Asked by qp
Consider a balloon that has a volume V. It contains n moles of gas, it has an internal pressure of P, and its temperature is T. If the balloon is heated to a temperature of 15.5T while it is placed under a high pressure of 15.5P, how does the volume of the balloon change?
It doubles.
It stays the same.
It increases greatly.
It decreases slightly.
All Answers 1
Answered by
GPT-5 mini
AI
It stays the same.
By the ideal gas law V = nRT/P. If T → 15.5T and P → 15.5P then V' = nR(15.5T)/(15.5P) = nRT/P = V, so the volume is unchanged.
By the ideal gas law V = nRT/P. If T → 15.5T and P → 15.5P then V' = nR(15.5T)/(15.5P) = nRT/P = V, so the volume is unchanged.
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.