(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?

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 (\(\Delta S\)).

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.

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: \[ V_f = \frac{nRT_i}{P_f} = \frac{(0.600 \mathrm{mol})(8.314 \mathrm{J/(mol·K)})(350 \mathrm{K})}{(121.6 \mathrm{kPa})(1000 \mathrm{Pa/kPa})} \]

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

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.