To find the total change in translational kinetic energy, we first need to know the final volume of the air after it has been warmed to 37°C in the lungs. We can use the ideal gas law for this:
PV = nRT
Where P is pressure, V is volume, n is the number of moles of the gas, R is the ideal gas constant, and T is the temperature in Kelvins. The ideal gas constant R has a value of 8.314 J/mol·K. Note that the gas law applies to both the initial and final states of the air, so we can write two equations:
P * V_initial = n * R * T_initial
P * V_final = n * R * T_final
Dividing the second equation by the first one, we get:
V_final / V_initial = T_final / T_initial
We are given V_initial = 0.50 L and T_initial = -10°C, which we need to convert to Kelvin: T_initial_K = -10 + 273.15 = 263.15 K. The final temperature is 37°C, which is equivalent to T_final_K = 37 + 273.15 = 310.15 K. Plugging in these values:
V_final / 0.50 L = 310.15 K / 263.15 K
Solving for V_final:
V_final = 0.50 L * (310.15 K / 263.15 K) = 0.589 L
Now we can find the change in translational kinetic energy. The formula for the (average) translational kinetic energy per molecule of an ideal gas is:
K.E. = (3/2) * k * T
Where k is the Boltzmann constant (1.38 * 10^(-23) J/K). The total translational kinetic energy of the gas is the product of the kinetic energy per molecule and the total number of molecules (N):
Total K.E. = N * (3/2) * k * T
We can relate the number of molecules (N) to the number of moles (n) through the Avogadro constant (NA = 6.022 * 10^(23) molecules/mol)
N = n * NA
Now we can write the total kinetic energy for the initial and final states:
Total K.E._initial = n * NA * (3/2) * k * T_initial_K
Total K.E._final = n * NA * (3/2) * k * T_final_K
The change in kinetic energy is the difference between the final and initial values:
ΔK.E. = Total K.E._final - Total K.E._initial
Factoring out the common terms (n * NA * (3/2) * k), we get:
ΔK.E. = n * NA * (3/2) * k * (T_final_K - T_initial_K)
In order to determine n, we need to rearrange the ideal gas law for the initial state:
n = (P * V_initial) / (R * T_initial_K)
Plugging in the known values, we have:
n = (101 kPa * 0.50 L) / (8.314 J/mol·K * 263.15 K)
Note that 1 kPa = 1000 Pa = 1000 J/m³, and 1 L = 0.001 m³. With this conversion, we have:
n = (101 * 1000 J/m³ * 0.50 * 0.001 m³) / (8.314 J/mol·K * 263.15 K)
n ≈ 0.0232 mol
Now, we can plug in the values for the change in kinetic energy:
ΔK.E. = 0.0232 mol * 6.022 * 10^(23) molecules/mol * (3/2) * 1.38 * 10^(-23) J/K * (310.15 K - 263.15 K)
ΔK.E. ≈ 1856.8 J
The total change in translational kinetic energy of the air you inhaled is approximately 1857 J.
On a cold day, you take a breath, inhaling 0.50 L of air whose initial temperature
is −10°C. In your lungs, its temperature is raised to 37°C. Assume that the
pressure is 101 kPa and that the air may be treated as an ideal gas. What is the
total change in translational kinetic energy of the air you inhaled?
1 answer