11. A steel cylinder with a capacity of 6 liters contains oxygen at 0^circ C and a pressure of 30.4 MPa . What volume of oxygen can be obtained from the cylinder under normal conditions?

To solve this problem, we can use the ideal gas law, which relates the pressure, volume, temperature, and number of moles of a gas. The ideal gas law is given by the equation:<br /><br />\[ PV = nRT \]<br /><br />where:<br />- \( P \) is the pressure of the gas,<br />- \( V \) is the volume of the gas,<br />- \( n \) is the number of moles of the gas,<br />- \( R \) is the ideal gas constant,<br />- \( T \) is the temperature of the gas in Kelvin.<br /><br />Given:<br />- The volume of the steel cylinder is 6 liters (which is the initial volume of the gas),<br />- The initial pressure of the gas is 30.4 MPa (which is the initial pressure of the gas),<br />- The initial temperature of the gas is \( 0^{\circ}C \) (which is the initial temperature of the gas).<br /><br />We need to find the volume of the gas under normal conditions, which means we need to find the final volume of the gas when the pressure is 1 MPa and the temperature is \( 25^{\circ}C \) (which is the normal temperature and pressure).<br /><br />First, we need to convert the initial temperature to Kelvin:<br /><br />\[ T_{\text{initial}} = 0^{\circ}C + 273.15 = 273.15 \, K \]<br /><br />Next, we need to convert the initial pressure to Pa:<br /><br />\[ P_{\text{initial}} = 30.4 \, MPa = 30.4 \times 10^6 \, Pa \]<br /><br />We also need to convert the initial volume to cubic meters:<br /><br />\[ V_{\text{initial}} = 6 \, L = 0.006 \, m^3 \]<br /><br />We can now use the ideal gas law to find the number of moles of gas in the cylinder:<br /><br />\[ n = \frac{P_{\text{initial}} V_{\text{initial}}}{R T_{\text{initial}}} \]<br /><br />Using the value of the ideal gas constant \( R = 8.314 \, J/(mol \cdot K) \), we can calculate the number of moles of gas:<br /><br />\[ n = \frac{(30.4 \times 10^6 \, Pa)(0.006 \, m^3)}{(8.314 \, J/(mol \cdot K))(273.15 \, K)} \]<br /><br />\[ n \approx 0.082 \, mol \]<br /><br />Now, we need to find the final volume of the gas when the pressure is 1 MPa and the temperature is \( 25^{\circ}C \). We can use the ideal gas law again:<br /><br />\[ V_{\text{final}} = \frac{nRT_{\text{final}}}{P_{\text{final}}} \]<br /><br />Converting the final temperature to Kelvin:<br /><br />\[ T_{\text{final}} = 25^{\circ}C + 273.15 = 298.15 \, K \]<br /><br />Converting the final pressure to Pa:<br /><br />\[ P_{\text{final}} = 1 \, MPa = 1 \times 10^6 \, Pa \]<br /><br />Substituting the values into the equation, we can calculate the final volume of the gas:<br /><br />\[ V_{\text{final}} = \frac{(0.082 \, mol)(8.314 \, J/(mol \cdot K))(298.15 \, K)}{1 \times 10^6 \, Pa} \]<br /><br />\[ V_{\text{final}} \approx 0.207 \, m^3 \]<br /><br />Therefore, the volume of oxygen that can be obtained from the cylinder under normal conditions is approximately 0.207 cubic meters.