Question

A closed system consisting of \(5 \mathrm{~kg}\) of a gas undergoes a process during which the relationship between pressure and specific volume is \(p v^{1.3}=\) constant. The process begins with \(p_{1}=1\) bar, \(v_{1}=0.2 \mathrm{~m}^{3} / \mathrm{kg}\) and ends with \(p_{2}=0.25\) bar. Determine the final volume, in \(\mathrm{m}^{3}\), and plot the process on a graph of pressure versus specific volume.

Short answer

The final volume is approximately 2.2 m³. The process graph shows a curve from (0.2, 1) to (0.44, 0.25).

Step-by-step solution

Understand the Given Information

Identify the given values for the process. Initial pressure, \(p_1\), is 1 bar and the initial specific volume, \(v_1\), is 0.2 m^3/kg. The final pressure, \(p_2\), is 0.25 bar. The relationship between pressure and specific volume is given by \(p v^{1.3} = \text{constant}\).

Apply the Process Relation

Use the relationship \(p v^{1.3} = \text{constant}\) to set up an equation for the initial state: \[ p_1 v_1^{1.3} = \text{constant} \]. Substitute the known values: \[ 1 \times (0.2)^{1.3} = \text{constant} \].

Calculate the Constant

Compute the constant: \[ (0.2)^{1.3} \approx 0.091 \]. Therefore, \[ \text{constant} = 1 \times 0.091 = 0.091 \].

Apply the Constant to Final State

Use the constant to find the final specific volume \(v_2\). The equation is \[ p_2 v_2^{1.3} = 0.091 \]. Substitute \(p_2 = 0.25\) bar: \[ 0.25 v_2^{1.3} = 0.091 \].

Solve for Final Specific Volume

Isolate \(v_2\) by dividing both sides by 0.25: \[ v_2^{1.3} = \frac{0.091}{0.25} = 0.364 \]. Take the 1.3rd root: \[ v_2 = (0.364)^{\frac{1}{1.3}} \approx 0.44 \text{ m}^3/\text{kg} \].

Calculate the Final Volume

To find the final volume \(V_2\), multiply \(v_2\) by the mass \(m\): \[ V_2 = v_2 \times m = 0.44 \times 5 \approx 2.2 \text{ m}^3 \].

Plot the Process on a Graph

The graph should be a plot of pressure (y-axis) versus specific volume (x-axis). Plot the initial state at \( (v_1 = 0.2, p_1 = 1) \) and the final state at \( (v_2 = 0.44, p_2 = 0.25) \). Draw a curve between the two points that follows the relationship \( p v^{1.3} = \text{constant} \).

Key concepts

specific volume

In thermodynamics, specific volume is the volume occupied by a unit mass of a substance. It is often denoted as \(v\) and measured in \(\mathrm{m^3/kg}\). Specifically, specific volume is the reciprocal of density, highlighting how much space one kilogram of a substance takes up. In the given problem, the initial specific volume \(v_1\) is given as \(0.2\ \mathrm{m^3/kg}\). By understanding the concept of specific volume, you can relate changes in volume to changes in pressure within the closed system framework. This relationship is crucial for solving problems involving gas laws and state changes in thermodynamic processes.

pressure-volume relationship

The pressure-volume (P-V) relationship is fundamental in understanding how gases behave under different conditions. For this exercise, the relationship is given by \( pv^{1.3} = \text{constant} \). This means that the product of pressure \(p\) and specific volume \(v\) raised to the 1.3 power remains constant throughout the process. This type of relationship is known as a polytropic process, where \( pv^n = \text{constant} \), with \(n\) being the polytropic index. It's essential to familiarize yourself with this type of relationship as it can describe various thermodynamic processes, such as isothermal (\(n = 1\)), adiabatic (\(n\) depends on the specific heat), and others.

closed system thermodynamics

A closed system, in thermodynamics, is one where mass remains constant while energy can be exchanged with its surroundings. This exercise deals with a closed system containing 5 kg of a gas. This means during the process, although the pressure and specific volume change, the mass of the gas does not. In a closed system, understanding the initial and final states is key, often requiring the use of thermodynamic tables, equations of state, or specific process relationships like \( pv^{1.3} = \text{constant} \) in this case.

graphing thermodynamic processes

Graphing the thermodynamic processes helps visualize how the process evolves between different states. In this exercise, you would plot the process on a pressure vs. specific volume (P-v) diagram. Start by plotting the initial state at \( (v_1 = 0.2, p_1 = 1) \) and the final state at \( (v_2 = 0.44, p_2 = 0.25) \). Draw a curve connecting these points that adheres to the given relationship \( pv^{1.3} = \text{constant} \). This curve will typically bow towards either the pressure or specific volume axis, depending on your process.