Problem 27 - Chemistry: The Central Science in SI Units, Global Edition

Question

(a) What sign for \(\Delta S\) do you expect when the volume of 0.200 mol of an ideal gas at \(27^{\circ} \mathrm{C}\) is increased isothermally from an initial volume of \(10.0 \mathrm{~L} ?\) (b) If the final volume is \(18.5 \mathrm{~L},\) calculate the entropy change for the process. (c) Do you need to specify the temperature to calculate the entropy change?

Step-by-Step Solution

Answer

(a) The sign of \(\Delta S\) is positive when the volume of an ideal gas increases isothermally. (b) The entropy change for the given process is approximately 1.6628 J/K. (c) No, the temperature does not need to be specified to calculate the entropy change for an isothermal process involving an ideal gas.

Step 1: In this step, we will think conceptually about the entropy change when the volume of an ideal gas increases at a constant temperature. Entropy is a measure of the randomness or disorder of a system. When the volume of ideal gas increases, the particles have more space to occupy, and thus the randomness of the system increases. Therefore, when the volume increases isothermally for an ideal gas, the entropy change (\(\Delta S\)) is positive. #Step 2: Plug the given values into the formula for entropy change in part (b)#

To calculate the entropy change for part (b), we will use the formula \(\Delta S = nR\ln\frac{V_f}{V_i}\), where: \(n\) = number of moles = 0.200 mol \(R\) = gas constant = 8.314 J/(mol·K) \(V_i\) = initial volume = 10.0 L \(V_f\) = final volume = 18.5 L Keep in mind that the volumes should be converted to m³, so we need to multiply the liters by 0.001 to get m³. #Step 3: Calculate the entropy change in part (b)#

Step 2: Using the values from step 2, we can calculate the entropy change: \[\Delta S = (0.200 \,\text{mol}) \times (8.314 \,\text{J/(mol·K)}) \times \ln\frac{(18.5 \times 0.001 \,\text{m³})}{(10.0 \times 0.001 \,\text{m³})}\] \[\Delta S \approx 0.200 \times 8.314 \times \ln\frac{18.5}{10.0}\] \[\Delta S \approx 1.6628\,\text{J/K}\] The entropy change is positive, and its value is approximately 1.6628 J/K. #Step 4: Answer part (c) regarding the temperature requirement#

In part (c), we are asked if the temperature needs to be specified to calculate the entropy change. Looking back at the formula for the entropy change, we see that: \[\Delta S = nR\ln\frac{V_f}{V_i}\] The formula does not include the temperature directly, so we do not need to specify the temperature when calculating the entropy change for an isothermal process involving an ideal gas. However, having a constant temperature is essential for the process, as it ensures that the entropy change formula applies only to the isothermal case.

Key Concepts

Ideal Gas Law

The ideal gas law is a fundamental equation in understanding the behavior of gases. It combines several individual gas laws, including Boyle's, Charles's, and Avogadro's laws, into one single equation. This equation is expressed as \( PV = nRT \), where:

• \(P\) is the pressure of the gas
• \(V\) is the volume
• \(n\) is the number of moles
• \(R\) is the ideal gas constant, approximately 8.314 J/(mol·K)
• \(T\) is the temperature in Kelvin

One of the essential aspects of the ideal gas law is that it assumes gas behavior under ideal conditions, meaning no interactions between particles and that the volume of the gas particles themselves is negligible. This law is often used as a good approximation when dealing with real gases under many conditions, especially at high temperatures and low pressures where gas molecules behave closer to ideal.

Isothermal Process

An isothermal process is one in which the temperature remains constant. This constancy can be achieved by allowing the system to exchange heat with its surroundings to compensate for changes in internal energy. In the context of an ideal gas, an isothermal process means that while the gas might expand or contract, the temperature \(T\) does not change.

How isothermal processes relate to the ideal gas law:

During an isothermal process involving an ideal gas, the relationship between pressure and volume becomes crucial. According to the law, since \(T\) is constant, the product of pressure and volume \(PV\) is also constant. If volume increases, pressure must decrease, and vice versa, to maintain the equation \(PV = nRT\). This behavior is essential because it underpins the calculation of changes in other properties like entropy.

Calculation of Entropy

Entropy is a concept that describes the degree of disorder or randomness in a system. When calculating the change in entropy for a process involving an ideal gas, especially an isothermal one, we use the formula:\[ \Delta S = nR \ln \frac{V_f}{V_i} \]Where:

• \( \Delta S \) is the change in entropy
• \(n\) is the number of moles of gas
• \(R\) is the ideal gas constant
• \(V_f\) is the final volume
• \(V_i\) is the initial volume

This formula shows that the change in entropy only depends on the volume change and mole number, not the temperature directly, under isothermal conditions. During an isothermal expansion of an ideal gas, there is an increase in entropy because the gas molecules have more space to spread out, increasing randomness. Thus, the entropy change for such a process is usually positive, aligning with the second law of thermodynamics.