A scuba diver 60 ft below the ocean surface inhales 55.0 mL of compressed air from a scuba tank at an initial pressure of 2.60 atm and temperature of 8 ∘C.
What is the final pressure of air, in atmo-spheres, in the lungs when the gas expands to 155.0 mL at a body temperature of 37 ∘C, and the amount of gas remains constant?
2.60*55/281=P2*155.00/310= 1.268 which is not correct can someone please lead me in the right direction?
It is correct if you punch in the right numbers. Your set us is right; the answer is wrong.
To solve this problem, you will need to use the combined gas law formula, which relates the initial and final conditions of a gas sample.
The combined gas law formula is:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 = initial pressure
V1 = initial volume
T1 = initial temperature
P2 = final pressure
V2 = final volume
T2 = final temperature
Using the given values:
P1 = 2.60 atm
V1 = 55.0 mL
T1 = 8 °C = 8 + 273.15 = 281.15 K
V2 = 155.0 mL
T2 = 37 °C = 37 + 273.15 = 310.15 K
Now, we can rearrange the formula to solve for P2:
P2 = (P1 * V1 * T2) / (V2 * T1)
Plugging in the values:
P2 = (2.60 atm * 55.0 mL * 310.15 K) / (155.0 mL * 281.15 K)
P2 = 4.460142 atm
The final pressure of air in the lungs when the gas expands to 155.0 mL at a body temperature of 37 °C is approximately 4.460 atm.
To solve this problem, we can use the combined gas law equation, which relates the initial and final pressure, volume, and temperature of a gas sample.
The combined gas law equation is:
(P1 * V1)/T1 = (P2 * V2)/T2
Where P1, V1, and T1 are the initial pressure, volume, and temperature respectively, and P2, V2, and T2 are the final pressure, volume, and temperature respectively.
Let's solve the problem step by step:
Step 1: Convert the initial and final temperatures from Celsius to Kelvin. Kelvin temperature scale starts at absolute zero, so to convert from Celsius to Kelvin, you add 273.15 to the Celsius temperature.
Initial temperature (T1) = 8 + 273.15 = 281.15 K
Final temperature (T2) = 37 + 273.15 = 310.15 K
Step 2: Convert the initial volume from milliliters (mL) to liters (L). There are 1000 mL in 1 L.
Initial volume (V1) = 55.0 mL = 55.0/1000 = 0.055 L
Final volume (V2) = 155.0 mL = 155.0/1000 = 0.155 L
Step 3: Substitute the known values into the combined gas law equation and solve for P2, the final pressure.
(P1 * V1)/T1 = (P2 * V2)/T2
(2.60 atm * 0.055 L)/281.15 K = (P2 * 0.155 L)/310.15 K
Now, cross multiply and solve for P2:
(2.60 * 0.055 * 310.15) / (281.15 * 0.155) = P2
After performing the calculation, P2 comes out to be approximately 2.85 atm.
Therefore, the final pressure of the air in the lungs when the gas expands to 155.0 mL at a body temperature of 37 ∘C is approximately 2.85 atm.