The volume of a gas is 605 liters at 27.0^circ C The new temperature is -3.0^circ C What is the new volume? (Remember to convert temperatures to the Kelvin scale.) square

To find the new volume of the gas when the temperature changes, we can use Charles's Law. Charles's Law states that the volume of a gas is directly proportional to its temperature in Kelvin, provided the pressure and the amount of gas remain constant. The formula for Charles's Law is:<br /><br />\[<br />\frac{V_1}{T_1} = \frac{V_2}{T_2}<br />\]<br /><br />where:<br />- \( V_1 \) is the initial volume,<br />- \( T_1 \) is the initial temperature in Kelvin,<br />- \( V_2 \) is the final volume,<br />- \( T_2 \) is the final temperature in Kelvin.<br /><br />First, convert the temperatures from Celsius to Kelvin:<br /><br />\[<br />T_1 = 27.0^{\circ}C + 273.15 = 300.15 \, K<br />\]<br /><br />\[<br />T_2 = -3.0^{\circ}C + 273.15 = 270.15 \, K<br />\]<br /><br />Now, plug the values into Charles's Law equation:<br /><br />\[<br />\frac{605 \, \text{L}}{300.15 \, \text{K}} = \frac{V_2}{270.15 \, \text{K}}<br />\]<br /><br />Solve for \( V_2 \):<br /><br />\[<br />V_2 = \frac{605 \, \text{L} \times 270.15 \, \text{K}}{300.15 \, \text{K}}<br />\]<br /><br />\[<br />V_2 \approx \frac{163470.75 \, \text{L} \cdot \text{K}}{300.15 \, \text{K}}<br />\]<br /><br />\[<br />V_2 \approx 544.7 \, \text{L}<br />\]<br /><br />Therefore, the new volume of the gas at \(-3.0^{\circ}C\) is approximately \(544.7\) liters.