Question

Predict using Boyle's law, what will happen to a balloon that an ocean diver takes to a pressure of 202 kPa.

Answer 1

The volume of the balloon will halve

Explanation:

Boyle's law states that for an ideal gas kept at constant temperature, the pressure of the gas is proportional to its volume. Mathematically,

$pV=const.$

where

p is the gas pressure

V is the volume

The equation can also be rewritten as

$p_1 V_1 = p_2 V_2$

And if we apply it to the gas inside the balloon in this problem (assuming its temperature is constant), we have:

$p_1 = 101 kPa$ is the initial pressure at sea level (the atmospheric pressure)

$V_1$ is the initial volume

$p_2 = 202 kPa$ is the final pressure

$V_2$ is the final volume

Substituting into the equation, we find:

$V_2 = \frac{p_1 V_1}{p_2}=\frac{(101)V_1}{202}=\frac{V_1}{2}$

Which means that the volume of the balloon will halve.

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