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?

Short answer

The entropy change (\(\Delta S\)) for the process is -51.1 J/K, indicating a decrease in entropy when the ideal gas undergoes an isothermal compression process. The temperature must be specified in order to accurately calculate the entropy change.

Step-by-step solution

Determine the sign of \(\Delta S\) by analyzing the given conditions

When the pressure of an ideal gas is increased isothermally, the gas has to be compressed. Since the process is isothermal, there is no temperature change. Therefore, the final volume (\(V_f\)) must be smaller than the initial volume (\(V_i\)). In the formula for entropy change: \[ \Delta S = nR \ln\frac{V_f}{V_i} \] since \(\frac{V_f}{V_i} < 1\), the natural logarithm of a number smaller than 1 is always negative. So, we can expect that \(\Delta S\) will be negative, which means the entropy of the system decreases.

Use the Ideal Gas Law to find the initial and final volumes

The initial and final volumes can be determined using the ideal gas law: \[ PV = nRT \] For the initial state: \[ V_i = \frac{nRT}{P_i} = \frac{(0.600 \, mol)(8.31 \frac{J}{mol \cdot K})(350 \, K)}{(0.750 \, atm)(101.3 \frac{J}{L \cdot atm})} = 0.105 \, m^3 \] For the final state: \[ V_f = \frac{nRT}{P_f} = \frac{(0.600 \, mol)(8.31 \frac{J}{mol \cdot K})(350 \, K)}{(1.20 \, atm)(101.3 \frac{J}{L \cdot atm})} = 0.0658 \, m^3 \]

Calculate the entropy change using the formula

Now that we have both the initial and final volumes, we can compute the entropy change: \[ \Delta S = nR \ln\frac{V_f}{V_i} = (0.600 \, mol)(8.31 \frac{J}{mol \cdot K}) \ln\left(\frac{0.0658 \, m^3}{0.105 \, m^3}\right) = -51.1 \, J/K \]

Discuss the relevance of specifying the temperature

To calculate the entropy change, we indeed needed to specify the temperature during the process. The reason is that the entropy change is dependent on the volumes, which in turn are dependent on the temperature through the ideal gas law. If the temperature were different (but the pressures remain the same), the initial and final volumes would also change, which would subsequently change the entropy change for the process. In conclusion, the entropy change (\(\Delta S\)) is negative (-51.1 J/K), indicating a decrease in entropy when the ideal gas undergoes an isothermal compression process. The temperature is indeed required to calculate the entropy change for the process.

Key concepts

Ideal Gas Law

Understanding the Ideal Gas Law is crucial for grasping how pressure, volume, and temperature interact in a gas system. It’s represented by the equation \( PV = nRT \), where \(P\) stands for pressure, \(V\) is volume, \(n\) indicates the number of moles, \(R\) is the ideal gas constant, and \(T\) signifies temperature. This law implies that if the number of moles and temperature remain constant (an isothermal process), any increase in pressure will result in a corresponding decrease in volume, and vice versa. This relationship helps us understand the compression of gases and how their entropy changes during such processes. For example, in the exercise problem, the Ideal Gas Law is used to determine the initial and final volumes of the gas when experiencing a change in pressure at a constant temperature.

Isothermal Process

An isothermal process is characterized by a constant temperature throughout the entire system. It’s a key concept when discussing thermal dynamics and entropy. During an isothermal process involving an ideal gas, heat may flow into or out of the system, but the overall internal energy of the system remains unchanged. This means that the process is reversible and the net work done can be calculated precisely. The exercise highlights this process by manipulating the pressure of the gas without altering its temperature. Knowing the process is isothermal is vital because it allows us to assume the temperature remains the same throughout, simplifying the calculation of entropy change. Isothermal conditions also ensure that the formula for entropy change involves only a comparison of volumes before and after the process.

Entropy and Temperature

Entropy, represented by \(S\), measures the disorder or randomness in a system; temperature, on the other hand, is an intensive property reflecting the average kinetic energy of particles. The relationship between entropy and temperature is intricate, as temperature can have a profound effect on the entropy of a system. For instance, when the temperature increases, particles move more vigorously and the disorder typically increases, leading to a rise in entropy. Conversely, a decrease in temperature generally correlates with a decrease in entropy. However, during an isothermal compression, as seen in the exercise problem, the entropy can decrease without a change in temperature due to the reduction in volume, which corresponds to a decrease in the number of accessible states or positions the gas particles can occupy. This nuanced understanding of entropy change is at the heart of thermodynamics and is essential for comprehensively analyzing thermal processes.