Many gases are shipped in high-pressure containers. Consider a steel tank whose volume is \(210.0 \mathrm{~L}\) that contains \(\mathrm{O}_{2}\) gas at a pressure of \(16,500 \mathrm{kPa}\) at \(23^{\circ} \mathrm{C}\). (a) What mass of \(\mathrm{O}_{2}\) does the tank contain? (b) What volume would the gas occupy at STP? (c) At what temperature would the pressure in the tank equal $15.2 \mathrm{MPa} ?\( (d) What would be the pressure of the gas, in \)\mathrm{kPa}$, if it were transferred to a container at \(24^{\circ} \mathrm{C}\) whose volume is \(55.0 \mathrm{~L} ?\)

First, we need to convert the provided information into appropriate units. The tank volume is given in Liters, the pressure is given in kPa, and the temperature is given in Celsius. The values we need are: \(V = 210.0 L\) \(P = 16,500 kPa = 16,500,000 Pa\) (since 1kPa = 1000Pa) \(T = 23^{\circ} C = 296 K\) (since T(K) = T(°C) + 273) Now, we can use the ideal gas law formula to solve for the moles of O2 gas: \(PV = nRT\) \(n = \frac{PV}{RT}\) \(n = \frac{16,500,000 Pa\times210 L}{8.314 J/(mol K)\times296 K}\) \(n = 15526.7/2454\) \(n \approx 6.33 mol\) Finally, to find the mass of O2, we multiply the moles by the molar mass of O2: Mass = moles × molar mass of O2 Mass = \(6.33 mol \times 32 g/mol\) Mass = \(202.56 g\) Therefore, the steel tank contains approximately 202.56 g of O2.

STP (standard temperature and pressure) is defined as a temperature of 273 K and a pressure of 100 kPa. To find the volume the gas would occupy at STP, we can use the ideal gas law formula again. This time we need to consider the number of moles and the constant for ideal gases given as the constant for the ideal gases in \(L\cdot kPa/(mol\cdot K)\), \(R=8.31 L\cdot kPa/(mol\cdot K)\): \(V = \frac{nRT}{P}\) \(V = \frac{6.33 mol\times8.31 L\cdot kPa/(mol\cdot K)\times273 K}{100 kPa}\) \(V \approx 137.10 L\) So, the volume of the gas at STP would be approximately 137.1 L.

We have been given the pressure as 15.2 MPa, which is equal to 15,200 kPa. We will use the ideal gas law formula to find the temperature at this particular pressure, considering the number of moles, volume, and ideal gas constant given: \(T = \frac{PV}{nR}\) \(T = \frac{15,200 kPa\times210 L}{6.33 mol\times8.31 L\cdot kPa/(mol\cdot K)}\) \(T \approx 452.47 K\) The temperature at which the pressure in the tank equals 15.2 MPa is approximately 452.47 K.

The volume of the container is given as 55.0 L and the temperature is given in Celsius as 24°C which corresponds to 297 K. We can use the ideal gas law formula to find the pressure under these conditions: \(P = \frac{nRT}{V}\) \(P = \frac{6.33 mol\times8.31 L\cdot kPa/(mol\cdot K)\times297 K}{55.0 L}\) \(P \approx 51,099 kPa\) The pressure of the gas in a 55.0 L container at 24°C would be approximately 51,099 kPa.