Question

Entropy changes don't depend on a process pathway. For an ideal gas, the entropy is \(S=N k \ln V\) (see above). (a) Express \(\Delta S_{V}=S_{2}\left(V_{2}\right)-S_{1}\left(V_{1}\right)\), the entropy change upon changing the volume from \(V_{1}\) to \(V_{2}\), at fixed particle number \(N\). (b) Express \(\Delta S_{N}=S_{2}\left(N_{2}\right)-S_{1}\left(N_{1}\right)\), the entropy change upon changing the particle number from \(N_{1}\) to \(N_{2}\), at fixed volume \(V\). (c) Write an expression for the entropy change, \(\Delta S\), for a two-step process \(\left(V_{1}, N_{1}\right) \rightarrow\left(V_{2}, N_{1}\right) \rightarrow\left(V_{2}, N_{2}\right)\) in which the volume changes first at fixed particle number, then the particle number changes at fixed volume. (d) Show that the entropy change \(\Delta S\) above is exactly the same as for the two-step process in reverse order: changing the particle number first, then the volume.

Short answer

The entropy change is \( \Delta S = N_1 k \ln \left( \frac{V_2}{V_1} \right) + (N_2 - N_1) k \ln V_2 \) and it remains the same regardless of the order of the steps.

Step-by-step solution

- Express the Entropy Change for Volume Change

Given the entropy formula for an ideal gas, \[ S = N k \ln V \], we need to express the entropy change \( \Delta S_V \) when the volume changes from \( V_1 \) to \( V_2 \), while keeping the particle number \(N\) constant. This is calculated as follows: \[ \Delta S_V = S_2(V_2) - S_1(V_1) = N k \ln V_2 - N k \ln V_1 \]. Using the properties of logarithms, this simplifies to: \[ \Delta S_V = N k \ln \left( \frac{V_2}{V_1} \right) \].

- Express the Entropy Change for Particle Number Change

Next, we need to express the entropy change \( \Delta S_N \) when the particle number changes from \( N_1 \) to \( N_2 \), while keeping the volume \(V\) constant. Again, using the entropy formula, we have: \[ \Delta S_N = S_2(N_2) - S_1(N_1) = N_2 k \ln V - N_1 k \ln V \]. Since \( k \ln V \) is a common factor, this simplifies to: \[ \Delta S_N = (N_2 - N_1) k \ln V \].

- Entropy Change for Two-Step Process

Now, consider the two-step process where the volume changes first while the particle number is fixed (\( V_1, N_1 \rightarrow V_2, N_1 \)), and then the particle number changes while the volume is fixed (\( V_2, N_1 \rightarrow V_2, N_2 \)). The total entropy change \( \Delta S \) for this process is the sum of the entropy changes from each step: \[ \Delta S = \Delta S_V + \Delta S_N = N_1 k \ln \left( \frac{V_2}{V_1} \right) + (N_2 - N_1) k \ln V_2 \].

- Reversing the Order of the Two-Step Process

Let's show that the entropy change is the same if we reverse the order of the steps. First, change the particle number while the volume is fixed (\( V_1, N_1 \rightarrow V_1, N_2 \)), then change the volume while the particle number is fixed (\( V_1, N_2 \rightarrow V_2, N_2 \)). The total entropy change \( \Delta S \) for this process is: \[ \Delta S = \Delta S_N' + \Delta S_V' \], where \[ \Delta S_N' = (N_2 - N_1) k \ln V_1 \] and \[ \Delta S_V' = N_2 k \ln \left( \frac{V_2}{V_1} \right) \]. Adding these together gives: \[ \Delta S = (N_2 - N_1) k \ln V_1 + N_2 k \ln \left( \frac{V_2}{V_1} \right) \].

- Comparing the Results

Compare the entropy changes from both orders of the two-step process. Notice that the terms involving \( \ln V_1 \) and \( \ln V_2 \) combine similarly in both cases: \[ \Delta S = N_1 k \ln \left( \frac{V_2}{V_1} \right) + (N_2 - N_1) k \ln V_2 \] and \[ \Delta S = (N_2 - N_1) k \ln V_1 + N_2 k \ln \left( \frac{V_2}{V_1} \right) \]. Thus, we can see that the entropy change \( \Delta S \) is identical regardless of the order of the steps.

Key concepts

entropy

Entropy is a measure of the disorder or randomness in a system. The concept of entropy is central to the second law of thermodynamics, which states that the total entropy of an isolated system can never decrease over time. When we deal with ideal gases, the entropy can be calculated using specific formulas. For instance, the entropy of an ideal gas is given by \[ S = N k \ln V \] where:

• S is the entropy.
• N is the number of particles in the system.
• k is the Boltzmann constant.
• V is the volume of the gas.

If the volume changes while the number of particles stays constant, the change in entropy, \(\Delta S\), can be found by: \[ \Delta S_V = N k \ln \left( \frac{V_2}{V_1} \right) \]This equation shows how entropy varies with changes in volume while particle number remains the same.

ideal gas

An ideal gas is a theoretical gas composed of many randomly moving point particles that interact only through elastic collisions. The concept of an ideal gas is useful because it simplifies the study of gases under various conditions without the complications that arise from intermolecular forces and the volume of individual molecules.For an ideal gas, several equations of state and other relationships hold true:

• The Ideal Gas Law: \[ PV = NkT \] where P is the pressure, V is the volume, N is the number of particles, k is the Boltzmann constant, and T is the temperature.
• The entropy formula: \[ S = N k \ln V \].

These equations help us understand how gases will behave under different conditions. In the entropy change examples, we see how changes in particle number or volume impact the system's entropy.

thermodynamic process

A thermodynamic process involves the transfer of energy from one form to another and often includes changes in the physical state or properties of a system. These processes can be described in terms of changes in variables such as pressure, volume, temperature, and entropy.In the context of the problem, we examined two specific thermodynamic processes:

• Changing the volume while keeping the particle number constant, which leads to a certain change in entropy: \[ \Delta S_V = N k \ln \left( \frac{V_2}{V_1} \right) \]
• Changing the particle number while keeping the volume constant, leading to another entropy change: \[ \Delta S_N = (N_2 - N_1) k \ln V \]

When combining these two steps into a single process, regardless of the order of steps, we end up with the same total change in entropy. This illustrates an essential principle in thermodynamics: the total change in entropy is independent of the path taken, depending only on the initial and final states.