an ocean depth of 10.0 m, a drivers lung capacity is 2.40 l. the air temperture is 32.0n c and the pressure is 101.30 kpa.what is the volume of the drivers lungs at the same depth, at a temp of 21.0c and a pressure of 141.20 kpa?
(p1v1/t1) = (p2v2/t2)
To find the volume of the driver's lungs at the same depth but under different temperature and pressure conditions, we can use the principles of Boyle's Law and Charles's Law. Boyle's Law states that the volume of a gas at constant temperature is inversely proportional to its pressure, while Charles's Law states that the volume of a gas at constant pressure is directly proportional to its temperature.
Let's break down the given information:
- Initial volume of the driver's lungs: 2.40 L
- Initial temperature: 32.0°C
- Initial pressure: 101.30 kPa
- Final temperature: 21.0°C
- Final pressure: 141.20 kPa
1. Convert the temperatures from Celsius to Kelvin:
- Initial temperature: 32.0°C + 273.15 = 305.15 K
- Final temperature: 21.0°C + 273.15 = 294.15 K
2. Apply Charles's Law to calculate the volume at the final temperature while keeping the pressure constant:
V1 / T1 = V2 / T2
V2 = (V1 * T2) / T1
V2 = (2.40 L * 294.15 K) / 305.15 K
V2 ≈ 2.32 L (rounded to two decimal places)
3. Apply Boyle's Law to calculate the volume at the final pressure while keeping the temperature constant:
P1 * V1 = P2 * V2
V2 = (P1 * V1) / P2
V2 = (101.30 kPa * 2.32 L) / 141.20 kPa
V2 ≈ 1.67 L (rounded to two decimal places)
Therefore, the volume of the driver's lungs at the same depth, with a temperature of 21.0°C and a pressure of 141.20 kPa, is approximately 1.67 liters.