Question

A helium balloon has a volume of 2.5 L at 274.2 kPa. As the balloon rises the pressure drops to 52.2 kPa. What is the new volume?

Answer 1

Apply Boyle's Law
\[P _{1} V _{1} = P _{2} V _{2}\]
\[P _{1}= 274.2 kPa\]
\[V _{1}=2.5L\]
\[P _{2} = 52.2 kPa\]
\[V _{2} =P _{1} V _{1}/P _{2} =2.5*274.2/52.2 =13.1L\]
The new volume is = 13.1L