Final answer:
Using the combined gas law, the pressure of the air in the lungs when it expands to 165.0 ml at a body temperature of 37 °C, while the amount of gas remains constant, is calculated to be 1.273 atm.
Step-by-step explanation:
The student is asking about the behavior of gases using the combined gas law, which relates the pressure, volume, and temperature of a given amount of gas. In the scenario provided, a scuba diver inhales air from a scuba tank at a certain pressure, volume, and temperature (P1, V1, T1). The question then asks what the pressure will be when the volume and temperature of the air change to new values (P2, V2, T2) as it expands in the diver's lungs. To solve this, we will use the combined gas law equation, which is P1 * V1 / T1 = P2 * V2 / T2, where the temperatures are in Kelvin.
Firstly, we need to convert the temperatures from Celsius to Kelvin by adding 273.15. So, the initial temperature T1 = 7 °C = 280.15 K, and the final temperature T2 = 37 °C = 310.15 K. After substituting the given values into the combined gas law and solving for P2, we get P2 = (P1 * V1 / T1) * (T2 / V2).
P2 = (3.40 atm * 65.0 ml / 280.15 K) * (310.15 K / 165.0 ml) = 1.273 atm
Therefore, the pressure of the air in the lungs when it expands to 165.0 ml at a body temperature of 37 °C will be 1.273 atm, assuming the amount of gas remains constant.