Question

Consider a gas mixture whose apparent molecular weight is 33 , initially at 3 bar and \(300 \mathrm{~K}\), and occupying a volume of \(0.1 \mathrm{~m}^{3}\). The gas undergoes an expansion during which the pressure-volume relation is \(p V^{1.3}=\) constant and the energy transfer by heat to the gas is \(3.84 \mathrm{~kJ}\). Assume the ideal gas model with \(c_{v}=0.6+\left(2.5 \times 10^{-4}\right) T\), where \(T\) is in \(\mathrm{K}\) and \(c_{\mathfrak{v}}\) has units of \(\mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\). Neglecting kinetic and potential energy effects, determine (a) the final temperature, in \(\mathrm{K}\). (b) the final pressure, in bar. (c) the final volume, in \(\mathrm{m}^{3}\). (d) the work, in \(\mathrm{kJ}\).

Short answer

Find the constant using \[ p V^{1.3} \], calculate final volume, use Ideal Gas Law for final temperature, and apply thermodynamics to find work done.

Step-by-step solution

- Determine the final volume

Use the pressure-volume relation given: \[ p V^{1.3} = \text{constant} \]Since the initial pressure (\[ p_1 = 3 \text{ bar} \]) and initial volume (\[ V_1 = 0.1 \text{ m}^3 \]) are known, calculate the constant: \[ p_1 V_1^{1.3} = 3 \times (0.1)^{1.3} \]Let's call the constant \[ K \]. So, \[ K = 3 \times (0.1)^{1.3} \].

- Find the final pressure using the gas relation

Rearrange the equation to solve for the final volume (\[ V_2 \]) at final pressure (\[ p_2 \]): \[ p_2 V_2^{1.3} = K \] and we know: \[ p_2 V_2^{1.3} = 3 \times (0.1)^{1.3} \]The relationship between \[ p_1 \] and \[ V_1 \] will help us solve for \[ V_2 \] when \[ p_2 \] is at another known value. Substitute into the ideal gas law to determine \[ V_2 \].

- Determine the final temperature

Apply the ideal gas law to find the final temperature (\[ T_2 \]) using the given specific heats (\[ c_v \]). The ideal gas law is \[ pV = nRT \], given the initial and final conditions, substitute \[ p_2 \] and \[ V_2 \] to determine \[ T_2 \].

- Calculate the work done

The work done (\[ W \]) can be determined from the first law of thermodynamics: \[ \text{Heat Added } (Q) = \text{Work Done } (W) + \text{Change in Internal Energy } (\Delta U)\]Given \[ Q = 3.84 \text{kJ} \], \[ \Delta U \] can be calculated using \[ c_v \] and rearranged to solve for \[ W \].

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation used to relate the pressure, volume, and temperature of a gas. It is expressed as \( pV = nRT \). In this equation, \( p \) stands for pressure, \( V \) for volume, \( n \) for the number of moles of the gas, \( R \) for the universal gas constant, and \( T \) for the absolute temperature in Kelvins.
Understanding this law is crucial because it allows you to calculate any one of these variables if the others are known. For example, for our problem, the initial conditions of the gas are given: 3 bar pressure, 0.1 m³ volume, and 300 K temperature.
Using these values and the Ideal Gas Law, we can determine how other variables will change when the gas undergoes transformations such as expansion or compression. This forms the basis for further calculations, such as finding the final volume and temperature after expansion.
It's essential to convert all units appropriately, like pressure to Pascals or volume to cubic meters, as needed, to ensure consistency in calculations.

Pressure-Volume Relationship

In the given problem, the gas follows a specific pressure-volume relationship: \( pV^{1.3} = \text{constant} \). This is an example of a polytropic process where the exponent (1.3 in this case) impacts the state variables' relationship.
To use this relationship, start by calculating the constant \( K \) from the initial conditions. For instance, given \( p_1 = 3 \text{ bar} \) and \( V_1 = 0.1 \text{ m}^3 \), you can find \( K \) as: \( K = 3 \times (0.1)^{1.3} \).
Once you have the constant, any change in volume will inversely affect the pressure according to the formula. If the volume increases, the pressure will decrease and vice versa. This relationship is key to predicting how the gas will behave under varying conditions and is necessary for determining the final volume and pressure after expansion.
To find the final pressure or volume, rearrange the equation: \( p_2 V_2^{1.3} = K \). This straightforward yet powerful relationship helps unravel many complexities in gas behavior under different thermodynamic processes.

First Law of Thermodynamics

The First Law of Thermodynamics, also known as the law of energy conservation, is pivotal in thermodynamics problems. It states: \( Q = W + \triangle U \). Here, \( Q \) is the heat added to the system, \( W \) is the work done by the system, and \( \triangle U \) is the change in internal energy.
In the given exercise, heat is added to the gas: \( Q = 3.84 \text{kJ} \). To determine the work done (\( W \)), you must consider the change in internal energy. The internal energy change can be calculated if you know the specific heat capacity at constant volume (\( c_v \)) and the temperature change.
From the problem statement, \( c_v = 0.6 + (2.5 \times 10^{-4}) T \). Using the initial and final temperatures, you can calculate \( \triangle U \). Once \( \triangle U \) is known, rearrange the First Law equation to find \( W \). This forms the foundation of energy balance within the system and ensures that the calculations align with physical principles.

Specific Heat Capacity

Specific heat capacity is the amount of heat required to change the temperature of a unit mass of a substance by one degree. In the context of our problem, the specific heat capacity at constant volume (\( c_v \)) is given as a function of temperature: \( c_v = 0.6 + (2.5 \times 10^{-4}) T \).
This relationship indicates that \( c_v \) changes with temperature, which is crucial for accurately calculating the change in internal energy (\( \triangle U \)). As the gas undergoes expansion, its temperature changes, affecting \( c_v \).
To find \( \triangle U \), integrate \( c_v \) over the temperature range. That provides the energy change per kilogram of gas. Knowing \( c_v \) and the temperature difference helps you compute the total internal energy change, which is key to solving the First Law of Thermodynamics.
Understanding specific heat capacity enables precise energy calculations and improves the accuracy of the thermodynamic analysis. That ensures a thorough comprehension of how much energy is stored or released during thermal processes.