If 21 mL of gas are subjected to a temperature change from 10.0^circ C to 120^circ C and a pressure change from 1.0 atm to 15 atm, the new volume is: (Be sure to use the correct number of significant figures.) 440 mL 1.0 mL 0.53 mL 1.9 mL

To solve this problem, we can use the combined gas law, which relates the pressure, volume, and temperature of a gas:<br /><br />\[<br />\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}<br />\]<br /><br />where:<br />- \( P_1 \) and \( P_2 \) are the initial and final pressures,<br />- \( V_1 \) and \( V_2 \) are the initial and final volumes,<br />- \( T_1 \) and \( T_2 \) are the initial and final temperatures in Kelvin.<br /><br />First, convert the temperatures from Celsius to Kelvin:<br />\[<br />T_1 = 10.0^\circ C + 273.15 = 283.15 \, K<br />\]<br />\[<br />T_2 = 120^\circ C + 273.15 = 393.15 \, K<br />\]<br /><br />Now, plug in the known values into the combined gas law equation:<br />\[<br />\frac{1.0 \, \text{atm} \times 21 \, \text{mL}}{283.15 \, \text{K}} = \frac{15 \, \text{atm} \times V_2}{393.15 \, \text{K}}<br />\]<br /><br />Solve for \( V_2 \):<br />\[<br />V_2 = \frac{1.0 \, \text{atm} \times 21 \, \text{mL} \times 393.15 \, \text{K}}{15 \, \text{atm} \times 283.15 \, \text{K}}<br />\]<br /><br />Calculate \( V_2 \):<br />\[<br />V_2 = \frac{21 \times 393.15}{15 \times 283.15}<br />\]<br /><br />\[<br />V_2 = \frac{8256.15}{4247.25} \approx 1.944 \, \text{mL}<br />\]<br /><br />Considering significant figures, the initial volume (21 mL) has two significant figures, so the final answer should also have two significant figures:<br /><br />\( V_2 \approx 1.9 \, \text{mL} \)<br /><br />Therefore, the new volume is approximately 1.9 mL.