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.
Answers
Answer from: helpmeplz33
It stays the same
Explanation:
From the gas law:
P1V1/T1 = P2V2/T2
P1 = P, V1 = V, T1 = T and P2 = 15.5P, V2 = ? , T2 = 15.5T
Substituting the various variables into the gas law equation:
PV/T = 15.5PV2/15.5T
Make V2 the subject of the formula:
V2 = 15.5TPV/15.5PT
V2 = V.
V1 = V and V2 = V, hence, the volume of the balloon stays the same.
Answer from: Quest
the answer would be a
Answer from: Quest
it is the most commonly accepted one
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