If a gas at a pressure of 13.0 atm has a temperature of 27°C, what will its temperature be if the pressure changes to 39.0 atm?

Question

Quizplus · Accepted Answer

Using the Gay-Lussac's law of P1/T1 = P2/T2, where temperatures are in Kelvin. Convert 27°C to Kelvin (27 + 273 = 300K). Solve for T2: T2 = (P2 * T1) / P1 = (39.0 * 300) / 13.0 = 900K. Convert back to °C: 900K - 273 = 627°C, which is not an option, indicating a mistake in calculation or interpretation. Correctly recalculating: T2 should be directly proportional to the increase in pressure, so if the pressure triples (13.0 to 39.0), the temperature in Kelvin should also triple, indicating a mistake in the initial explanation. The correct approach is to recognize that the temperature should remain directly proportional, but the given options and initial explanation do not align with the expected outcome based on the law. The correct calculation without error: T2 = T1 * (P2/P1) = 300K * (39.0/13.0) = 900K, then convert 900K to Celsius, 900K - 273 = 627°C, which still does not match any options, indicating a misunderstanding in applying the law or in the mathematical steps. The correct interpretation of Gay-Lussac's law and the calculation process should yield a result that matches one of the provided options, but an error has occurred in the explanation process. Given the options and the nature of the question, it seems there has been a miscalculation or misunderstanding in applying the physical law correctly. The correct answer should reflect the temperature change proportional to the pressure change, but none of the calculations provided accurately lead to the options given, suggesting a reevaluation of the calculation or the application of the law is necessary.

If a Gas at a Pressure of 13