Question

What is the entropy change when \(1.32 \mathrm{~g}\) of propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) at \(0.100\) atm pressure is compressed by a factor of five at a constant temperature of \(20^{\circ} \mathrm{C}\) ? Assume that propane behaves as an ideal gas.

Short answer

The entropy change is approximately -0.399 J/K.

Step-by-step solution

Understand Given Values and Conditions

We are told that propane \( \mathrm{C}_3\mathrm{H}_8 \), weighs 1.32 g, initially at 0.100 atm, is compressed by a factor of 5, and the temperature remains constant at 20°C. We need to find the entropy change during this process.

Convert Temperature to Kelvin

Convert the temperature from Celsius to Kelvin. \[ T = 20 + 273.15 = 293.15 \, \text{K} \]

Calculate the Initial Pressure and Final Pressure

The initial pressure is given as 0.100 atm. Since the gas is compressed by a factor of 5, the final pressure is:\[ P_f = 5 \times 0.100 \text{ atm} = 0.500 \text{ atm} \]

Calculate Number of Moles of Propane

The molar mass of propane is \[ (3 \times 12.01) + (8 \times 1.01) = 44.09 \, \text{g/mol} \]Using the given mass, calculate the number of moles \[ n = \frac{1.32 \, \text{g}}{44.09 \, \text{g/mol}} \approx 0.0299 \, \text{mol} \]

Use Ideal Gas Law for Entropy Change Formula

For an isothermal process of an ideal gas, the entropy change \( \Delta S \) can be calculated using the formula:\[ \Delta S = nR \ln \left( \frac{V_f}{V_i} \right) = -nR \ln \left( \frac{P_f}{P_i} \right) \]where \( R = 8.314 \, \text{J/mol K} \) is the ideal gas constant. Since volume times pressure is a constant for isothermal processes, \( V_f/V_i = P_i/P_f \).

Calculate Entropy Change

Substitute known values into the formula:\[ \Delta S = -0.0299 \, \text{mol} \times 8.314 \, \text{J/mol K} \times \ln \left( \frac{0.500 \, \text{atm}}{0.100 \, \text{atm}} \right) \]\[ \Delta S = -0.0299 \, \text{mol} \times 8.314 \, \text{J/mol K} \times \ln (5) \]\[ \Delta S = -0.0299 \, \text{mol} \times 8.314 \, \text{J/mol K} \times 1.609 \]\[ \Delta S \approx -0.399 \, \text{J/K} \]

Interpret the Result

Since \( \Delta S \) is negative, it indicates a decrease in entropy during the compression, which is expected as compression typically decreases the system's disorder.

Key concepts

Ideal Gas Behavior

An ideal gas is a useful model in thermodynamics due to its simplified nature. Essentially, it's a theoretical gas where the molecules do not have any interactions with each other and occupy no volume—even if they are moving. This makes calculations easier and is often a good approximation for real gases under certain conditions.

The concept of an ideal gas assumes that the gas molecules are in constant, random motion and that the only interactions are during perfectly elastic collisions. This means no energy is lost when the molecules collide with each other or the walls of the container.

One key formula involving ideal gases is the ideal gas law, expressed mathematically 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 (8.314 J/mol K).
• \( T \) is the temperature in Kelvin.

Using this equation, we can find various properties of gases under the assumption they behave ideally. In the context of the given problem, propane is assumed to behave as an ideal gas during the isothermal compression.

Isothermal Process

An isothermal process is one in which the temperature remains constant throughout the change. For a gas, this means that any work done on or by the system is balanced by heat transfer, keeping the overall temperature steady.

In an isothermal process involving an ideal gas, pressure and volume have an inverse relationship. If we compress a gas isothermally, the pressure increases as the volume decreases. For ideal gases, isothermal processes allow us to explore the relation between entropy and volume or pressure changes without worrying about temperature variations.

The entropy change \( \Delta S \) in an isothermal process can be calculated using:
\( \Delta S = nR \ln \left( \frac{V_f}{V_i} \right) = -nR \ln \left( \frac{P_f}{P_i} \right) \).
This equation shows how entropy can change when an ideal gas is compressed isothermally. Since both pressure and volume changes can impact entropy equally in an isothermal condition due to their relation in the ideal gas law, we use these transformations to understand the behavior of gases.

Molar Mass and Its Importance

Molar mass is a critical concept in chemistry and is defined as the mass of one mole of a substance, typically in grams per mole. It links the mass of a substance to the amount of substance by providing the mass of 6.022 × 10²³ molecules (Avogadro's number) or atoms of that substance.

For propane, represented as \( \text{C}_3\text{H}_8 \), the molar mass is calculated by adding the atomic masses of its constituent elements:
\((3 \times 12.01) + (8 \times 1.01) = 44.09 \, \text{g/mol}\).

Knowing the molar mass allows us to convert between the mass of a gas and the number of moles, which is crucial for calculations involving the ideal gas law or entropy change. In our problem, by dividing the given mass of propane (1.32 g) by its molar mass, we determine that there are approximately 0.0299 moles of propane involved. Understanding molar mass helps navigate between microscopic (moles) and macroscopic (grams) worlds, which is essential in real-world applications and experiments.