TICKET 1-30 1. The gas mixture consists of oxygen and nitrogen and has the following composition: r_(O2)= 0.1;r_(N2)=0.9 The mixture is contained in a cylinder with a volume of 400 I, at a pressure of P_(abs)=0.9 MPa and a temperature of 27^circ C Find the mass of this mixture.

To find the mass of the gas mixture, we can use the ideal gas law equation:<br /><br />\[ PV = nRT \]<br /><br />where:<br />- \( P \) is the pressure of the gas mixture,<br />- \( V is of the gas mixture,<br />- \( n \) is the number of moles of mixture,<br />- \( R \) is the universal gas constant,<br />- \( T \) is the temperature of the gas mixture in Kelvin.<br /><br />Given:<br />- \( P_{abs} = 0.9 \) MPa = \( 0.9 \times 10^6 \) Pa<br />- \( V = 400 \) L = \( 0.4 \) m³<br />- \( T = 27^{\circ}C = 300.15 \) K<br /><br />The ideal gas law equation can be rearranged to solve for the number of moles (\( n \)):<br /><br />\[ n = \frac{PV}{RT} \]<br /><br />Substituting the given values:<br /><br />\[ n = \frac{(0.9 \times 10^6 \, \text{Pa})(0.4 \, \text{m}^3)}{(8.314 \, \text{J/(mol·K)})(300.15 \, \text{K})} \]<br /><br />\[ n \approx 113.1 \, \text{mol} \]<br /><br />Now, we can calculate the mass of the gas mixture by multiplying the number of moles by the molar mass of the gas mixture. The molar mass of the gas mixture can be calculated by taking the weighted average of the molar masses of oxygen and nitrogen, based on their respective mole fractions in the gas mixture.<br /><br />The molar mass of oxygen (\( M_{O2} \)) is approximately \( 32 \, \text{g/mol} \), and the molar mass of nitrogen (\( M_{N2} \)) is approximately \( 28 \, \text{g/mol} \).<br /><br />\[ M_{\text{mixture}} = r_{O2} \times M_{O2} + r_{N2} \times M_{N2} \]<br /><br />\[ M_{\text{mixture}} = 0.1 \times 32 \, \text{g/mol} + 0.9 \times 28 \, \text{g/mol} \]<br /><br />\[ M_{\text{mixture}} = 28.4 \, \text{g/mol} \]<br /><br />Finally, we can calculate the mass of the gas mixture by multiplying the number of moles by the molar mass of the gas mixture:<br /><br />\[ \text{Mass} = n \times M_{\text{mixture}} \]<br /><br />\[ \text{Mass} \approx 113.1 \, \text{mol} \times 28.4 \, \text{g/mol} \]<br /><br />\[ \text{Mass} \approx 3200.44 \, \text{g} \]<br /><br />Therefore, the mass of the gas mixture is approximately \( 3200.44 \, \text{g} \).