Problem 59 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

A quantity of \(1.6 \mathrm{~g}\) helium gas is expanded adiabatically \(3.0\) times and then compressed isobarically to the initial volume. Assume ideal behaviour of gas and both the processes reversible. The entropy change of the gas in this process is \((\ln 3=1.1)\) (a) \(-1.1 \mathrm{cal} / \mathrm{K}\) (b) \(+1.1 \mathrm{cal} / \mathrm{K}\) (c) \(-2.2 \mathrm{cal} / \mathrm{K}\) (d) \(+2.2 \mathrm{cal} / \mathrm{K}\)

Step-by-Step Solution

Answer

The calculated entropy change is approximately \( -0.9 \text{cal/K} \) which is not listed in the options, indicating a possible error.

Step 1: Understand the adiabatic expansion

In an adiabatic process, there is no heat exchange with the surroundings, implying the entropy change within the system is zero.

Step 2: Characterize the isobaric compression

During the isobaric (constant pressure) compression, entropy change can occur as there is heat exchange with the surroundings. To calculate the entropy change, use the formula \(\Delta S = \frac{Q}{T} \), where \( Q \) is the heat added to the system and \(T\) is the temperature at which it occurs.

Step 3: Determine the entropy change for isobaric compression

As the gas is compressing back to its original volume, the entropy change during this step will be the negative of the entropy change for expansion. Considering an ideal gas and constant temperature, the pressure-volume work done on the gas is equal but opposite to the adiabatic expansion. Therefore, the entropy change for the isobaric compression will be \( -n R \ln V_f/V_i = -n R \ln 3\).

Step 4: Calculate the numerical value of the entropy change

Substitute the values of \(n\), which is the number of moles of helium gas, and \(R\), the ideal gas constant in calories (\(R = 1.987 \text{cal/mol}\cdot\text{K}\)). For helium, \( n = \frac{1.6}{4} = 0.4 \text{mol}\), since the molar mass of helium is 4 g/mol. Now calculate the entropy change: \(\Delta S = -0.4 \times 1.987 \times 1.1\) cal/K.

Step 5: Obtain the final answer

Perform the multiplication to obtain the final value of the entropy change. \(\Delta S = -0.4 \times 1.987 \times 1.1 = -0.87164\) cal/K. Since the entropy change must be expressed to one significant digit as in the multiple-choice answers, it rounds to \( -0.9 \text{cal/K} \) which is not listed in the options, indicating an error in the calculation or the given options.

Key Concepts

Adiabatic Expansion

Adiabatic expansion is a fascinating concept in physical chemistry, which refers to the process where a gas expands without exchanging heat with its surroundings. This might seem counterintuitive, as we're used to the idea that interactions between systems often involve some energy transfer. In an adiabatic process, however, the system is perfectly insulated, so all the work done on the gas results in a change in the internal energy, but not the temperature.

During adiabatic expansion, even though a gas may do work and expand, its entropy remains unchanged because the process is insulated and no heat is exchanged. Remember that entropy is a measure of the spread of energy, and without heat transfer, this 'spread' doesn't change.

Isobaric Compression

Isobaric compression is a term used to describe the compression of a gas at a constant pressure. Imagine squeezing a balloon gently enough that the pressure stays the same but the volume decreases. In real-life applications, this kind of process could happen in engines or pumps, where gases are compressed for various purposes.

Unlike adiabatic processes, isobaric processes allow for the exchange of heat with the surroundings. This impacts the system's entropy, because as the volume of a gas decreases under constant pressure, the system either loses or gains heat. In terms of calculating entropy change for such a process, we use the formula \(\Delta S = \frac{Q}{T}\), where \(Q\) represents the heat transferred and \(T\) is the absolute temperature of the gas.

Entropy Calculation

Entropy, symbolized by \(S\), is essentially a measure of disorder or randomness in a system. It's a core idea in thermodynamics that is crucial for understanding how systems evolve over time. Calculating the change in entropy, \(\Delta S\), involves assessing how much energy is spread out in a process or how much energy is shared with a system's surroundings at a particular temperature.

For the calculation in reversible processes, we often use the relationship \(\Delta S = \frac{Q_{rev}}{T}\), where \(Q_{rev}\) stands for the reversible heat exchanged and \(T\) indicates the absolute temperature. Significant to note is that when there is no heat exchange, as in an adiabatic process, the entropy change \(\Delta S\) is zero, reflecting that there is no increase in disorder.

Ideal Gas Behavior

The behavior of an ideal gas is governed by the Ideal Gas Law \(PV=nRT\), which connects the pressure \(P\), volume \(V\), number of moles \(n\), ideal gas constant \(R\), and absolute temperature \(T\). This law works under the assumption that the gas particles do not attract or repel each other and that the particles themselves take up no space.

Under these assumptions, the Ideal Gas Law can be a powerful tool in understanding and predicting how gases will behave under varying conditions of temperature and pressure. It’s important when considering adiabatic expansion or isobaric compression because it helps us to quantify changes in various properties of the gas, such as volume or temperature, and it also facilitates the calculation of work done by or on the gas during a process.