Thinking & Inquiry (4 marks) Do the following calculations carefully. Use the space beneath the multiple-choice answer sheet on the main test page to do your rough work.It will not be marked. 15. Oxygen gas, initially at a pressure of 300.0kPa , is allowed to expand at constant temperature until its volume has increased from 100.0 mL to 225.0 mL. What is the final pressure? a) 133.3 kPa c) 675.0 kPa e) 0.001485 kPa b) 75.00 kPa d) 0.007500kPa 16. Determine the final pressure when a gas contained in a steel cylinder at 95 kPa is heated from 20.0^circ C to 30.0^circ C a) 63 kPa c) 98 kPa e) 6.3 kPa b) 92 kPa d) 140 kPa

To solve these problems, we will use the ideal gas law principles. <br /><br />**Question 15:**<br /><br />For a gas expanding at constant temperature (isothermal process), Boyle's Law applies:<br /><br />\[ P_1 V_1 = P_2 V_2 \]<br /><br />Where:<br />- \( P_1 = 300.0 \, \text{kPa} \)<br />- \( V_1 = 100.0 \, \text{mL} \)<br />- \( V_2 = 225.0 \, \text{mL} \)<br /><br />We need to find \( P_2 \).<br /><br />Rearranging the formula gives:<br /><br />\[ P_2 = \frac{P_1 V_1}{V_2} \]<br /><br />Substitute the known values:<br /><br />\[ P_2 = \frac{300.0 \, \text{kPa} \times 100.0 \, \text{mL}}{225.0 \, \text{mL}} \]<br /><br />\[ P_2 = \frac{30000 \, \text{kPa} \cdot \text{mL}}{225.0 \, \text{mL}} \]<br /><br />\[ P_2 = 133.3 \, \text{kPa} \]<br /><br />The correct answer is **a) 133.3 kPa**.<br /><br />**Question 16:**<br /><br />For a gas heated at constant volume, Gay-Lussac's Law applies:<br /><br />\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]<br /><br />Where:<br />- \( P_1 = 95 \, \text{kPa} \)<br />- \( T_1 = 20.0^{\circ} \mathrm{C} = 293.15 \, \text{K} \) (convert Celsius to Kelvin by adding 273.15)<br />- \( T_2 = 30.0^{\circ} \mathrm{C} = 303.15 \, \text{K} \)<br /><br />We need to find \( P_2 \).<br /><br />Rearranging the formula gives:<br /><br />\[ P_2 = P_1 \times \frac{T_2}{T_1} \]<br /><br />Substitute the known values:<br /><br />\[ P_2 = 95 \, \text{kPa} \times \frac{303.15 \, \text{K}}{293.15 \, \text{K}} \]<br /><br />\[ P_2 = 95 \, \text{kPa} \times 1.0341 \]<br /><br />\[ P_2 \approx 98.24 \, \text{kPa} \]<br /><br />The closest answer is **c) 98 kPa**.