Question

a helium balloon has a volume of 2.00 L at 101 kPa. As the balloon rises, the pressure drops to 97.0 kPa. what is the new volume?​

Answer 1

Answer: The new volume is 2.1 L.

Explanation:

Use the formula Boyle's law.

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

Here, $P_{1}, P_{2}$ are the initial pressure and the final pressure and $V_{1},V_{2}$ are the initial and the final volumes.

As, it is given in the problem, a helium balloon has a volume of 2.00 L at 101 kPa. As the balloon rises, the pressure drops to 97.0 kPa.

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

Put P_{1}=101 kPa,V_{1}=2.00L and P_{2}=97.0 kPa.

$(101 )(2.00)=(97.0)V_{2}$

$V_{2}=\frac{(101)(2.00)}{97}\\V_{2}=2.1 L$

Therefore, the new volume is 2.1 L.