1. Given: N_(2(g)) P_(2):992kPa V_(1):68.4L V_(2):98.438L

To solve this problem, we can use the ideal gas law equation:<br /><br />\[ PV = nRT \]<br /><br />where:<br />- \( P \) is the pressure,<br />- \( V \) is the volume,<br />- \( n \) is the number of moles,<br />- \( R \) is the ideal gas constant,<br />- \( T \) is the temperature.<br /><br />Given:<br />- \( N_{2(g)} \) (Nitrogen gas)<br />- \( P_{2} = 992 \, \text{kPa} \)<br />- \( V_{1} = 68.4 \, \text{L} \)<br />- \( V_{2} = 98.438 \, \text{L} \)<br /><br />We need to find the relationship between the initial and final states of the gas.<br /><br />Using the ideal gas law for both states:<br /><br />\[ P_{1}V_{1} = nRT_{1} \]<br />\[ P_{2}V_{2} = nRT_{2} \]<br /><br />Since the number of moles \( n \) and the gas constant \( R \) are the same for both states, we can equate the two equations:<br /><br />\[ P_{1}V_{1} = P_{2}V_{2} \]<br /><br />Substitute the given values:<br /><br />\[ P_{1} \times 68.4 = 992 \times 98.438 \]<br /><br />Solve for \( P_{1} \):<br /><br />\[ P_{1} = \frac{992 \times 98.438}{68.4} \]<br /><br />\[ P_{1} \approx 1440.6 \, \text{kPa} \]<br /><br />So, the initial pressure \( P_{1} \) is approximately 1440.6 kPa.