Problem 28 - Chemistry: The Central Science in SI Units, Global Edition

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 \(76.0 \mathrm{kPa} ?(\mathbf{b})\) If the final pressure on the gas is \(121.6 \mathrm{kPa}\), 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) For an isothermal process where the pressure of an ideal gas increases and the volume decreases, we expect a negative sign for the entropy change (\(\Delta S\)). (b) To calculate the entropy change for the process, first find the initial and final volume using the ideal gas law. Then, use the entropy change formula for an ideal gas: \(\Delta S = nR\ln\frac{V_f}{V_i}\). Plug in the values and solve for \(\Delta S\). (c) Yes, specifying the temperature is necessary for calculating the entropy change, as it is required to determine the initial and final volume of the ideal gas.

Step 1: Analyze the isothermal process

In an isothermal process, the temperature of the gas remains constant which means the internal energy of the system also remains constant. When the pressure of an ideal gas increases, its volume decreases, which leads to a decrease in the entropy of the gas. So, in this case, we expect a negative sign for the entropy change (\(\Delta S\)).

Step 2: Entropy change formula for an ideal gas

The entropy change for an ideal gas can be represented by the following equation: \[ \Delta S = nR\ln\frac{V_f}{V_i} \] where \(\Delta S\) is the entropy change, \(n\) is the number of moles, \(R\) is the gas constant (8.314 J/mol·K), \(V_f\) is the final volume, and \(V_i\) is the initial volume.

Step 3: Calculate the initial and final volume

First, we need to find the initial and final volume of the ideal gas using the ideal gas law: \[ PV = nRT \] For the initial state, we have \(P_i = 76.0 \mathrm{kPa}\), \(n = 0.600 \mathrm{mol}\), \(R = 8.314 \mathrm{J/(mol·K)}\), and \(T_i = 350 \mathrm{K}\). Converting the pressure to Pascals (1 kPa = 1000 Pa), we can calculate the initial volume: \[ V_i = \frac{nRT_i}{P_i} = \frac{(0.600 \mathrm{mol})(8.314 \mathrm{J/(mol·K)})(350 \mathrm{K})}{(76.0 \mathrm{kPa})(1000 \mathrm{Pa/kPa})} \] And for the final state, we have \(P_f = 121.6 \mathrm{kPa}\) and the same \(n\) and \(T_i\). Calculating the final volume: \[ V_f = \frac{nRT_i}{P_f} = \frac{(0.600 \mathrm{mol})(8.314 \mathrm{J/(mol·K)})(350 \mathrm{K})}{(121.6 \mathrm{kPa})(1000 \mathrm{Pa/kPa})} \]

Step 4: Calculate the entropy change

Now that we have the initial and final volume, we can use the entropy change formula for an ideal gas from Step 2: \[ \Delta S = nR\ln\frac{V_f}{V_i} = (0.600 \mathrm{mol})(8.314 \mathrm{J/(mol·K)})\ln\frac{V_f}{V_i} \]

Step 5: Answer the question about specifying the temperature

The temperature is essential for determining the initial and final volume of the ideal gas, which in turn is necessary for calculating the entropy change. Therefore, specifying the temperature is indeed necessary for calculating the entropy change.

Key Concepts

Isothermal Process

An isothermal process occurs when a system changes without any variation in temperature. For an ideal gas, this means the internal energy stays the same.
In this process, if pressure increases, volume decreases and vice versa. Because the temperature remains constant, it’s important to understand how this impacts entropy, a measure of disorder.
In an isothermal compression, where pressure increases and volume decreases, entropy tends to decrease. This happens because the gas molecules have less space to disperse, leading to less randomness.

Ideal Gas Law

The ideal gas law is fundamental to understanding gas behavior in physics and chemistry. It combines several gas laws and is represented by the equation: \[ PV = nRT \]where:

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

This equation helps directly link pressure, volume, and temperature. In isothermal processes, because temperature remains constant, changes in pressure lead to inversely proportional changes in volume.

Entropy Formula

Entropy is a key concept in thermodynamics, representing the degree of randomness or disorder. The change in entropy (\(\Delta S\)) for an ideal gas can be calculated using:\[\Delta S = nR\ln\frac{V_f}{V_i}\]where:

• \(n\) is the number of moles,
• \(R\) is the universal gas constant,
• \(V_f\) is the final volume,
• \(V_i\) is the initial volume.

This formula shows how volume changes affect entropy in isothermal processes. The logarithmic factor indicates that even small changes in volume can significantly impact entropy.

Pressure and Volume Relationship

In thermodynamics, the relationship between pressure and volume is crucial, especially in isothermal transformations. When the pressure on a gas increases, the volume tends to decrease if the temperature stays constant. This inverse relation can be explained by Boyle's Law: \[ P_1 V_1 = P_2 V_2 \]During an isothermal compression:

• The molecules are pushed closer together,
• The volume decreases as pressure increases,
• The system's disorder (entropy) typically decreases.

This interaction highlights how environmental changes like pressure affect the internal state of gases, impacting their entropy.