Let's break down the problem: **Given:** - Initial volume, \( V_1 = 150 \) mL - Initial pressure, \( P_1 = 1.000 \) atm - Initial temperature, \( T_1 \) (not given, but let's assume standard temperature, \( 0^\circ C = 273 \) K, unless otherwise specified) - Final temperature, \( T_2 = 53^\circ C = 53 + 273 = 326 \) K - Final pressure, \( P_2 = 1.000 \) atm (assuming pressure is constant, since only temperature changes) **Asked:** - Final volume, \( V_2 \) at \( T_2 = 53^\circ C \) ### **Step 1: Use Charles's Law** Charles's Law states: \[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \] ### **Step 2: Plug in the values** Assume \( T_1 = 273 \) K (if not given, typically standard temperature): \[ \frac{150 \text{ mL}}{273 \text{ K}} = \frac{V_2}{326 \text{ K}} \] ### **Step 3: Solve for \( V_2 \)** \[ V_2 = 150 \text{ mL} \times \frac{326 \text{ K}}{273 \text{ K}} \] \[ V_2 = 150 \times 1.194 \] \[ V_2 \approx 179.1 \text{ mL} \] --- ### **Final Answer:** **The total volume of butane at \( 53^\circ C \) is approximately \( \boxed{179 \text{ mL}} \) (rounded to 3 significant figures).** --- **If you have a different initial temperature, replace \( 273 \) K with the correct value in step 2.**