Question

An ideal gas sample containing 1.75 moles for which \(C_{V, m}=5 R / 2\) undergoes the following reversible cyclical process from an initial state characterized by \(T=275 \mathrm{K}\) and \(P=1.00\) bar: a. It is expanded reversibly and adiabatically until the volume triples. b. It is reversibly heated at constant volume until \(T\) increases to \(275 \mathrm{K}\) c. The pressure is increased in an isothermal reversible compression until \(P=1.00\) bar. Calculate \(q, w, \Delta U, \Delta H,\) and \(\Delta S\) for each step in the cycle, and for the total cycle.

Short answer

During the reversible cyclical process, the ideal gas undergoes three steps: 1. Adiabatic expansion: \(q_1=0\), \(w_1=-\Delta U_1\), \(\Delta S_1=0\). 2. Isochoric heating: \(q_2=\Delta U_2=nC_V(T_f-T_i)\), \(w_2=0\), \(\Delta S_2= n C_{V} \ln \frac{T_f}{T_i}\), and \(\Delta H_2 = n C_{V}(T_{f}-T_{i})\). 3. Isothermal compression: \(q_3=-w_3= nRT \ln\frac{V_f}{V_i}\), \(w_3=-nRT \ln\frac{V_f}{V_i}\), \(\Delta S_3 = nR \ln\frac{V_f}{V_i}\), and \(\Delta H_3 = 0\). Total cycle calculations: \(\Delta U_{total} = 0\), \(\Delta H_{total} = 0\), \(\Delta S_{total} = \Delta S_{1} + \Delta S_{2} + \Delta S_{3}\), \(q_{total} = q_{1} + q_{2} + q_{3}\), \(w_{total} = w_{1} + w_{2} + w_{3}\).

Step-by-step solution

Adiabatic Expansion

During an adiabatic expansion, there's no heat interaction with the surroundings (\(q=0\)). The work (\(w\)) done by the system can be calculated using the equation: \[ w = -\Delta U \] And the change in internal energy (\(\Delta U\)) can be determined using the equation: \[ \Delta U=nC_{V}(T_{f}-T_{i}) \] Where \(T_{f}\) and \(T_{i}\) are the final and initial temperatures of the system, respectively. Given the initial state and the adiabatic exponent \(\gamma = \frac{C_{P}}{C_{V}}\) (where \(C_{P} = C_{V} + R\)), the final temperature can be found using the relation: \[ T_{f}=T_{i} \times \left(\frac{V_{i}}{V_{f}}\right)^{\gamma-1} \] All these quantities can be calculated using the provided values and these equations. The change in entropy (\(\Delta S\)) during an adiabatic process is zero, as there's no heat interaction with the surroundings.

Isochoric Heating

During isochoric heating, the volume of the system remains constant so no work is performed (\(w=0\)). The heat added (\(q\)) to the system can be calculated as: \[ q = \Delta U = n C_{V} (T_{f}-T_{i}) \] This is also the change in the internal energy (since \(q = \Delta U\) for a constant volume process). The change in entropy (\(\Delta S\)) for this process can be calculated using the formula: \[ ΔS= n C_{V} \ln \left(\frac{T_f}{T_i}\right) \] Enthalpy (\(H\)) is a state function and at constant volume, the change in enthalpy (\(\Delta H\)) is equal to the heat added to the system: \[ \Delta H = q = \Delta U = n C_{V} (T_{f}-T_{i}) \]

Isothermal Compression

In the process of isothermal compression, the temperature of the system remains constant, so \(\Delta U = 0\) (as internal energy is a function of temperature for an ideal gas). The work done on the system can be calculated using the equation: \[w = -nRT \ln\left(\frac{V_f}{V_i}\right)\] Since \(\Delta U = q + w\) and \(\Delta U = 0\), the heat exchanged (\(q\)) can be found as: \[ q = -w = nRT \ln\left(\frac{V_f}{V_i}\right) \] The change in entropy (\(\Delta S\)) for this process can be calculated using the formula: \[ \Delta S = nR \ln\left(\frac{V_f}{V_i}\right) \] The change in enthalpy \(\Delta H\) for an isothermal process in an ideal gas is zero, since H is a function of temperature only.

Total Cycle Calculations

Since all quantities calculated are state functions (except for \(q\) and \(w\)), their total change over the cycle can be determined by adding up the changes through each step of the cycle: \[ \Delta U_{total} = \Delta U_{1} + \Delta U_{2} + \Delta U_{3} \] \[ \Delta H_{total} = \Delta H_{1} + \Delta H_{2} + \Delta H_{3} \] \[ \Delta S_{total} = \Delta S_{1} + \Delta S_{2} + \Delta S_{3} \] Similarly, total heat \(q_{total}\) and work \(w_{total}\) can be calculated: \[ q_{total} = q_{1} + q_{2} + q_{3} \] \[ w_{total} = w_{1} + w_{2} + w_{3} \] Due to the cyclical nature of the process described, \(\Delta U_{total}\) and \(\Delta H_{total}\) should both equal zero, which could act as checks on the calculations.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in thermodynamics that describes the behavior of an ideal gas. It is presented as:
\( PV = nRT \)
where:

• \(P\) is the pressure of the gas
• \(V\) is the volume of the gas
• \(n\) is the number of moles
• \(R\) is the ideal gas constant
• \(T\) is the temperature in Kelvin

This law assumes ideal conditions, where interactions between molecules are negligible and the volume of the molecules themselves is much smaller than the total volume of the gas.
In the context of thermodynamic cycles like the one described in the exercise, the Ideal Gas Law helps us relate different properties such as pressure, volume, and temperature at various points. This is crucial for calculating changes during expansions or compressions, like in adiabatic and isothermal processes.

Adiabatic Processes

Adiabatic processes occur when a gas expands or compresses without any heat exchange with its surroundings. In these processes, \( q = 0 \), since no heat is added or removed. The key for adiabatic processes is the conservation of energy:
\[ w = -\Delta U \]
This equation shows that the work done during the process comes entirely from the internal energy of the gas.
In the exercise, the adiabatic expansion involves the volume of the gas tripling. The adiabatic condition implies that:\[ PV^{\gamma} = \text{constant} \]
where \( \gamma \) (gamma) is the heat capacity ratio \( \frac{C_P}{C_V} \). The temperature change can be deduced using \[ T_f = T_i \times \left(\frac{V_i}{V_f}\right)^{1-\gamma} \],relating initial and final states without heat exchange. Understanding these principles is crucial for predicting outcomes in adiabatic processes within cyclic thermodynamic paths.

Isothermal Processes

In isothermal processes, the temperature remains constant while the gas undergoes expansion or compression. For ideal gases, the internal energy change \( \Delta U \) is zero because internal energy is solely dependent on temperature.
The work done in an isothermal process is given by:\[ w = -nRT \ln\left(\frac{V_f}{V_i}\right) \]
Since the temperature does not change, you can evaluate the work simply by examining the change in volume. This equation shows how the internal energy balance is maintained through heat exchange, \( q = -w \), to keep temperature constant.
Isothermal processes are crucial in examples like the isothermal compression phase of the given cycle, where the goal is often to return pressure to an initial state. Comprehending these isothermal transformations allows calculation of other state function changes like entropy in thermodynamic cycles.

Entropy Changes

Entropy is a measure of disorder or randomness in a system. In thermodynamics, it quantifies the energy dispersion in a process. Each part of a thermodynamic cycle can have an entropy change, \( \Delta S \), calculated differently based on the conditions.

• In adiabatic processes, \( \Delta S = 0 \) because there is no heat exchange.
• During isochoric (constant volume) processes, it can be calculated as \[ \Delta S = n C_{V} \ln\left(\frac{T_f}{T_i}\right) \] given temperature changes occur without volume change.
• In isothermal processes, it's given by \[ \Delta S = nR \ln\left(\frac{V_f}{V_i}\right) \] , which reflects energy reorganization as volume changes while maintaining temperature.

The total entropy change of a reversible cycle is zero, signifying a return to the initial state with no net energy dispersion. Understanding entropy changes is essential for evaluating energy transformations and efficiencies in thermodynamic systems.