Question

A weather balloon has a volume of 1750 L at 103 kPa. The balloon is then released to the atmosphere. What is the pressure if at its highest point above the ground the volume of the balloon is 5,150 L


I also need help with this... chemistry is so difficult

Answer 1

Answer:

35.0 kPa

Explanation:

As pressure decreases, the volume of a gas increases at a given temperature., so since the balloon got bigger, the new pressure must be less than 103kPa

Assuming the temperature does not change, use Boyles Law

P1V1 = P2V2

(103kPa) (1750L) = P2 (5150L)

P2 = (103)(1750) / 5150

Answer 2

The pressure of the gas is inversely proportional to the volume of the gas. The pressure of the balloon at 5150 L is 35.0 kPa.

What is Boyle's Law?

Boyle's Law states the relationship between the initial pressure and volume to the final pressure and the volume. It is given as,

$\rm P_{1}V_{1} = \rm P_{2}V_{2}$

Given,

The initial pressure of the weather balloon = 10.3 kPa

The initial volume of the weather balloon = 1750 L

The final pressure of the weather balloon =?

The final volume of the weather balloon = 5150 L

Substituting values in the above equation:

$\begin{aligned} \rm P_{2} &= \rm \dfrac{P_{1}V_{1}}{V_{2}}\\\\&= \dfrac{103 \times 1750}{5150}\\\\&= 35 \;\rm kPa\end{aligned}$

Therefore, the final pressure decreases with an increase in volume.

Learn more about Boyle's Law here:

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