A mass of \(0.532 \mathrm{kg}\) of molecular oxygen is contained in a cylinder at a pressure of \(1.0 \times 10^{5} \mathrm{Pa}\) and a temperature of \(0.0^{\circ} \mathrm{C} .\) What volume does the gas occupy?

To convert mass to moles, use the molecular weight of molecular oxygen (O2) and the given mass. The molecular weight of molecular oxygen (O2) = 2 × molecular weight of oxygen (O) = 2 × 16 = 32 g/mol Given mass of O2 = 0.532 kg = 532 g Moles of O2 = \(\frac{mass}{molecular\ weight} = \frac{532 \mathrm{g}}{32 \frac{\mathrm{g}}{\mathrm{mol}}} = 16.625 \mathrm{mol}\)

The given temperature is in Celsius. To use it in the ideal gas law, we must convert it to Kelvin. Temperature in Kelvin = Temperature in Celsius + 273.15 \(T_K = 0.0^{\circ} \mathrm{C} + 273.15 = 273.15 \mathrm{K}\)

The ideal gas law is defined as follows, \(PV=nRT\) Where, P = Pressure V = Volume n = Moles of gas R = Ideal Gas Constant (8.314 J/(mol·K)) T = Temperature in Kelvin We are given the pressure P = \(1.0 \times 10^{5} \mathrm{Pa}\), moles of O2 n = \(16.625 \mathrm{mol}\), and temperature T = \(273.15 \mathrm{K}\), and we want to find the volume V. Rearranging the ideal gas law formula for volume, \(V = \frac{nRT}{P}\) Substitute the given values, \(V = \frac{(16.625 \mathrm{mol})(8.314 \frac{\mathrm{J}}{\mathrm{mol}\cdot\mathrm{K}})(273.15 \mathrm{K})}{1.0 \times 10^{5} \mathrm{Pa}}\) Calculate the volume, \(V = \frac{(16.625)(8.314)(273.15)}{1.0 \times 10^{5}} = 37.798 \mathrm{m^3}\) Therefore, the volume occupied by the molecular oxygen gas in the cylinder is 37.798 \(\mathrm{m^3}\).