Question

A constant volume tank filled with 2 kg of air rejects heat to a heat reservoir at \(300 \mathrm{K}\). During the process the temperature of the air in the tank decreases to the reservoir temperature. Determine the expressions for the entropy changes for the tank and reservoir and the total entropy change or entropy generated of this isolated system. Plot these entropy changes as functions of the initial temperature of the air. Comment on your results. Assume constant specific heats for air at \(300 \mathrm{K}\).

Short answer

Answer: As the initial temperature decreases, the entropy change of the air in the tank (\(\Delta S_{tank}\)), the entropy change of the heat reservoir (\(\Delta S_{res}\)), and the total entropy change (\(\Delta S_{total}\)) all tend to approach zero.

Step-by-step solution

Find the initial and final temperatures of the air in the tank

We are given the mass of the air (\(m = 2 \, kg\)) and the final temperature of the air (\(T_2 = 300 \, K\)), and we are asked to find the expressions for the entropy change as functions of the initial temperature (\(T_1\)). Since the volume is constant, the pressure will change during the process, but we don't need to calculate it for this problem.

Calculate the entropy change for the air in the tank

For a constant volume process with constant specific heats, the entropy change for an ideal gas can be calculated using the following formula: $$ \Delta S_{tank} = m \, c_v \, \ln\frac{T_2}{T_1} $$ where \(c_v\) is the specific heat at constant volume. We are given the mass of the air (\(m = 2 \, kg\)), the final temperature (\(T_2 = 300 \, K\)), and we know we're treating the air with constant specific heats at \(300 \, K\). Therefore, \(c_v\) will also be a constant value for this problem.

Calculate the entropy change for the heat reservoir

The heat reservoir is at a constant temperature, meaning that any heat it receives or rejects, \(Q_{res}\), can be dissipated or absorbed without changing its temperature, \(T_{res}=T_2=300 \, K\). The entropy change of the heat reservoir can be calculated by the formula: $$ \Delta S_{res} = \frac{-Q_{res}}{T_{res}} $$ where \(Q_{res}\) is the amount of heat rejected by the air to the heat reservoir, given by: $$ Q_{res} = m \, c_v \, (T_1 - T_2) $$ Then, we can plug in the expression for \(Q_{res}\) into the entropy change equation: $$ \Delta S_{res} = \frac{-m \, c_v \, (T_1 - T_2)}{T_{res}} $$

Calculate the total entropy change for the isolated system

The total entropy change of the isolated system is the sum of the entropy changes of the tank and the heat reservoir: $$ \Delta S_{total} = \Delta S_{tank} + \Delta S_{res} $$ Plugging in the expressions from Steps 2 and 3 into this equation, we get: $$ \Delta S_{total} = m \, c_v \, (\ln\frac{T_2}{T_1} - \frac{T_1-T_2}{T_{res}}) $$

Plot the entropy changes as functions of the initial temperature

To plot the entropy changes as functions of the initial temperature, we can use the expressions we derived in Steps 2, 3, and 4. We can use software like Desmos, Matlab, or Python to create these plots by plugging in a range of initial temperatures (\(T_1\)) and observing how \(\Delta S_{tank}\), \(\Delta S_{res}\), and \(\Delta S_{total}\) change accordingly.

Comment on the results

After plotting the entropy changes, the following results can be observed: 1. The entropy change of the air in the tank (\(\Delta S_{tank}\)) is negative and becomes less negative or approaches zero as the initial temperature decreases. This is due to the fact that as the air temperature approaches the reservoir temperature, there is less change in the air entropy during the process. 2. The entropy change of the heat reservoir (\(\Delta S_{res}\)) is positive and becomes less positive or approaches zero as the initial temperature decreases. This is because the reservoir gains less heat as the initial temperature gets closer to the reservoir temperature. 3. The total entropy change (\(\Delta S_{total}\)) is always positive or equal to zero. This indicates that the process is irreversible, and the entropy is generated during the process. When the initial temperature approaches the reservoir temperature, the total entropy change approaches zero, meaning that the process becomes more reversible. 4. It can be observed that the entropy change for the air in the tank (\(\Delta S_{tank}\)) and the entropy change for the heat reservoir (\(\Delta S_{res}\)) tend to have opposite trends, ensuring that the total entropy change of the system stays positive according to the second law of thermodynamics.

Key concepts

Isolated System Entropy

Understanding the concept of entropy within an isolated system provides crucial insight into the nature of thermodynamic processes. An isolated system does not exchange matter or energy with its surroundings. According to the second law of thermodynamics, the entropy of an isolated system can remain constant or increase, but it can never decrease.

In the context of the exercise, the combined tank and heat reservoir system is treated as isolated. Therefore, no heat or matter is exchanged with the outside environment. As air within the tank cools down and rejects heat to the reservoir, it undergoes an entropy decrease while the reservoir experiences an entropy increase. Notably, the overall change in entropy, or entropy generation, within this isolated system is never negative—highlighting a fundamental principle that in real-world processes, irreversibilities lead to entropy production.

When analyzing such systems, it's important to recognize that the total entropy change can serve as an indicator of irreversibility. A non-zero increase in total entropy signifies irreversible processes, while a zero change might indicate a reversible process or a system in equilibrium.

Constant Volume Process

A constant volume process occurs when a system undergoes a thermodynamic process without changing its volume. In this scenario, the pressure and temperature within the system can change while the volume remains unchanged.

In our textbook case, we focus on air within a tank that conducts heat to the heat reservoir while maintaining constant volume. During this process, the equation \[\Delta S_{tank} = m \cdot c_v \cdot \ln\frac{T_2}{T_1}\] expresses the change in entropy of the air. Here, the initial and final temperature play a significant role. As the temperature of air in the tank (\(T_1\) to \(T_2\)) approaches the constant temperature of the reservoir, the entropy change \(\Delta S_{tank}\) diminishes. This characteristic is rooted in the logarithmic relationship, illustrating how sensitivity to temperature variation changes depending on the initial state.

Graphically depicted, one would see a curve that flattens as the initial temperature narrows toward the reservoir's temperature, conveying the decrease in entropy change – a clear claimant of the temperature's impact in a constant volume process.

Specific Heats at Constant Volume

The concept of specific heats at constant volume (\(c_v\)) plays a pivotal role in thermodynamic processes. Specific heats represent the amount of heat required to change the temperature of a unit mass of a substance by one degree while keeping the volume constant.

In our exercise, the specific heat at constant volume for air is assumed to be constant at the reservoir temperature of 300 K. This assumption greatly simplifies our calculations, as \(c_v\) remains unchanged and can be factored out of the expressions for entropy change. The specific heat provides the link between the amount of thermal energy absorbed or released by the substance and the resultant temperature change.

By using the constant \(c_v\), we infer that all heat exchanged in the process is fully utilized in adjusting the internal energy of the air, which, in turn, influences the entropy change. Consequently, specific heats are essential for quantifying thermodynamic properties and understanding how substances respond to thermal interactions within constant volume processes.