An airplane passenger has 150 ml of air in his stomach just before the plane takes off from a sea level airport where the pressure is 1 atm. What volume will this air have at a cruising altitude where the pressure in the man’s stomach has equalized with the cabin pressure 75 kPa? Assume that the temperature of the air in his stomach remains constant at 37 ◦C.

72

75kPa=75000Pa

1atm= 101325pa
P1V1 = P2V2
150ml*101325=75000V2
V2=202.65ml
=202ml

To solve this problem, we can use Boyle's Law, which states that the volume of a gas is inversely proportional to its pressure, assuming constant temperature. The equation for Boyle's Law is:

P1 * V1 = P2 * V2

Where:
P1 = initial pressure (in this case, the pressure at sea level, 1 atm)
V1 = initial volume (the initial volume of air in the passenger's stomach, 150 ml)
P2 = final pressure (the cabin pressure at a cruising altitude, 75 kPa)
V2 = final volume (the volume of air in the passenger's stomach at cruising altitude)

We need to convert the given pressure values to the same unit before applying the equation. 1 atm is equivalent to 101.325 kPa. Therefore, the equation becomes:

(101.325 kPa) * (150 ml) = (75 kPa) * (V2)

Now, let's solve for V2:

(101.325 kPa * 150 ml) / 75 kPa = V2

To convert ml to liters (L), we divide ml by 1000. So:

(101.325 kPa * 150 ml) / (75 kPa * 1000) = V2

Simplifying:

(15198.75 kPa * ml) / 75000 kPa = V2

Cancelling out the units:

ml / kPa = V2

So, the volume of air in the passenger's stomach at cruising altitude is simply 15198.75 / 75000 ml, or 0.20265 L (rounded to 5 decimal places).

Therefore, the volume of air in the passenger's stomach at cruising altitude is approximately 0.20265 L.