Given a gas is at 1 L and 291 K, what is the temperature of the gas when volume changes to 12 L? 3600 K 3000 K 3492 K 273 K

To solve this problem, we can use 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:<br /><br />V1 / T1 = V2 / T2<br /><br />Where:<br />V1 = Initial volume<br />T1 = Initial temperature<br />V2 = Final volume<br />T2 = Final temperature<br /><br />Given:<br />V1 = 1 L<br />T1 = 291 K<br />V2 = 12 L<br /><br />We need to find T2.<br /><br />Using the formula, we can rearrange it to solve for T2:<br /><br />T2 = (V2 * T1) / V1<br /><br />Plugging in the given values:<br /><br />T2 = (12 L * 291 K) / 1 L<br />T2 = 3492 K<br /><br />Therefore, the temperature of the gas when the volume changes to 12 L is 3492 K.