Given a gas is at 22 L and 297 K, what is the temperature of the gas when volume changes to 12 L? Round to one decimal place. square K

This question is asking for the new temperature of a gas when its volume changes, while assuming that the pressure and the amount of gas remain constant. This is a classic application of Charles's Law, which states that the volume of a gas is directly proportional to its temperature, provided the pressure and the amount of gas remain constant. The formula for Charles's Law is V1/T1 = V2/T2, where V1 and T1 are the initial volume and temperature, and V2 and T2 are the final volume and temperature. <br /><br />In this case, we are given that the initial volume V1 is 22 L and the initial temperature T1 is 297 K. The final volume V2 is given as 12 L, and we are asked to solve for the final temperature T2. <br /><br />Substituting the given values into the formula, we get:<br /><br />22 L / 297 K = 12 L / T2<br /><br />To solve for T2, we can cross-multiply and divide:<br /><br />T2 = (12 L * 297 K) / 22 L<br /><br />Calculating this gives us T2 = 158.73 K. However, the question asks us to round to one decimal place, so the final answer is T2 = 158.7 K.