Question 5 (1 point) You have 0.5 L of air at 203 K in an expandable container at constant pressure. You heat the container to 273 K. What is the volume of air? (V_(1))/(T_(1))=(V_(2))/(T_(2)) V_(1)=0.5L T_(1)=203K V_(2)=? T_(2)=273K Hint: there is an example problem in your student notes. (Charles Law) 0.67 L ) 0.37 L ) 0.25 L 1.5 L

To solve this problem, we will use Charles' Law, which states that the volume of a gas is directly proportional to its temperature when pressure is held constant. The formula for Charles' Law is:<br /><br />\[<br />\frac{V_1}{T_1} = \frac{V_2}{T_2}<br />\]<br /><br />Given:<br />- \( V_1 = 0.5 \) L<br />- \( T_1 = 203 \) K<br />- \( T_2 = 273 \) K<br /><br />We need to find \( V_2 \).<br /><br />Rearrange the formula to solve for \( V_2 \):<br /><br />\[<br />V_2 = V_1 \times \frac{T_2}{T_1}<br />\]<br /><br />Substitute the given values into the equation:<br /><br />\[<br />V_2 = 0.5 \, \text{L} \times \frac{273 \, \text{K}}{203 \, \text{K}}<br />\]<br /><br />Calculate the value:<br /><br />\[<br />V_2 = 0.5 \times 1.3448 \approx 0.6724 \, \text{L}<br />\]<br /><br />Rounding to two decimal places, we get:<br /><br />\[<br />V_2 \approx 0.67 \, \text{L}<br />\]<br /><br />Therefore, the volume of air at 273 K is approximately 0.67 L.<br /><br />The correct answer is:<br />0.67