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## Q. 5.5

An inflated helium balloon with a volume of 0.55 L at sea level (1.0 atm) is allowed to rise to a height of 6.5 km, where the pressure is about 0.40 atm. Assuming that the temperature remains constant, what is the final volume of the balloon?

Strategy The amount of gas inside the balloon and its temperature remain constant, but both the pressure and the volume change. What gas law do you need?

## Verified Solution

$\frac{P_{1}V_{1} }{n_{1} T_{1}} =\frac{P_{2}V_{2} }{n_{2}T_{2} }$

Because $n_{1} = n_{2} and T_{1} = T_{2}$,

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

which is Boyle’s law [see Equation (5.2)]. The given information is tabulated:

$\begin{matrix} Initial Conditions & & Final Conditions \\\hline \\P_{1}=1.0 atm & & P_{2}=0.40 atm \\ V_{1}=0.55 L & & V_{2}= ? \end{matrix}$

Therefore,

$V_{2}=V_{1} \times \frac{P_{1}}{P_{2}}$

$=0.55 L \times \frac{1.0 atm}{0.40 atm}$

=1.4  L

Check When pressure applied on the balloon is reduced (at constant temperature), the helium gas expands and the balloon’s volume increases. The final volume is greater than the initial volume, so the answer is reasonable.