Problemas
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
Roztwór
Daniel
élite · Tutor durante 8 años
4.3
(289 Votos)
Respuesta
158.7 K
Explicación
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. 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. Substituting the given values into the formula, we get:22 L / 297 K = 12 L / T2To solve for T2, we can cross-multiply and divide:T2 = (12 L * 297 K) / 22 LCalculating 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.