Question

(II) You buy an “airtight” bag of potato chips packaged at sea level and take the chips on an airplane flight. When you take the potato chips out of your “carry-on”bag, you notice it has noticeably “puffed up.” Airplane cabins are typically pressurized at 0.75 atm, and assuming the temperature inside an airplane is about the same as inside a potato chip processing plant, by what percentage has the bag “puffed up” in comparison to when it was packaged?

Short answer

The bag had puffed up\(33\% \) in comparison to when it was packaged.

Step-by-step solution

Concepts 

The ideal gas law is \(PV = nRT\) for n mole of ideal gas.

Here, you have to use Bayle’s law for this problem.

Given data 

The initial pressure inside the bag is \({P_1} = 1.00\;{\rm{atm}}\).

The final pressure inside the bag on the airplane is \({P_2} = 0.75\;{\rm{atm}}\).

Let \({V_1}\) and \({V_2}\)bethe initial and final volume of the bag, respectively.

You can assume the temperature does not change.

Calculation 

Now, according to Boyle’s law, you get the following:

\(\begin{array}{c}{P_1}{V_1} = {P_2}{V_2}\\\frac{{{V_1}}}{{{V_2}}} = \frac{{{P_2}}}{{{P_1}}}\\\left( {\frac{{{V_2} - {V_1}}}{{{V_1}}}} \right) \times 100\% = \left( {\frac{{{P_1} - {P_2}}}{{{P_2}}}} \right) \times 100\% \end{array}\)

Now, substituting the values into the above equation, you getthe following:

\(\begin{array}{c}\left( {\frac{{{V_2} - {V_1}}}{{{V_1}}}} \right) \times 100\% = \left( {\frac{{{P_1} - {P_2}}}{{{P_2}}}} \right) \times 100\% \\ = \left( {\frac{{1.00\;{\rm{atm}} - 0.75\;{\rm{atm}}}}{{0.75\;{\rm{atm}}}}} \right) \times 100\% \\ = 33\% \end{array}\)

Hence, the bag had puffed up\(33\% \) in comparison to when it was packaged.