Question

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.

Answer 1

It stays the same

Step-by-step explanation:

Edge 2020

Answer 2

It stays the same

Step-by-step 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.