(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.

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.

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.

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.