Question

An airplane passenger has $100 \textrm{ cm}^3$ of air in his stomach just before the plane takes off from a sea-level airport. What volume will the air have at cruising altitude if cabin pressure drops to $7.50 \times 10^4 \textrm{ N/m}^2$?

Final Answer

$135 \textrm{ cm}^3$

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Video Transcript

This is College Physics Answers with Shaun Dychko.
At sea level this passenger on the plane will be experiencing atmospheric pressure and will have some initial volume of air in their stomach of 100 cubic centimeters and call that

*V1*. And that equals the number of moles of air times the gas constant times the temperature at sea level. And when the plane goes up to cruising altitude, it will have a different pressure*P2*that we're given and we'll need to find out what is this volume two in their stomach. And that equals the same number of moles of gas in their stomach times the same gas constant times temperature two but I think we can make an assumption that the cabin of the plane will be designed to keep the same temperature for the comfort of the passengers. So we'll say that*T2*and*T1*are the same. And we'll write it as just*T*with no subscripts on this line here. So we're going to divide the two equations here, the blue and the black one. And we have*P2V2*over*P*atmosphere*V1*equals*NRT2*but I just wrote*T*with no subscript there since*T2*is the same as*T1*, divided by*NRT1*but just*T*here. And this makes the number one. And we can solve for*V2*by multiplying both sides by*P*atmosphere*V1*over*P2*and we're multiplying this one by that, as well. So*V2*is*V1*times atmospheric pressure divided by the pressure at cruising altitude. That's 100 cubic centimeters times 1.013 times ten to the five Newtons per square meter divided by 7.5 times ten to the four Newtons per square meter giving a volume of 135 cubic centimeters at cruising altitude.