Answer:
4.8L ( i.e 4.8 x 10^-3 m3)
Explanation:
Step 1:
Data obtained from the question.
Initial volume (V1) = 4.2L
Initial temperature (T1) = 0°C
Final temperature (T2) = 37°C
Final volume (V2) =?
Step 2:
Conversion of celsius temperature to Kelvin temperature. This is illustrated below
K = °C + 273
T1 = 0°C = 0°C + 273 = 273K
T2 = 37°C = 37°C + 273 = 310K
Step 3:
Determination of the final volume.
Since the pressure is constant,
Charles' Law equation will be applied as shown below:
V1 /T1 = V2/T2
4.2/273 = V2 /310
Cross multiply to express in linear form
273 x V2 = 4.2 x 310
Divide both side by 273
V2 = (4.2 x 310)/273
V2 = 4.8L ( i.e 4.8 x 10^-3 m3)
Therefore, the volume of the air in the lungs at that point is 4.8L ( i.e 4.8 x 10^-3 m3)
