Problem 30 - Chemistry The Central Science

Question

(a) What sign for \(\Delta S\) do you expect when the pressure on 0.600 mol of an ideal gas at \(350 \mathrm{~K}\) is increased isothermally from an initial pressure of 0.750 atm? (b) If the final pressure on the gas is 1.20 atm, calculate the entropy change for the process. (c) Do you need to specify the temperature to calculate the entropy change? Explain.

Step-by-Step Solution

Answer

(a) The sign of ΔS is expected to be negative, as the pressure increase leads to a decrease in volume and thus a negative ln(Vf / Vi) value. (b) Calculate initial and final volumes using ideal gas equation and then find ΔS using \( \Delta S = nR \ln \frac{V_{f}}{V_{i}} \). (c) Yes, specifying the temperature is essential for determining the entropy change, as it appears in the ideal gas equation and affects initial and final volumes.

Step 1: Part (a): Determine the sign of ΔS

As the pressure of the system increases, the volume will decrease since the temperature is kept constant. From the equation for the entropy change, ΔS is proportional to the natural logarithm of the ratio of final volume to initial volume (Vf / Vi). Since the volume decreases when the pressure increases, Vf < Vi, and thus, ln(Vf / Vi) will be negative. Therefore, the entropy change, ΔS, will be negative when the pressure is increased isothermally.

Step 2: Part (b): Calculate the entropy change for the process

First, we use the ideal gas equation, \( PV = nRT\), to find the initial and final volumes of the ideal gas. Recall that we have: - Initial pressure, \(P_{i} = 0.750 \:\text{atm}\), - Final pressure, \(P_{f} = 1.20 \:\text{atm}\), - Number of moles of gas, \(n = 0.600 \:\text{mol}\), - Temperature, \(T = 350 \:\text{K}\), - and the ideal gas constant, \(R = 0.08206 \frac{\text{L} \cdot \text{atm}}{\text{K} \cdot \text{mol}}\). Calculate the initial volume, \(V_{i}\): \( V_{i} = \frac{nRT}{P_{i}} \) Calculate the final volume, \(V_{f}\): \( V_{f} = \frac{nRT}{P_{f}} \) Next, we can calculate the entropy change, ΔS: \( \Delta S = nR \ln \frac{V_{f}}{V_{i}} \) Plugging in the calculated values for initial volume and final volume, find ΔS.

Step 3: Part (c): Do you need to specify the temperature to calculate the entropy change?

Yes, you need to specify the temperature to calculate the entropy change in this process. The temperature appears in the ideal gas equation, which is used to relate the pressure and volume changes in the system. In the isothermal process, the temperature is constant, and it affects the initial and final volumes as well as the entropy change. As a result, specifying the temperature is essential for determining the entropy change.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation that relates four key properties of a gas: pressure (\(P\)), volume (\(V\)), the number of moles (\(n\)), and temperature (\(T\)). This equation is expressed as \( PV = nRT \), where \(R\) is the ideal gas constant. This concept simplifies the behavior of gases under various conditions and serves as a good approximation for many gases under a range of conditions.

• The pressure is the force exerted by the gas particles per unit area.
• The volume is the space that the gas occupies.
• The number of moles measures the amount of gas present.
• The temperature affects the kinetic energy of gas particles.

By keeping these parameters in balance, the Ideal Gas Law helps predict how a gas will behave when any of these conditions change. When the temperature and other factors remain constant, adjustments to one of these variables (e.g., pressure or volume) will cause a change in the other corresponding variable, as seen in many gas-related calculations.

Isothermal Process

An isothermal process is a thermodynamic process in which the temperature of a system remains constant while other parameters such as pressure and volume might change. Since the temperature does not vary, the internal energy of an ideal gas remains constant through the process.
For an ideal gas, an isothermal process is often simplified using the Ideal Gas Law. If a gas undergoes isothermal expansion or compression, any change in its volume will inversely affect its pressure. This characteristic is crucial for determining the entropy change (\(\Delta S\)) during the process. Since the entropy change for an isothermal process can be calculated using volume ratios, the constancy in temperature provides a straightforward way to delineate other variables using known equations from thermodynamics.

• In an isothermal process, \( \Delta T = 0\).
• Energy exchange occurs mainly as work done by or on the system.
• The calculation of entropy in such a process often involves volume ratios.

Thus, understanding isothermal processes is crucial for solving problems involving ideal gases, especially regarding energy conservation and entropy calculations.

Pressure-Volume Relationship

The Pressure-Volume Relationship, often referred to as Boyle’s Law in the context of an ideal gas, describes how pressure and volume react inversely to changes when temperature is held constant. In simpler terms, if the volume of a gas increases, the pressure decreases, and vice versa, provided temperature and amount of gas remain constant.
In mathematical terms, this relationship is given by \( P_1V_1 = P_2V_2 \) in an isothermal process. This relation is foundational for understanding how gases behave under compression or expansion.

• If the volume reduces, pressure increases as gas particles collide more frequently.
• If the volume increases, pressure decreases as particles have more space to move.
• This principle is vital for calculating changes in state functions like entropy.

The interplay between pressure and volume is critical for solving problems involving entropy change in isothermal processes, as the decrease in volume directly affects the calculations for entropy, emphasizing its connection to the pressure-volume relationship.