A fixed quantity of gas at \(25^{\circ} \mathrm{C}\) exhibits a pressure of $99 \mathrm{kPa}\( and occupies a volume of \)4.00 \mathrm{~L}$ (a) Calculate the volume the gas will occupy if the pressure is increased to \(202.6 \mathrm{kPa}\) while the temperature is held constant. (b) Calculate the volume the gas will occupy if the temperature is increased to \(100^{\circ} \mathrm{C}\) while the pressure is held constant.

We are given the initial conditions of a fixed quantity of gas: pressure \(P_1 = 99 \text{kPa}\), volume \(V_1 = 4.00 \text{L}\), and temperature \(T_1 = 25^{\circ} \text{C}\). We need to calculate the final volume \(V_2\), under two different conditions: (a) The pressure \(P_2 = 202.6 \text{kPa}\), temperature held constant. (b) The temperature increased to \(T_2 = 100^{\circ} \text{C}\), pressure held constant.

To use the ideal gas law or Boyle's law equations, we need to convert the temperature values from Celsius to Kelvin. This can be done by adding \(273.15\) to each temperature value: For \(T_1\): \[T_1 = 25^{\circ} \text{C} + 273.15\text{K} = 298.15 \text{K}\] For \(T_2\): \[T_2 = 100^{\circ} \text{C} + 273.15\text{K} = 373.15 \text{K}\]

For part (a), where the pressure is changed and the temperature is held constant, we use Boyle's law to calculate the volume: \[P_1V_1 = P_2V_2\] We will solve for \(V_2\) by isolating the variable: \[V_2 = \frac{P_1V_1}{P_2}\] Then plug in the given values: \[V_2 = \frac{99 \text{kPa} \times 4.00 \text{L}}{202.6 \text{kPa}}\] And compute the final volume \(V_2\): \[V_2 \approx 1.96 \text{L}\] Thus, the volume the gas would occupy when the pressure is increased to \(202.6 \text{kPa}\), while the temperature is held constant, is approximately \(1.96 \text{L}\).

For part (b), where the temperature is increased and the pressure is held constant, we have to find the final volume \(V_2\) using the ideal gas law equation: \[\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\] As the pressure remains constant (\(P_1 = P_2\)), we can simplify the equation: \[\frac{V_1}{T_1} = \frac{V_2}{T_2}\] Next, we isolate \(V_2\): \[V_2 = \frac{V_1 \times T_2}{T_1}\] Plugging in the given values: \[V_2 = \frac{4.00 \text{L} \times 373.15 \text{K}}{298.15 \text{K}}\] And compute the final volume \(V_2\): \[V_2 \approx 5.01 \text{L}\] Thus, the volume the gas would occupy when the temperature is increased to \(100^{\circ} \text{C}\), while the pressure is held constant, is approximately \(5.01 \text{L}\).