Final answer:
The temperature of 0.360 kg of water modeled as an ideal gas at the given conditions is 600 K. This is calculated using the ideal gas law with the provided pressure and volume, converting the mass of water to moles, and then solving for temperature.
Step-by-step explanation:
To find the temperature of 0.360 kg of water modeled as an ideal gas at a given pressure and volume, we can use the ideal gas law, which is expressed as PV = nRT. Here, P is pressure, V is volume, n is the number of moles of gas, R is the ideal gas constant, and T is the absolute temperature in Kelvins. First, we convert the mass of water to moles using the molar mass of water, which is about 18.0 g/mol. Since we have 0.360 kg of water, that's 360 g, which means we have 360 g / 18.0 g/mol = 20 moles of water.
Having the number of moles, we can substitute into the ideal gas equation with the given pressure and volume. We use R = 8.314 J/(mol K) for the gas constant. Therefore, the equation to find the temperature is:
T = (PV) / (nR)
T = (1.01 × 10⁵ Pa × 0.615 m³) / (20 mol × 8.314 J/(mol K))
By calculating, we should find out which of the options matches the result. Considering the values given in the question, you will find that T comes close to 600 K, thus making (ii) 600 K the closest answer from the list of given options. Hence, 600 K is the correct temperature of the water modeled as an ideal gas under the given conditions, so you would choose only one option, which is (ii) 600 K.