Question

One-fourth kg of a gas contained within a piston-cylinder assembly undergoes a constant-pressure process at 5 bar beginning at \(v_{1}=0.20 \mathrm{~m}^{3} / \mathrm{kg}\). For the gas as the system, the work is \(-15 \mathrm{~kJ}\). Determine the final volume of the gas, in \(\mathrm{m}^{3}\).

Short answer

The final volume of the gas is 0.02 m³.

Step-by-step solution

Define the Given Variables

The mass of the gas, m, is 0.25 kg. The initial specific volume, v1, is 0.20 m³/kg. The constant pressure, P, is 5 bar (which can be converted to 500 kPa since 1 bar = 100 kPa). The work done by the gas, W, is -15 kJ.

Formulate the Work Equation for a Constant-Pressure Process

For a constant-pressure process, the work done by the gas is given by the relationship: \[ W = P(V_2 - V_1) \] where V1 and V2 are the initial and final volumes of the gas, respectively.

Convert Work into Consistent Units

The work given is -15 kJ, which is -15000 J (since 1 kJ = 1000 J). Also, convert the pressure into pascals (Pa): \[ P = 5 \text{ bar} = 500 \times 10^3 \text{ Pa} \] since 1 bar = 100 kPa and 1 kPa = 1000 Pa.

Calculate the Initial Volume

The initial volume, V1, is the product of the specific volume and mass: \[ V_1 = v_1 \times m \] Substituting the given values: \[ V_1 = 0.20 \text{ m}^3/\text{kg} \times 0.25 \text{ kg} = 0.05 \text{ m}^3 \]

Use the Work Equation to Find the Final Volume

Rearrange the work equation to solve for the final volume, V2: \[ -15000 \text{ J} = 500 \times 10^3 \text{ Pa} (V_2 - 0.05 \text{ m}^3) \] Solving for V2: \[ V_2 - 0.05 \text{ m}^3 = \frac{-15000 \text{ J}}{500 \times 10^3 \text{ Pa}} \] \[ V_2 - 0.05 \text{ m}^3 = -0.03 \text{ m}^3 \] \[ V_2 = -0.03 \text{ m}^3 + 0.05 \text{ m}^3 \] \[ V_2 = 0.02 \text{ m}^3 \]

Key concepts

thermodynamics

Thermodynamics is the branch of physics that deals with heat, work, and energy. It especially looks at how these quantities interact and change within systems. One major principle within thermodynamics is the Law of Energy Conservation, which basically states that energy can neither be created nor destroyed, only changed from one form to another.
In our exercise, we deal specifically with a process happening at constant pressure, which is one type of thermodynamic process. The first law of thermodynamics tells us how the internal energy of a system changes when heat is added to it or work is done by it. It’s an essential concept when dealing with piston-cylinder assemblies and many other mechanical systems. Understanding this allows us to better calculate and predict the outcomes of thermodynamic processes, such as the one in this problem.

piston-cylinder assembly

A piston-cylinder assembly is a common setup in many engines and mechanical systems. It consists of a piston enclosed in a cylinder that can move back and forth. The piston’s movement often affects the volume and pressure of any gas or fluid inside the cylinder.
In our problem, we are examining how a gas, when subjected to work under constant pressure, interacts within the piston-cylinder assembly. As the piston moves, it changes the volume inside the cylinder. This is represented by our variables such as initial and final volumes (V_1 and V_2). The movement of the piston under constant pressure allows us to use the known formulas and laws to find unknown values like the final volume in this exercise.

work done by gas

Work done by a gas in a thermodynamic context refers to the energy transferred when the gas either expands or contracts. In our exercise, we have a constant-pressure process, meaning the pressure does not change even though the volume does. The work done by or on the gas (W) is related to the change in volume and is calculated using the formula:
[ ... W = P(V_2 - V_1) ... ] Note that work can be positive or negative, depending on whether the gas is doing work on the surroundings (expanding) or work is being done on the gas (compressing). In our problem, the work done is -15 kJ (-15000 J), indicating that the gas is being compressed. This directly impacts our calculations to find the final volume of the gas.

specific volume

Specific volume is an important property in thermodynamics and refers to the volume occupied by a unit mass of a substance. It is typically expressed in m³/kg. In our exercise, the specific volume ( v_1 ) helps us determine the initial volume of the gas.
Specific volume can essentially give us insight into how dense the substance is under certain conditions. By understanding how specific volume interacts with mass to give us total volume, we can solve many types of practical problems in engineering and physics. For this problem, the initial specific volume was 0.20 m³/kg, and with a mass of 0.25 kg, the initial volume was calculated to help us solve for the final volume after work was done on the gas.