Question

An airplane passenger has 1.00 L 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 0.80 atm?

a) (0.50 L)
b) (1.00 L)
c) (1.25 L)
d) (1.50 L)

Answer 1

Using Boyle's Law, the volume of the air in the stomach at cruising altitude when the cabin pressure drops to 0.80 atm will be 1.25 L. Sea level pressure is 1 atm, and Boyle's Law formula P1V1 = P2V2 helps solve for the final volume, which is 1.25 L. The correct answer is c) (1.25 L).

Step-by-step explanation:

To calculate the volume of air in the passenger's stomach at cruising altitude, we can use Boyle's Law, which states that the product of pressure and volume is a constant for a given mass of confined gas as long as the temperature is constant (P1V1 = P2V2).

Let's represent the initial pressure as P1 and the initial volume as V1, and the pressure at cruising altitude as P2 and the unknown volume as V2. Sea level pressure is typically 1 atm, and the pressure in the stomach is 1 atm before takeoff. At cruising altitude, the cabin pressure drops to 0.80 atm.

Inputting the known values into the equation, we get:

P1V1 = P2V2

(1.00 atm)(1.00 L) = (0.80 atm)(V2)

Solving for V2 gives:

V2 = (1.00 L × 1.00 atm) / 0.80 atm = 1.25 L

Therefore, the volume of the air in the passenger's stomach at cruising altitude would be 1.25 L, which corresponds to option (c).