when breathing slowly a person inhales air at room temperature and exhales it at body temperature. We may assume that the lungs maintain ordinary atmospheric pressure as they expand. If a resting person inhales 0.6L of air at 24 degrees what volume does the air occupy when it is exhaled at 37 degrees and what mass is that? ans 0.626L and 7.74 * 10^-4 kg. Given density of air to be 1.29 kg/m3.

Question

when breathing slowly a person inhales air at room temperature and exhales it at body temperature. We may assume that the lungs maintain ordinary atmospheric pressure as they expand. If a resting person inhales 0.6L of air at 24 degrees what volume does the air occupy when it is exhaled at 37 degrees and what mass is that? ans 0.626L and 7.74 * 10^-4 kg. Given density of air to be 1.29 kg/m3.

Answer 1

Use PV=nRT

where
P=pressure
V=volume
T=temperature in Kelvin (Celsius+273.15°)
n=number of moles (quantity of substance)
R=gas constant

In the given case, the quantity of air remain the same, so n is a constant. The pressure is assumed constant, so the equation is reduced to
V1/T1=V2/T2
where 1 and 2 represent initial and final states.
V1=0.6L
T1=273.15+24°
T2=273.15+37°
V2=(V1/T1)*T2
=0.6L*(310.15/297.15)

At 0°C, air weighs 1.29 kg/m².
So the volume of 0.6L at 24° has to be transformed to the volume at 0° for the calculation of mass.

Also,
1.29 kg-m^-3
=1.29 g-l^-1

Volume at 0°
=(V1/T1)*T2
=0.6L*(273.15/297.15)
=0.552L
Mass of air = 0.552*1.29 g
=0.712 g

Note: the given answer (7.74*10^-4 kg) for the mass assumes that air weighs 1.29 kg/m^3 at 24°C.

Answer 2

To find the volume of air and the mass when it is exhaled, we can use the ideal gas law. The ideal gas law equation is given by:

PV = nRT

Where:
P = pressure (constant at atmospheric pressure)
V = volume of gas
n = number of moles of gas
R = ideal gas constant
T = temperature in Kelvin

First, let's convert the temperatures to Kelvin:
Initial temperature (T1) = 24 degrees Celsius + 273.15 = 297.15 K
Final temperature (T2) = 37 degrees Celsius + 273.15 = 310.15 K

Since pressure (P) and the number of moles (n) remain constant, we can rewrite the ideal gas law as:

V1/T1 = V2/T2

Solving for V2 (volume of exhaled air):
V2 = (V1 * T2) / T1

Given:
V1 = 0.6 L
T1 = 297.15 K
T2 = 310.15 K

Substituting the values into the equation:
V2 = (0.6 * 310.15) / 297.15
V2 ≈ 0.626 L

Therefore, the volume of air when exhaled is approximately 0.626 L.

To find the mass, we can use the density of air formula:

density = mass / volume

Rearranging the formula to solve for mass:
mass = density * volume

Given:
density of air = 1.29 kg/m^3 (which is equivalent to 1.29 * 10^3 g/L)

Substituting the values into the equation:
mass = (1.29 * 10^3 g/L) * 0.626 L
mass ≈ 7.74 * 10^-4 kg

Therefore, the mass of the exhaled air is approximately 7.74 * 10^-4 kg.