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Chapter 15: Problem 5

There are \(n_{1}\) moles of an ideal gas at pressure \(P_{1}\) and temperature \(T\) in one compartment of an insulated container. In an adjoining compartment separated by a partition are \(n_{2}\) moles of an ideal gas at pressure \(P_{2}\) and temperature \(T .\) When the partition is removed: (a) Calculate the final pressure of the mixture. (b) Calculate the entropy change when the gases are identical. (c) Calculate the entropy change when the gases are different. (d) Prove that the entropy change in part \((c)\) is the same as that which would be produced by two independent free expansions.

Short Answer

Expert verified
Final pressure: \( P_{f} = \frac{P_{1}V_{1} + P_{2}V_{2}}{V_{1} + V_{2}} \)Entropy change (identical gases): 0.Entropy change (different gases): \( \triangle S_{total} = -n_{1}R\ln\left(\frac{V_{1}}{V_{1} + V_{2}}\right) - n_{2}R\ln\left(\frac{V_{2}}{V_{1} + V_{2}}\right) \)Entropy change is same for free expansions.

Step by step solution

01

Understand the initial setup

There are two compartments of an insulated container. Each compartment contains an ideal gas at the same temperature but different pressures and amounts.
02

Apply the Ideal Gas Law (Part a)

Using the ideal gas law, we have for each compartment: Compartment 1: \[ P_{1}V_{1} = n_{1}RT \]Compartment 2: \[ P_{2}V_{2} = n_{2}RT \] Here, \(V_{1} \) and \(V_{2}\) are the volumes of the compartments.
03

Calculate the final volume (Part a)

After the partition is removed, the gases mix and fill the combined volume, \(V = V_{1} + V_{2}\). Because the process is adiabatic and temperature remains constant, the final volume is: \[ V_{f} = V_{1} + V_{2} \]
04

Calculate the final pressure (Part a)

Using the combined volumes and pressures: \[ P_{1}V_{1} + P_{2}V_{2} = P_{f}(V_{1} + V_{2}) \]Therefore, the final pressure is: \[ P_{f} = \frac{P_{1}V_{1} + P_{2}V_{2}}{V_{1} + V_{2}} \]
05

Calculate the entropy change for identical gases (Part b)

The change in entropy for identical gases is zero because the mixing of identical particles is irreversible: \[ \triangle S_{identical} = 0 \]
06

Calculate the entropy change for different gases (Part c)

For different gases, the change in entropy for each gas mixing freely can be computed as follows: \[\triangle S_{mixing} = -n_{1}R\ln\left(\frac{V_{1}}{V_{1} + V_{2}}\right) - n_{2}R\ln\left(\frac{V_{2}}{V_{1} + V_{2}}\right) \] Thus, the total entropy change: \[ \triangle S_{total} = \triangle S_{1} + \triangle S_{2} \]
07

Prove entropy change for independent free expansions (Part d)

For independent free expansions, each gas expands to fill the entire volume independently. So for gas 1: \[\triangle S_{1} = n_{1} R\ln\left(\frac{V_{1} + V_{2}}{V_{1}}\right) \]and for gas 2: \[\triangle S_{2} = n_{2} R\ln\left(\frac{V_{1} + V_{2}}{V_{2}}\right) \]Thus, the combined entropy change matches the earlier expression for different gases: \[\triangle S_{total} = \triangle S_{1} + \triangle S_{2} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ideal Gas Law
The Ideal Gas Law is a fundamental equation in chemistry and physics, described by the formula:
  • \[ PV = nRT \]
Here,
P is the pressure of the gas,
V is its volume,
n is the number of moles of the gas,
R is the gas constant, and
T is the absolute temperature. The law shows how these properties of the gas are related. In the given exercise, each compartment in the insulated container contains gas at different pressures but the same temperature.
For Compartment 1:
\( P_1V_1 = n_1RT \)
For Compartment 2:
\( P_2V_2 = n_2RT \).
When the partition between the compartments is removed, the gases mix, maintaining the temperature, but the volumes add up as:
\[ V_{final} = V_1 + V_2 \]
Adiabatic Process
An adiabatic process is one where no heat is transferred into or out of the system. In simpler terms, it's a process where the energy is only transferred by work.
In the context of the exercise, when the partition is removed and the gases mix, the process is considered adiabatic because the container is insulated, meaning no heat exchange occurs with the surroundings.
Importantly, the temperature stays constant while the volume changes.
The final pressure of the mixed gases can be calculated using the initial pressures and volumes:
\[ P_{final} = \frac{P_1V_1 + P_2V_2}{V_1 + V_2} \].
Entropy Change
Entropy is a measure of disorder or randomness in a system. When gases mix, entropy usually increases, reflecting the greater disorder.
When the partition is removed, and the gases mix, the entropy change calculation depends on whether the gases are identical or different.
  • For identical gases, energy states and positional probabilities remain unchanged, meaning the entropy change is zero:
    \[ \triangle S_{identical} = 0 \]
  • For different gases, the entropy change is computed as:
    \[ \triangle S_{mixing} = -n_1R \text{ln}\frac{V_1}{V_1 + V_2} - n_2R\text{ln}\frac{V_2}{V_1 + V_2} \].
This calculation accounts for the increased randomness as the gases distribute more uniformly in the new volume.
Irreversible Processes
Irreversible processes are those where the system cannot return to its initial state without leaving a lasting impact. Mixing different gases is an example of this.
For part (c), the entropy change of two different gases mixing freely matches the entropy produced by two independent free expansions.
Each gas in the free expansions increases its volume to fill the entire container:
  • For gas 1:
    \[ \triangle S_1 = n_1 R\text{ln}\frac{V_1 + V_2}{V_1} \]
  • For gas 2:
    \[ \triangle S_2 = n_2 R\text{ln}\frac{V_1 + V_2}{V_2} \]
The total entropy change:
\[ \triangle S_{total} = \triangle S_1 + \triangle S_2 \] proves that mixing different gases and independent free expansions produce equivalent entropy changes, emphasizing the irreversible nature of the process.

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Most popular questions from this chapter

Suppose that we have \(1 \mathrm{~mol}\) of a monatomic gas \(B_{1}\), whose nuclei are in their lowest energy state, and \(1 \mathrm{~mol}\) of a monatomic gas \(B_{2}\) consisting of exactly the same atoms as \(B_{1}\), except that the nuclei are in an excited state whose energy \(\epsilon\) is much larger than \(k T\) and whose lifetime is much larger than the time for diffusion to occur. Both gases are at the same pressure and are maintained at the same temperature \(T\) by a heat reservoir. (a) Immediately after the nuclei of gas \(B_{2}\) have been excited, diffusion takes place. Calculate the entropy change of the universe. (b) After the nuclei of gas \(B_{2}\) have been excited, a time much larger than the lifetime of the excited state is allowed to elapse, and then diffusion takes place. Calculate the entropy change of the universe. (c) Show that the answer to part \((b)\) is larger than that to part \((a)\). (This problem is due to M. J. Klein.)

There are \(n_{1}\) moles of an ideal monatomic gas at temperature \(T_{1}\) and pressure \(P\) in one compartment of an insulated container. In an adjoining compartment separated by an insulating partition are \(n_{2}\) moles of another ideal monatomic gas at temperature \(T_{2}\) and pressure \(P .\) When the partition is removed: (a) Show that the final pressure of the mixture is \(P\). (b) Calculate the entropy change when the gases are identical. (c) Calculate the entropy change when the gases are different.

Consider a uniform substance of variable mass. Suppose that, at any moment, there are \(n\) moles of substance and that the system undergoes an infinitesimal reversible process in which \(n\) changes by \(d n\). For any 1 mol of substance, $$ \mathrm{d} q=T d s=d u+P d \nu $$ and the total heat transfer \(d Q=n\) ? \(q=n T d s\). (a) Prove that $$ \mathrm{d} Q=d U+P d V-h d n $$ where \(U\) and \(V\) refer to the entire system, but \(h\) is the molar enthalpy. (b) From Prob. \(15.11(a)\) show that $$ T d S=d U+P d V-g d n $$ (c) How does \(T\) dS differ from dQ? (d) Show that, if \(T\) and \(P\) remain constant, then ? \(Q=0\) but \(T d S\) does not.

Consider a hydrostatic system of constant mass maintained in thermal and mechanical equilibrium. It is not in chemical equilibrium, however, because of chemical reactions and transport of matter between phases. Under these circumstances, the system undergoes an irreversible cycle. Prove that $$ \oint \frac{\mathrm{d} Q}{T}<0 $$

Show that, if the Helmholtz function of a phase is expressed as a function of \(T, V, n_{1}, n_{2}, \ldots, n_{c}\), then: (a) \(d A=-S d T-P d V+\sum \mu_{j} d n_{j}, \quad\) where \(\mu_{j}=\left(\partial A / \partial n_{j}\right)_{T, V, \text { other } n^{\prime} s^{\circ}}\) (b) \(A=-P V+\sum \mu_{j} n_{j}\) (c) \(-S d T+V d P=\sum n_{j} d \mu_{j}\)
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