Question

A helium balloon containing 12m³ of gas at a pressure of 120kPa is released into the air. Assuming that the temperature is constant, calculate the volume the balloon would have when the pressure inside the balloon has fallen to 40kPa. Answer in m³.

Answer 1

Answer:

36 m³

Explanation:

From gas laws, at a constant temperature, the product of volume and pressure are directly proportional hence expressed as p1v1=p2v2 where p is pressure while v represent volume. Subscripts 1 and 2 represent the initial and respectively. Making v2 the subject then

$v2=\frac {p1v1}{p2}$

Substituting 120 kpa for P1, 12m³ for V1 and 40 kpa for p2 then

$v2=\frac {120 Kpa\times 12}{40 kpa}=36 m^{3}$

Therefore, the volume will be 36 m³

Answer 2

The volume that the balloon would have when the pressure inside the balloon falls is 36 cubic meters.

Given the following data:

• Initial volume = 12 $m^3$

• Initial pressure = 120 kPa

• Final pressure = 40 kPa

To calculate the volume the balloon would have when the pressure inside the balloon falls, we would apply Boyle's law:

Mathematically, Boyle's law is given by the formula;

$P_1V_1 = P_2V_2$

Where;

• $P_1$ is the original (initial) pressure.

• $P_2$ is the final pressure.

• $V_1$ is the original (initial) volume.

• $V_2$ is the final volume.

Substituting the given parameters into the formula, we have;

$120 \times 12 = 40 \times V_2\\\\1440 = 40V_2\\\\V_2=\frac{1440}{ 40} \\\\V_2= 36 \;m^3$

Final volume = 36 cubic meters.

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