\) to a final state of \(0.5 \mathrm{m}^{3} / \mathrm{kg}\) and \(80^{\circ} \mathrm{C}\). Assuming the surroundings to be at \(25^{\circ} \mathrm{C}\) and \(100 \mathrm{kPa}\) determine the increase in the useful work potential of the helium during this process.

Short answer

The decrease in the useful work potential (exergy) of the 8 kg mass of helium during the process is 55,579.2 J (Joules).

Step-by-step solution

Convert given temperatures to Kelvin

Before proceeding with the calculations, we need to convert the given temperatures from Celsius to Kelvin. To do this, simply add 273.15 to each temperature: \(T_1 = 15^\circ C + 273.15 = 288.15 \,\text{K}\) \(T_2 = 80^\circ C + 273.15 = 353.15 \,\text{K}\) \(T_0 = 25^\circ C + 273.15 = 298.15 \,\text{K}\)

Calculate the initial and final pressures of the helium

Using the Ideal Gas Law, \(PV = nRT\), and the molar mass of helium (\(M_{He} = 4.0026 \,\text{g/mol}\)), we can find the pressures at the initial and final states: \(P_1 = \frac{m R T_1}{M_{He} v_1}\) \(P_2 = \frac{m R T_2}{M_{He} v_2}\) Where \(R\) is the universal gas constant (\(R = 8.314 \,\text{J/mol K}\)) and \(m = 8 \,\text{kg}\). Calculate \(P_1\) and \(P_2\): \(P_1 = \frac{8 \times 8.314 \times 288.15}{4.0026 \times 3} = 199.92 \,\text{kPa}\) \(P_2 = \frac{8 \times 8.314 \times 353.15}{4.0026 \times 0.5} = 1957.93 \,\text{kPa}\)

Calculate the initial and final specific internal energies

For an ideal gas, the specific internal energy (\(u\)) can be calculated as: \(u = C_v T\) Where \(C_v\) is the specific heat at constant volume. For helium, \(C_v = 3/2 R/M_{He} = 12.471 \,\text{J/molK}\). Calculate the initial and final specific internal energies: \(u_1 = C_v T_1 = 12.471 \times 288.15 = 3594.31 \,\text{J/mol}\) \(u_2 = C_v T_2 = 12.471 \times 353.15 = 4402.87 \,\text{J/mol}\)

Calculate the initial and final specific entropies

For an ideal gas, the specific entropy (\(s\)) can be calculated as: \(s = C_v \ln{\frac{T_2}{T_1}} - R \ln{\frac{P_2}{P_1}}\) Calculate the initial and final specific entropies: \(s_1 = 0\) (We can choose any reference point for the entropy, so let it be zero for \(T_1\) and \(P_1\)) \(s_2 = C_v \ln{\frac{T_2}{T_1}} - R \ln{\frac{P_2}{P_1}} = 12.471 \ln{\frac{353.15}{288.15}} - 8.314 \ln{\frac{1957.93}{199.92}} = 30.28 \,\text{J/molK}\)

Calculate the exergy change

Using the formula for exergy change, we can now calculate the increase in useful work potential: \(\Delta Ex = m(u_2 - u_1) + mP_0(v_2 - v_1) - T_0m(s_2 - s_1)\) Plug in the values and calculate the exergy change: \(\Delta Ex = 8 \times (4402.87 - 3594.31) + 8 \times 100 \times (0.5 - 3) - 298.15 \times 8 \times (30.28 - 0) = -55579.2 \,\text{J}\) The negative sign indicates that the useful work potential of the helium has decreased during this process. The decrease in the useful work potential of the helium during this process is \(55579.2 \,\text{J}\).

Key concepts

Ideal Gas Law

Understanding the Ideal Gas Law is crucial for analyzing the behavior of gases under various conditions. It's an equation of state that relates pressure (P), volume (V), temperature (T), and the amount of gas in moles (n), typically written as PV=nRT. For practical calculations, we adjust the formula to include the mass and specific volume (volume per unit mass, v) of the gas, which is expressed as PV=mRT/M, where M is the molar mass of the gas.

The Ideal Gas Law allows us to determine unknown properties of a gas when others are known, such as calculating the pressure of helium at different temperatures and volumes in the exercise. It assumes that the gas particles are in constant random motion, do not attract or repel each other, and that they occupy no volume. While these assumptions aren't perfectly true, the Ideal Gas Law provides a good approximation for most gases under normal conditions.

Specific Internal Energy

Specific internal energy, often represented as 'u,' is a measure of the energy contained within a substance, exclusive of any kinetic or potential energy it may possess due to its motion or position. For an ideal gas, this is directly proportional to its temperature, with the relationship given by the equation u = CvT, where Cv is the specific heat capacity at constant volume.

In thermodynamics problems, like our textbook example, understanding specific internal energy is crucial when evaluating the energy changes within a system during a process. It is noteworthy that the specific internal energy of an ideal gas does not depend on the pressure or specific volume but only on temperature, which simplifies calculations significantly. The specific heat capacity (Cv) value is unique to each gas and is essential to calculate the internal energy accurately.

Specific Entropy

Specific entropy, designated as 's,' is a fundamental concept in thermodynamics representing a system's disorder level at a microscopic scale. It's a measure of energy dispersal or spread at a specified temperature. For an ideal gas, the change in specific entropy between two states can be calculated using the equation s = Cv ln(T2/T1) - R ln(P2/P1), considering constant heat capacities.

In our example, the computation of entropy change is pivotal as it reflects the irreversibility of the process the helium undergoes. Irreversibility can be linked to the inefficiencies in a thermodynamic cycle, signifying energy that cannot be converted into work and is instead lost, often as heat to the surroundings. By selecting a reference point for the initial entropy, we can calculate the change in specific entropy, which plays a role in determining the exergy change of a system.

Exergy Change

Exergy change is a measure of the maximum useful work obtainable from a system as it reaches equilibrium with a reference environment, often linked to the concept of 'useful energy.' The exergy change formula, \(\Delta Ex = m(u_2 - u_1) + mP_0(v_2-v_1) - T_0m(s_2-s_1)\), encapsulates this by accounting for changes in internal energy, flow work, and entropy relative to the surroundings, at temperature T0 and pressure P0.

In our helium example, the negative exergy change value suggests a decrease in the gas’s ability to perform work - an important consideration in energy systems, such as power plants or refrigeration cycles, where maximizing efficiency and work output is key. This calculation requires the prior calculation of internal energies and entropies, as well as an understanding of how volume and pressure are related to the surroundings.