Question

A system consists of \(2 \mathrm{~m}^{3}\) of hydrogen gas \(\left(\mathrm{H}_{2}\right)\), initially at \(35^{\circ} \mathrm{C}, 215 \mathrm{kPa}\), contained in a closed rigid tank. Energy is transferred to the system from a reservoir at \(300^{\circ} \mathrm{C}\) until the temperature of the hydrogen is \(160^{\circ} \mathrm{C}\). The temperature at the system boundary where heat transfer occurs is \(300^{\circ} \mathrm{C}\). Modeling the hydrogen as an ideal gas, determine the heat transfer, in \(\mathrm{kJ}\), the change in entropy, in \(\mathrm{kJ} / \mathrm{K}\), and the amount of entropy produced, in \(\mathrm{kJ} / \mathrm{K}\). For the reservoir, determine the change in entropy, in \(\mathrm{kJ} / \mathrm{K}\). Why do these two entropy changes differ?

Short answer

Heat transfer: 42.5 kJ; Entropy change of system: 0.925 kJ/K; Entropy change of reservoir: -0.074 kJ/K; Entropy produced: 0.851 kJ/K. The difference in entropy changes is due to the irreversibility of the process.

Step-by-step solution

State the properties and initial conditions

Given:- Initial volume: \(V = 2 \text{ m}^3\)- Initial temperature: \(T_1 = 35 ^\circ \text{C} = 308.15 \text{ K}\)- Initial pressure: \(P_1 = 215 \text{ kPa}\)- Final temperature: \(T_2 = 160 ^\circ \text{C} = 433.15 \text{ K}\)- Reservoir temperature: \(T_{res} = 300 ^\circ \text{C} = 573.15 \text{ K}\)Model the hydrogen as an ideal gas.

Calculate the number of moles of hydrogen

Use the ideal gas law: \(PV = nRT\)\[ n = \frac{P_1 V}{R T_1} \]where \(R = 8.314 \, \text{J} / (\text{mol} \, \text{K})\)\[ n = \frac{215 \times 10^3 \times 2}{8.314 \times 308.15} \approx 168.4 \text{ moles} \]

Determine the heat transfer (Q)

For an ideal gas under constant volume, the heat transfer is:\[ Q = n C_v \Delta T \]- Find \(C_v\) for hydrogen gas, \( C_v \approx 20.18 \text{ J/(mol} \; \text{K)} \)- Calculate \( \, \Delta T = T_2 - T_1 = 433.15 - 308.15 = 125 \text{ K} \)\[ Q = 168.4 \times 20.18 \times 125 \approx 42518 \text{ J} = 42.5 \text{ kJ} \]

Find the change in entropy of the system (\(\Delta S_{sys}\))

For an ideal gas under constant volume:\[ \Delta S_{sys} = n C_v \ln \left( \frac{T_2}{T_1} \right) \]\[ \Delta S_{sys} = 168.4 \times 20.18 \times \ln \left( \frac{433.15}{308.15} \right) \approx 925.3 \text{ J/K} = 0.925 \text{ kJ/K} \]

Calculate the change in entropy of the reservoir (\(\Delta S_{res}\))

For the reservoir, entropy change is given by:\[ \Delta S_{res} = -\frac{Q}{T_{res}} \]\[ \Delta S_{res} = -\frac{42.5 \text{ kJ}}{573.15 \text{ K}} \approx -0.074 \text{ kJ/K} \]

Determine the total entropy produced in the process (\(S_{gen}\))

The total entropy produced (or the entropy generation) is the sum of the entropy changes of the system and the reservoir:\[ S_{gen} = \Delta S_{sys} + \Delta S_{res} \]\[ S_{gen} = 0.925 - 0.074 \approx 0.851 \text{ kJ/K} \]

Explain why the entropy changes differ

The difference in entropy changes arises because the process is not reversible. The entropy of the system increases by 0.925 kJ/K and the entropy of the reservoir decreases by 0.074 kJ/K, resulting in a net positive entropy generation of 0.851 kJ/K.

Key concepts

Ideal Gas Law

The ideal gas law is a key concept in thermodynamics. It's used to relate the pressure, volume, and temperature of a gas. The formula is given by:
\( PV = nRT \)
Here, \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is the temperature. By knowing some of these variables, you can easily find the others.

• In this problem, we have 2 \( \text{m}^3 \) of hydrogen gas at 215 kPa and 35°C.
• By rearranging the ideal gas law to solve for the number of moles (n), we find \( n \) = 168.4 moles.

This law makes it easier to understand the behavior of gases under different conditions and is fundamental for calculations in many thermodynamic problems.

Entropy Change

Entropy is a measure of disorder in a system. The change in entropy tells us how the disorder has changed. For an ideal gas at constant volume, the change in entropy can be calculated using:
\[ \Delta S_{sys} = n C_v \left( \ln \left( \frac{T_2}{T_1} \right) \right) \ \]

• In our example, hydrogen gas is heated from 35°C to 160°C, and we find \(\text{T}_2 = 433.15 \text{K } \text{ and } \text{T}_1 = 308.15 \text{K} \), which allows us to determine the entropy change.
• Using this formula, we calculate \(\text{\Delta S_{sys}} = 0.925 \text{ kJ/K} \).

Entropy change is crucial for understanding how energy is distributed in a system and indicates whether the process can be reversible or not.

Heat Transfer

Heat transfer refers to the movement of thermal energy from one place to another. It can occur through conduction, convection, or radiation. For an ideal gas under constant volume, the heat transfer (\text{Q}) can be determined using:
\[ Q = n C_v \Delta T \ \]

• \text{In our problem, there's a temperature change (\text{\Delta T}) from 35°C to 160°C, or 125 K.}
• By plugging into the formula, we calculate \text{Q} to be 42.5 kJ.

Understanding heat transfer is key to calculating energy exchanges in a variety of systems, making it an essential concept in thermodynamics.

Entropy Generation

Entropy generation (\text{S_{gen}}) is a measure of the irreversibility of a process. It indicates how much entropy is produced in the system due to inefficiencies or natural dispersion of energy. This can be calculated by adding the change in entropy of the system and the change in entropy of the reservoir:
\[ S_{gen} = \Delta S_{sys} + \Delta S_{res} \ \]

• The entropy change of the system was calculated to be 0.925 kJ/K.
• The reservoir experienced an entropy change of -0.074 kJ/K.
• Adding these, we find the entropy generation to be approximately 0.851 kJ/K.

This positive value tells us the process is not reversible and is guided by the second law of thermodynamics. It gives insight into the efficiency and natural progression of thermodynamic processes.