Question

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

Short answer

(a) The sign of the entropy change is negative because the volume decreases as the pressure increases isothermally. (b) The entropy change for the process is given by: \(\Delta S = (0.600 \mathrm{~mol})\times(8.314 \mathrm{~\frac{J}{mol \cdot K}})\times\ln{\frac{V_2}{V_1}}\) (c) Yes, temperature needs to be specified to calculate the entropy change, as it is needed to determine the initial and final volumes and plays a crucial role in understanding the energy distribution in the system.

Step-by-step solution

Determine the initial and final volumes

Use the Ideal Gas Law to find the initial and final volumes. The Ideal Gas Law states that: \(PV = nRT\) where P is pressure, V is volume, n is number of moles, R is the universal gas constant and T is temperature. The initial volume (V1): \(V_1 = \frac{nRT}{P_1}\) The final volume (V2): \(V_2 = \frac{nRT}{P_2}\) Using the given values: \(n = 0.600 \mathrm{~mol}\), \(T = 350 \mathrm{~K}\), \(P_1 = 0.750 \mathrm{~atm}\), and \(P_2 = 1.20 \mathrm{~atm}\)

Determine the sign of the entropy change

Entropy change will be positive if V2 > V1, and negative if V2 < V1. Compare V2 and V1 using the values obtained in Step 1. Note that since n, R, and T are the same for both V1 and V2, we can compare them by directly comparing P1 and P2. Since P1 < P2, we have V1 > V2. Therefore, the sign of the entropy change is negative.

Calculate the entropy change

For an ideal gas undergoing an isothermal process, the entropy change can be calculated using the following equation: \(\Delta S = nR\ln{\frac{V_2}{V_1}}\) Using the values we found for V1 and V2, we can calculate the entropy change: \(\Delta S = (0.600 \mathrm{~mol})\times(8.314 \mathrm{~\frac{J}{mol \cdot K}})\times\ln{\frac{V_2}{V_1}}\)

Determine if temperature needs to be specified

Yes, the temperature should be specified for calculating the entropy change, because it is needed to determine the initial and final volumes through the Ideal Gas Law. Additionally, temperature plays a crucial role in understanding and comparing the energy distribution in the system, which is what entropy measures. To summarize, (a) The sign of the entropy change is negative (b) The entropy change is given by: \(\Delta S = (0.600 \mathrm{~mol})\times(8.314 \mathrm{~\frac{J}{mol \cdot K}})\times\ln{\frac{V_2}{V_1}}\) (c) Yes, temperature needs to be specified to calculate the entropy change.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in chemistry that relates the pressure, volume, and temperature of an ideal gas. The equation is expressed as \(PV = nRT\).
Here’s what each variable stands for:

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

To solve for either volume or pressure, you can rearrange the equation accordingly. In our problem, we used it to find initial and final volumes by substituting the pressures and maintaining constant temperature. This is essential for predicting changes in states of gases, especially in isothermal processes.

Isothermal Process

An isothermal process occurs when the temperature of a system remains constant. For an ideal gas, this means that any change in pressure or volume must occur at a constant temperature.

Some key features are:

• The internal energy change is zero because energy change in an ideal gas depends only on temperature.
• Any work done by or on the gas results in heat exchange to maintain temperature.

In this problem, the isothermal condition allowed us to use the Ideal Gas Law to relate pressure and volume changes directly, emphasizing their inverse relationship. This process provided a straightforward way to understand entropy change: as the pressure increases, the volume must decrease, thus affecting entropy.

Thermodynamics Concepts

Thermodynamics is the study of heat, work, and energy in systems. In our context, we focused on understanding entropy, a measure of disorder or energy distribution in a system.

Key aspects include:

• Entropy increases with increased randomness or volume in gases.
• For an isothermal process, entropy change is determined by \(\Delta S = nR\ln{\frac{V_2}{V_1}}\).
• A negative entropy change suggests increased order or decreased volume.

In the given exercise, when pressure increased while maintaining temperature, the volume decreased, leading to a negative entropy change. Understanding such concepts helps in predicting how systems behave under certain conditions, providing valuable insights into natural processes.