Question

3.75 moles of an ideal gas with \(C_{V, m}=3 / 2 R\) undergoes the transformations described in the following list from an initial state described by \(T=298 \mathrm{K}\) and \(P=4.50\) bar. Calculate \(q, w, \Delta U, \Delta H,\) and \(\Delta S\) for each process. a. The gas undergoes a reversible adiabatic expansion until the final pressure is one third its initial value. b. The gas undergoes an adiabatic expansion against a constant external pressure of 1.50 bar until the final pressure is one third its initial value. c. The gas undergoes an expansion against a constant external pressure of zero bar until the final pressure is equal to one third of its initial value.

Short answer

For process a, we have the final temperature \(T_2 \approx 198.3 \mathrm{K}\), work done \(w \approx -2242 \ \mathrm{J}\), internal energy change \(\Delta U \approx -2242 \ \mathrm{J}\), enthalpy change \(\Delta H \approx -2804.08 \ \mathrm{J}\), and entropy change \(\Delta S \approx -21.47 \ \mathrm{J K^{-1}}\). For process b, we find the final temperature \(T_2 \approx 145.4 \mathrm{K}\), work done \(w \approx -1759.7 \ \mathrm{J}\), internal energy change \(\Delta U \approx -1759.7 \ \mathrm{J}\), enthalpy change \(\Delta H \approx -3519.4 \ \mathrm{J}\), and entropy change \(\Delta S \approx -10.98 \ \mathrm{J K^{-1}}\). For process c, we have the final temperature \(T_2 \approx 99.33 \mathrm{K}\), work done \(w = 0 \ \mathrm{J}\), internal energy change \(\Delta U = 0 \ \mathrm{J}\), enthalpy change \(\Delta H = 0 \ \mathrm{J}\), and entropy change \(\Delta S \approx -8.09 \ \mathrm{J K^{-1}}\).

Step-by-step solution

Calculate the final temperature

Using the given initial temperature and pressure, and final pressure, we can find the final temperature using the adiabatic process equation. We know that \(\gamma = \frac{C_P}{C_V} = \frac{C_V + R}{C_V} = \frac{5}{3}\). \[T_2 = T_1\left(\frac{P_1}{P_2}\right)^{\frac{\gamma-1}{\gamma}} = 298\left(\frac{4.5}{4.5/3}\right)^{\frac{2}{5}}\]

Calculate the work done

Since this is an adiabatic process, the heat is zero. Hence, using the first law of thermodynamics, we can find the work done: \(w = \Delta U\) First, we should calculate \(\Delta U = nC_{V}\Delta T = 3.75\cdot\left(\frac{3}{2}R\right)(T_2 - T_1)\)

Calculate the enthalpy change

Next, we will find the enthalpy change using \(\Delta H = \Delta U + P\Delta V\). We know that \(\Delta V = \frac{nRT_2}{P_2} - \frac{nRT_1}{P_1}\).

Calculate the entropy change

Finally, we will calculate the entropy change using \(\Delta S = nC_V \ln{\frac{T_2}{T_1}} + nR \ln{\frac{V_2}{V_1}}\). b. Adiabatic expansion against a constant external pressure

Calculate the final temperature

Similar to part (a), we can calculate the final temperature using the adiabatic process equation.

Calculate the work done

Since this is an adiabatic process, the heat is zero. Now, the work done is given by: \(w = -P_{ext} (V_2 - V_1)\)

Calculate the internal energy change and enthalpy change

Next, we should calculate the internal energy change using the first law of thermodynamics. After that, we can calculate the enthalpy change as in the previous scenario.

Calculate the entropy change

Finally, we calculate the entropy change using the same formula applied in part (a). c. Expansion against a constant external pressure of zero

Calculate the final temperature

Similar to the previous parts, we need to find the final temperature using the given pressure and adiabatic process equation.

Calculate the work done

Now, since the external pressure is zero, no work will be done during this process, i.e., \(w = 0\).

Calculate the internal energy change and enthalpy change

Since no work is done, the changes in internal energy and enthalpy will be zero in this case.

Calculate the entropy change

Finally, we will calculate the entropy change using the same formula applied in parts (a) and (b). These are the step-by-step solutions for all three processes given in the exercise.

Key concepts

Thermodynamics

Thermodynamics is the branch of science that deals with the relationships between heat and other forms of energy. More specifically, it studies how thermal energy is converted to and from other forms of energy and how it affects matter.

Understanding thermodynamics involves a few key concepts such as the system (a part of the universe we're focusing on), surroundings (everything outside the system), and states (described by quantities like temperature, pressure, and volume). Thermodynamically, processes can be categorized based on how they transfer heat and do work, with some examples being isothermal (constant temperature), isobaric (constant pressure), and adiabatic (no heat transfer).

In the exercise provided, an adiabatic process is emphasized, where an ideal gas expands without exchanging heat with the surroundings. This process is vital for understanding how systems behave when they are thermally isolated. An ideal gas is a hypothetical gas whose molecules don't interact and occupy no volume, which makes the calculations and understanding of thermodynamics much simpler.

First Law of Thermodynamics

The first law of thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed in an isolated system. The change in the internal energy of a system is equal to the heat added to the system minus the work done by the system on its surroundings.

In the context of the textbook exercise, this principal is used to determine work done by the gas during an adiabatic expansion. Mathematically, the first law is expressed as: \[\Delta U = q - w\]Here, \(\Delta U\) represents the change in internal energy, \(q\) is the heat exchanged, and \(w\) is the work done. Since the process is adiabatic, \(q=0\), which simplifies the equation to \(\Delta U = -w\). This relationship allows us to evaluate the change in the system’s internal energy by looking at the work done, even when no heat transfer occurs.

Entropy Change

Entropy is a measure of the disorder or randomness in a system, and it's a fundamental concept in the second law of thermodynamics. The entropy change, \(\Delta S\), in a system quantifies the change in this disorder during a process. For an ideal gas, the entropy change when the gas expands or contracts can be determined by considering both the temperature and volume changes.

The entropy change during an adiabatic process is particularly interesting because it can be nonzero even though no heat is transferred. In the solved exercise, the entropy change is calculated using the formula: \[\Delta S = nC_V \ln{\frac{T_2}{T_1}} + nR \ln{\frac{V_2}{V_1}}\]Where \(n\) is the number of moles, \(C_V\) is the molar heat capacity at constant volume, \(R\) is the universal gas constant, and \(T\) and \(V\) refer to temperature and volume, respectively, with subscripts 1 and 2 indicating initial and final states. Understanding how entropy changes allows students to grasp the direction of processes and the concept of reversible and irreversible processes.