an athlete has a large lung capacity of 7.0L. Assuming to be an ideal gas how many molecules of air are in the athletes lungs when the air temperature in the lungs is 37 degrees celcius under normal atmospheric pressure.
Calculate the number of moles, n, using
n = P V/(RT)
and multiply n by Avogadro's number.
T = 310 K
R = 0.08206 liter-atm/mole-K
P = pressure in atm
V = volume in liters
You many have seen other values of R that are used with SI units (m^3 and N/m^2). The value above will save you time doing the calculation.
It is also handy to remember that 1 mole of gas occupies 22.4 liters at 0 C (273 K) and 1 atm. You have
(7.0/22.4)*(273/310) = 0.275 moles
The capacity of lung in adult 500ml and mornal body temprutere37°cand one atomsphere pressure how moles full the lung
To determine the number of molecules in the athlete's lungs, we can use the ideal gas law, which states:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles of gas
R = ideal gas constant (8.314 J/(mol·K))
T = temperature in Kelvin
First, we need to convert the given temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 37 + 273.15 = 310.15 K
Next, we convert the lung capacity from liters to cubic meters:
V = 7.0 L = 7.0 * 0.001 m³ = 0.007 m³
Now, we can rearrange the ideal gas law equation to solve for the number of molecules (n):
n = (PV) / (RT)
Substituting the values:
n = (1 atm * 0.007 m³) / (8.314 J/(mol·K) * 310.15 K)
n ≈ 0.000003438 mol
Finally, we can convert the number of moles to the number of molecules using Avogadro's number, which is approximately 6.022 x 10^23 molecules/mol:
Number of molecules = n * Avogadro's number
Number of molecules ≈ 0.000003438 mol * 6.022 x 10^23 molecules/mol
Number of molecules ≈ 2.07 x 10^19 molecules
Therefore, there are approximately 2.07 x 10^19 molecules of air in the athlete's lungs under the given conditions.
To calculate the number of molecules of air in the athlete's lungs, we can use the Ideal Gas Law equation:
PV = nRT
Where:
P = pressure of the gas (atmospheric pressure)
V = volume of the gas (lung capacity)
n = number of moles of the gas (unknown)
R = ideal gas constant (8.314 J/(mol·K))
T = temperature of the gas (in Kelvin)
First, let's convert the temperature from Celsius to Kelvin:
37 degrees Celsius + 273.15 = 310.15 K
Since the pressure is given as normal atmospheric pressure, we can use 1 atmosphere (atm) as the value for P.
Now, we can rearrange the equation and solve for the number of moles (n):
n = (PV) / (RT)
n = (1 atm * 7.0 L) / (0.0821 atm·L/(mol·K) * 310.15 K)
The value for R has been converted to atm·L/(mol·K) to match the units of pressure and volume used.
Now, we can calculate the value of n by plugging in the values:
n = (7.0 atm·L) / (0.0821 atm·L/(mol·K) * 310.15 K)
n ≈ 0.285 mol
Finally, since 1 mole of gas contains 6.022 x 10^23 molecules (Avogadro's number), we can calculate the number of molecules of air in the athlete's lungs:
Number of molecules = n * (6.022 x 10^23 molecules/mol)
Number of molecules ≈ 0.285 mol * (6.022 x 10^23 molecules/mol)
Number of molecules ≈ 1.716 x 10^23 molecules
Therefore, there are approximately 1.716 x 10^23 molecules of air in the athlete's lungs.