A person takes in a breath of 27�C air and

holds it until it warms to 36.5�C. The air
has an initial volume of 0.39 L and a mass of
0.00077 kg.
Determine the work done by the air on
the lungs if the pressure remains constant at
1 atm.
Answer in units of J

To determine the work done by the air on the lungs, we can use the equation:

Work = Pressure * Change in Volume

In this case, the pressure remains constant at 1 atm, so we can calculate the work using only the change in volume.

However, the change in volume can be determined using the Ideal Gas Law equation:

PV = nRT

Where:
P = Pressure (in atm)
V = Volume (in liters)
n = Number of moles
R = Ideal Gas Constant (0.0821 L*atm/mol*K)
T = Temperature (in Kelvin)

First, we need to convert the temperatures from Celsius to Kelvin. We add 273 to each temperature:

Initial temperature, Ti = 27 + 273 = 300 K
Final temperature, Tf = 36.5 + 273 = 309.5 K

Next, let's calculate the number of moles of air in the initial volume using the ideal gas law equation:

PV = nRT
1 * 0.39 = n * 0.0821 * 300

Solving for n:
n = (1 * 0.39) / (0.0821 * 300)

n ≈ 0.0162 moles

Since we are dealing with a constant pressure, the moles of air remain constant throughout the process. Hence, the number of moles of air in the final volume is also 0.0162 moles.

Now, let's calculate the final volume (Vf) using the ideal gas law equation:

PV = nRT
1 * Vf = 0.0162 * 0.0821 * 309.5

Solving for Vf:
Vf = (0.0162 * 0.0821 * 309.5) / 1

Vf ≈ 0.405 L

Now that we have the initial and final volumes, we can calculate the change in volume (ΔV) as:

ΔV = Vf - Vi
ΔV = 0.405 - 0.39
ΔV ≈ 0.015 L

Finally, we can calculate the work done by the air on the lungs as:

Work = Pressure * ΔV
Work = 1 * 0.015
Work = 0.015 J

Therefore, the work done by the air on the lungs is approximately 0.015 J.